Prasanna – ICSE Solutions

OP Malhotra Class 9 Solutions – S Chand Class 9 Maths Solutions ICSE

October 15, 2024

ICSE Class 9 Maths Solutions S Chand – OP Malhotra Class 9 Maths Solutions

Units 1 Pure Arithmetic

OP Malhotra Class 9 Solutions Chapter 1 Rational and Irrational Numbers

Unit 2 Commercial Mathematics

ICSE Class 9 Maths Solutions S Chand Chapter 2 Compound Interest

Unit 3 Algebra

S Chand Class 9 Maths Solutions ICSE Chapter 3 Expansions

ICSE S Chand Maths Class 9 Solutions Chapter 4 Factorisation

OP Malhotra Class 9 Maths Solutions Chapter 5 Simultaneous Linear Equations in Two Variables

Class 9 ICSE Maths Solutions S Chand Chapter 6 Indices/Exponents

Class 9 ICSE Maths S Chand Solutions Chapter 7 Logarithms

Unit 4 Geometry

ICSE Class 9 Maths Solutions OP Malhotra Chapter 8 Triangles

ICSE Class 9 S Chand Maths Solution Chapter 9 Mid-Point and Intercept Theorems

S Chand ICSE Class 9 Maths Solutions Chapter 10 Pythagoras Theorem

S Chand ICSE Maths Class 9 Solutions Chapter 11 Rectilinear Figures

S Chand Class 9 ICSE Maths Solutions Chapter 12 Area Theorems

Chapter 12 Area Theorems Ex 12

ICSE Class 9 Maths S Chand Solutions Chapter 13 Circle

Unit 5 Statistics

S Chand Maths Class 9 ICSE Solutions Pdf Chapter 14 Statistics, Introduction, Data and Frequency Distribution

Chapter 14 Statistics, Introduction, Data and Frequency Distribution Ex 14

OP Malhotra Class 9 ICSE Solutions Chapter 15 Mean, Median and Frequency Polygon

Unit 6 Mensuration

S Chand Maths Class 9 Solutions ICSE Pdf Download Chapter 16 Area of Plane Figures

S Chand Class 9 Maths ICSE Solutions Chapter 17 Circle: Circumference and Area

S Chand Solutions Class 9 Maths Chapter 18 Surface Area and Volume of 3D Solids (Cuboid and Cube)

Unit 7 Trigonometry

S Chand Maths Class 9 Solutions Pdf Chapter 19 Trigonometrical Ratios

Unit 8 Coordinate Geometry

S Chand Class 9 Maths Solutions Chapter 20 Coordinates and Graphs of Simultaneous Linear Equations

Also Read:

ML Aggarwal Class 9 Solutions PDF

OP Malhotra Class 11 Solutions – S Chand Class 11 Maths Solutions ISC

September 24, 2024

ISC Class 11 S Chand Maths Solutions – OP Malhotra Maths Class 11 Solutions Pdf Free Download

Unit I Sets and Functions

OP Malhotra Class 11 Solutions Chapter 1 Sets

S Chand Class 11 Maths Solutions Chapter 2 Relations and Functions

Unit II Trigonometry

OP Malhotra Maths Class 11 Solutions Pdf Free Download Chapter 3 Angles and Arc Lengths

ISC Class 11 S Chand Maths Solutions Chapter 4 Trigonometrical Functions

Class 11 OP Malhotra Solutions Chapter 5 Compound and Multiple Angles

S Chand Maths Class 11 Pdf Free Download Chapter 6 Trigonometric Equations

ISC OP Malhotra Solutions Class 11 Chapter 7 Properties of Triangle

Chapter 7 Properties of Triangle Ex 7

Unit III Algebra

S Chand Maths Class 11 Solutions Pdf Chapter 8 Mathematical Induction

OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers

S Chand ISC Maths Class 11 Solutions Chapter 10 Quadratic Equations

ISC Class 11 Maths Solutions OP Malhotra Chapter 11 Inequalities

Class 11 ISC Maths S Chand Solutions Chapter 12 Permutations and Combinations

ISC Mathematics OP Malhotra Solutions Class 11 Chapter 13 Binomial Theorem

ISC Class 11 Maths Solutions S Chand Chapter 14 Sequence and Series

Unit IV Coordinate Geometry

Class 11 Maths S Chand Solutions Chapter 15 Basic Concepts of Points and their Coordinates

ISC Mathematics Class 11 OP Malhotra Solutions Chapter 16 The Straight Line

ISC Class 11 Maths S Chand Solutions Pdf Chapter 17 Circle

Unit V Calculus

Class 11 ISC Maths Solutions OP Malhotra Chapter 18 Limits

S Chand Class 11 ISC Maths Solutions Chapter 19 Differentiation

Unit VI Statistics and Probability

Class 11 ISC Maths OP Malhotra Solutions Chapter 20 Measures of Central Tendency

Unit VII Measures of Dispersion

ISC Mathematics Class 11 Solutions OP Malhotra Chapter 21 Measures of Dispersion

ISC Class 11 Maths OP Malhotra Solutions Chapter 22 Probability

Unit VIII Conic Section

Class 11 ISC OP Malhotra Solutions Chapter 23 Parabola

ISC Class 11 OP Malhotra Solutions Chapter 24 Ellipse

OP Malhotra ISC Class 11 Solutions Chapter 25 Hyperbola

Unit IX Introduction to 3-Dimensional Geometry

Class 11 Maths OP Malhotra Solutions Chapter 26 Points and their Coordinates in 3-Dimensions

ISC Maths Class 11 Solutions OP Malhotra Chapter 27 Mathematical Reasoning

Unit X Statistics

OP Malhotra Class 11 ISC Solutions Chapter 28 Statistics (Continued from Chapter 20)

OP Malhotra Maths Class 11 Pdf Free Download Chapter 29 Correlation Analysis

OP Malhotra Maths Class 11 Book Pdf Free Download Chapter 30 Index Numbers

OP Malhotra Maths Class 11 Solutions Chapter 31 Moving Average

OP Malhotra Class 12 Solutions – S Chand Class 12 Maths Solutions ISC

September 24, 2024

S Chand ISC Maths Class 12 Solutions – OP Malhotra Maths Class 12 Solutions

Unit I Relation and Functions

OP Malhotra Class 12 Solutions Chapter 1 Relations

Chapter 1 Relations Ex 1

S Chand Class 12 Maths Solutions Chapter 2 Functions

OP Malhotra Maths Class 12 Solutions Pdf Free Download Chapter 3 Binary Operations

S Chand ISC Maths Class 12 Solutions Chapter 4 Inverse Trigonometric Functions

Chapter 4 Inverse Trigonometric Functions Ex 4

Unit II Algebra

ISC Class 12 Maths Solutions OP Malhotra Chapter 5 Determinants

S Chand Maths Class 12 Solutions Pdf Free Download Chapter 6 Matrices

Unit III Calculus

OP Malhotra Solutions Class 12 Chapter 7 Continuity and Differentiability of Functions

Chapter 7 Continuity and Differentiability of Functions Ex 7

ISC S Chand Maths Class 12 Solutions Pdf Chapter 8 Differentiation (Continued from Book I)

Class 12 OP Malhotra Solutions Chapter 9 Indeterminate Forms of Limits

Class 12 S Chand Maths Solutions Chapter 10 Mean Value Theorems

ISC Class 12 OP Malhotra Solutions Chapter 11 Applications of Derivatives

ISC Class 12 S Chand Maths Solutions Chapter 12 Maxima and Minima

ISC OP Malhotra Solutions Class 12 Chapter 13 Indefinite Integral-1 (Standard Forms)

S Chand Class 12 ISC Maths Solutions Chapter 14 Indefinite Integral – 2 (Methods of Integration)

OP Malhotra ISC Class 12 Solutions Chapter 15 Indefinite Integral – 3 (Special Integrals)

ISC Class 12 Maths S Chand Solutions Chapter 16 Definite Integrals

Class 12 ISC Maths S Chand Solutions Chapter 17 Differential Equations

Unit IV Probability

OP Malhotra Class 12 ISC Solutions Chapter 18 Probability (Continued from Book I) (Laws of Probability)

ISC Class 12 Maths OP Malhotra Solutions Chapter 19 Baye’s Theorem

Chapter 19 Baye’s Theorem Ex 19

OP Malhotra Class 12 Solution Chapter 20 Theoretical Probability Distribution

Unit V Vectors

OP Malhotra Class 12 Maths Solutions Chapter 21 Vectors

Class 12 ISC OP Malhotra Solutions Chapter 22 Vectors (Continued)

Unit VI Three-Dimensional Geometry

ISC Maths Class 12 Solutions OP Malhotra Chapter 23 Three Dimensional Geometry

ISC Mathematics OP Malhotra Solutions Class 12 Chapter 24 The Plane

Unit VII Application of Integrals

Solutions of OP Malhotra Class 12 Chapter 25 Application of Integrals (Areas of a Curve)

Unit VIII Application of Calculus

Class 12 Maths OP Malhotra Solutions Chapter 26 Application of Calculus in Commerce and Economics

Unit IX Linear Regression

ISC Mathematics Class 12 OP Malhotra Solutions Pdf Chapter 27 Linear Regression

Chapter 27 Linear Regression Ex 27

Unit X Linear Programming

ISC Mathematics Class 12 Solutions OP Malhotra Chapter 28 Linear Programming

OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(k)

September 9, 2024

Students can track their progress and improvement through regular use of S Chand ISC Maths Class 12 Solutions Chapter 8 Differentiation Ex 8(k).

S Chand Class 12 ICSE Maths Solutions Chapter 8 Differentiation Ex 8(k)

Find the derivative of the following functions :

Question 1.
(x² + 2)³ (1 – x³)⁴
Solution:
Let y – (x² + 2)³ (1 – x³)⁴
Taking logarithm on both sides, we have
log y – log (x² + 2)³ (1 – x³)⁴
⇒ log y = 3 log(x² + 2) + 4log(1 – x³)
[∵ log ab = log a + log b and log a^b = b log a]
Diff. both sides w.r.t. x; we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(k) 1

Question 2.
\(\frac{x\left(1-x^2\right)^2}{\left(1+x^2\right)^{1 / 2}}\)
Solution:
Let y = \(\frac{x\left(1-x^2\right)^2}{\left(1+x^2\right)^{1 / 2}}\)
Taking logarithm on both sides, we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(k) 2

Question 3.
\(\frac{(x+1)^2 \sqrt{(x-1)}}{(x+4)^3 e^x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(k) 3

Question 4.
\(\sqrt{(x-1)(x-2)(x-3)(x-4)}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(k) 4

Question 5.
\(\frac{(x-a)(x-b)}{(x-p)(x-q)}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(k) 5

Question 6.
\(\frac{2(x-\sin x)^{3 / 2}}{\sqrt{x}}\)
Solution:
Let y = \(\frac{2(x-\sin x)^{3 / 2}}{\sqrt{x}}\)
Taking logarithm on both sides, we have
log y = log 2 + \(\frac { 3 }{ 2 }\)log (x – sin x) – \(\frac { 1 }{ 2 }\) log x
Diff. both sides w.r.t. x; we have
\(\frac{1}{y} \frac{d y}{d x}=\left[\frac{3}{2} \frac{1}{x-\sin x}[1-\cos x]-\frac{1}{2 x}\right]\)
Thus, \(\frac{d y}{d x}=\frac{2(x-\sin x)^{3 / 2}}{\sqrt{x}}\) \(\left[\frac{3}{2}\left(\frac{1-\cos x}{x-\sin x}\right)-\frac{1}{2 x}\right]\)

Question 7.
(i) x^1/x
(ii) \(x^{\sqrt{x}}\)
(iii) \(\left(\frac{1}{x}\right)^x\)
Solution:
(i) Let y = x^1/x ;
Taking logarithm on both sides, we have
log y = \(\frac { 1 }{ x }\) log x ;
DifF. both sides w.r.t. x; we have
\(\frac{1}{y} \frac{d y}{d x}=\frac{1}{x} \cdot \frac{1}{x}+\log x\left(-\frac{1}{x^2}\right)\)
⇒ \(\frac{d y}{d x}=\frac{x^{1 / x}}{x^2}(1-\log x)\)

(ii) Let y = \(x^{\sqrt{x}}\) ;
Taking logarithm on both sides, we have
log y = log \(x^{\sqrt{x}}\) = \(\sqrt{x}\) log x ;
Diff. both sides w.r.t. x; we have
\(\frac{1}{y} \frac{d y}{d x}=\frac{\sqrt{x}}{x}+\log x \frac{1}{2 \sqrt{x}}\) ;
⇒ \(\frac{d y}{d x}=y\left[\frac{1}{\sqrt{x}}+\frac{\log x}{2 \sqrt{x}}\right]\)
= \(x^{\sqrt{x}}\left[\frac{1}{\sqrt{x}}+\frac{\log x}{2 \sqrt{x}}\right]\)

(iii) \(\left(\frac{1}{x}\right)^x\) ;
Taking logarithm on both sides, we have
log y = log \(\left(\frac{1}{x}\right)^x\) = x log (\(\frac { 1 }{ x }\))
= x(- log x) = – x log x
Diff. both sides w.r.t. x; we have
\(\frac{1}{y} \frac{d y}{d x}=-\left[x \cdot \frac{1}{x}+\log x \cdot 1\right]=-(1+\log x)\) ;
⇒ \(\frac{d y}{d x}\) = – y(1 + log x)
= – \(\left(\frac{1}{x}\right)^x\) (1 + log x)

Question 8.
(sin x)^x
Solution:
Let y = (sin x)^x;
Taking logarithm on both sides, we have
log y = x log sin x ;
Diff. both sides w.r.t. x; we have
\(\frac{1}{y} \frac{d y}{d x}=\frac{x}{\sin x}\) cos x + log sinx
⇒ \(\frac{d y}{d x}\) = y[xcotx + log sin x]
= (sin x)^x[x cot x + log sin x]

Question 9.
x^{sin x}
Solution:
Let y = x^{sin x} ;
Taking logarithm on both sides, we have
log y = log x^{sin x} = sin x . log x ;
Diff. both sides w.r.t. x; we have
\(\frac{1}{y} \frac{d y}{d x}=\frac{\sin x}{x}\) cos x + logxcosx
⇒ \(\frac{d y}{d x}\) = y\(\left[\frac{\sin x}{x}+(\log x) \cos x\right]\)
= x^{sin x}\(\left[\frac{\sin x}{x}+\cos x \log x\right]\)

Question 10.
(sin x)^{tan x}
Solution:
Let y = (sin x)^{tan x} ;
Taking logoritum on both sides, we have
log y = log (sin x)^{tan x}
= tan x . log sin x ;
Diff. both sides w.r.t. x; we have
\(\frac{1}{y} \frac{d y}{d x}=\tan x \frac{1}{\sin x}\) cos x + log sin x sec² x
⇒ \(\frac{d y}{d x}\) = y[1 + sec² x log sin x]
= (sin x)^{tan x}[1 + sec² x log sin x]

Question 11.
(tan x)^{log x}
Solution:
Let y = (tan x)^{log x} ;
Taking logarithm on both sides, we have
log y = log (tan x)^{log x} – log x . log tan x ;
Diff. both sides w.r.t. x; we have
\(\frac{1}{y} \frac{d y}{d x}=\log x \frac{1}{\tan x} \sec ^2 x+\log \tan x \cdot \frac{1}{x} \Rightarrow \frac{d y}{d x}=y\left[\frac{\log x}{\sin x \cos x}+\frac{\log \tan x}{x}\right]\)
⇒ \(\frac{d y}{d x}=(\tan x)^{\log x}\left[\frac{\log x}{\sin x \cos x}+\frac{\log \tan x}{x}\right]\)

Question 12.
x^{log x}
Solution:
Let y = x^{log x} ;
Taking logarithm on both sides, we have
log y = log x^{log x} = log x . log x = (log x)² ;
Diff. both sides w.r.t. x; we have
\(\frac{1}{y} \frac{d y}{d x}=2 \log x \cdot \frac{1}{x} \Rightarrow \frac{d y}{d x}=y\left[\frac{2 \log x}{x}\right]=x^{\log x} \cdot \frac{2 \log x}{x}\)

Question 13.
(tan x)^{cos x}
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(k) 6

Question 14.
(log x)^x
Solution:
Let y = (log x)^x ;
Taking logarithm on both sides, we have
log y = log(log x)^x = x log(log x) ;
Diff. both sides w.r.t. x; we have
\(\frac{1}{y} \frac{d y}{d x}=\log (\log x) \cdot 1+x \frac{1}{\log x} \cdot \frac{1}{x}\)
⇒ \(\frac{d y}{d x}=y\left[\log (\log x)+\frac{1}{\log x}\right] \Rightarrow \frac{d y}{d x}=(\log x)^x\left[\log (\log x)+\frac{1}{\log x}\right]\)

Question 15.
\(\left(1+\frac{1}{x}\right)^x\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(k) 7

Question 16.
\(x^x \sqrt{x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(k) 8

Question 17.
(i) cos x^x
(ii) sin (x^x)
(iii) If y = \(\left(\tan \frac{\pi x}{4}\right)^{\frac{4}{\pi x}}\), find \(\frac { dy }{ dx }\) at x = 1.
Solution:
(i) Let y = cos x^x ;
Diff. both sides w.r.t. x; we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(k) 9

Question 18.
Find \(\frac { dy }{ dx }\) if
(i) y = log (x^x + cosec² x)
(ii) y = \(e^{\sin ^2 x}\left(2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right)\)
(ii) x^y = y^x
(iv) (cos x)^y = (sin y)^x
Solution:
(i) Given y = log (x^x + cosec² x); Diff both sides w.r.t. x; we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(k) 10

(ii) Given y = \(e^{\sin ^2 x}\left(2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right)\)
Taking logarithm on both sides; we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(k) 11

(iii) x^y = y^x;
Taking logarithm on both sides y log x = x log y ;
Diff. both sides w.r.t. x; we have
\(\frac{y}{x}+\log x \frac{d y}{d x}=\frac{x}{y} \frac{d y}{d x}\) + log y.1
⇒ \(\left(\log x-\frac{x}{y}\right) \frac{d y}{d x}=\log y-\frac{x}{y}\)
⇒ \(\frac{d y}{d x}=\frac{y(x \log y-y)}{x(y \log x-x)}\)

(iv) Given (cos x)^y = (sin y)^x;
Taking logarithm on both sides; we have
log (cos x)^y = log (sin y)^x
y log cos x = x log sin y;
Diff. both sides w.r.t. x; we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(k) 12

Question 19.
IF y = e^x-y, show that \(\frac { 1 }{ 2 }\).
Solution:
Given y = e^x-y ;
Taking logarithm on both sides; we have
log y = log e^x-y = x – y ;
Diff. both sides w.r.t. x; we have
⇒ \(\frac{1}{y} \frac{d y}{d x}=1-\frac{d y}{d x}\)
⇒ \(\left(\frac{1}{y}+1\right) \frac{d y}{d x}\) = 1 ⇒ \(\left(\frac{1+y}{y}\right) \frac{d y}{d x}\) = 1
⇒ \(\frac{d y}{d x}=\frac{y}{1+y}\)

Question 20.
If x^myⁿ = (x + y)^m+n, prove that \(\frac{d y}{d x}=\frac{y}{x}\).
Solution:
Given x^myⁿ = (x + y)^m+n ;
Taking logarithm on both sides; we have
log x^m + log yⁿ = log (x + y)^m+n
⇒ m log x + n log y = (m + n) log(x + y) [∵ log ab = log a + log b & log a^b = b log a]
Diff. both sides w.r.t. x; we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(k) 12a

Question 21.
If y = \(x^{x^{r \ldots \ldots \ldots \infty}}\), prove that \(\frac{d y}{d x}=\frac{y^2}{x(1-y \log x)}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(k) 13

Question 22.
If y = \(\sqrt{x}^{\sqrt{x}^{\sqrt{x} \ldots \ldots}}\) find \(\frac { dy }{ dx }\).
Solution:
Given y = \(\sqrt{x}^{\sqrt{x}^{\sqrt{x} \ldots \infty}}=(\sqrt{x})^y\)
Taking logarithm on both sides; we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(k) 14

Question 23.
If y = \(a^{x^{a^{x^{a^x \ldots \ldots \ldots . . \infty}}}}\) find \(\frac { dy }{ dx }\).
Solution:
Given y = \(a^{x^y}\)
Taking logarithm on both sides; we have
log y = log \(a^{x^y}\) = log a A
gain taking logarithm on both sides, we have
log log y = log(x^y log a) = y log x + log log a;
Diff. both sides w.r.t. x; we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(k) 15

Question 24.
If y = x^y, prove that x\(\frac{d y}{d x}=\frac{y^2}{(1-y \log x)}\).
Solution:
Given y = \(x^{x^{x \ldots \ldots \infty}}=x^y\)
Taking logarithm on both sides; we have
log y = y log x ;
Diff. both sides w.r.t. x; we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(k) 16

Question 25.
Find \(\frac { dy }{ dx }\) when x^y + y^x = c.
Solution:
Given x^y + y^x = c ⇒ u + v = c …(1)
where u = x^y …(2)
& v = y^x …(3)
Diff. eqn. (1) both sides w.r.t. x; we get
\(\frac{d u}{d x}+\frac{d v}{d x}=0\) … (4)
Taking logarithm on both sides of eqn. (2); we have
log u = y log x; diff. w.r.t. x, we have
\(\frac{1}{u} \frac{d u}{d x}=\frac{y}{x}+\log x \frac{d y}{d x}\)
⇒ \(\frac{d u}{d x}=x^y\left[\frac{y}{x}+\log x \frac{d y}{d x}\right]\) … (5)
Taking logaritum on both sides of eqn.(3); we have
log v = x log y
Diff. both sides w.r.t. x; we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(k) 17

Question 26.
If x^y = e^y-x, prove that \(\frac{d y}{d x}=\frac{2-\log x}{(1-\log x)^2}\).
Solution:
Given x^y = e^y-x;
Taking logarithm on both sides; we have
y log x = (y – x) log e – y – x
⇒ y(1 – log x) = x
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(k) 18

Question 27.
Differentiate (sin x)^x w.r.t. x².
Solution:
Let y = (sin x)^x …(1)
& z = x² …(2)
Taking logarithm on both sides of eqn. (1); we have
log y = x log sin x ; diff. w.r.t. x, we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(k) 19

OP Malhotra Class 12 Maths Solutions Chapter 24 The Plane Ex 24(f)

September 9, 2024

Access to comprehensive ISC Class 12 Maths OP Malhotra Solutions Chapter 24 The Plane Ex 24(f) encourages independent learning.

S Chand Class 12 ICSE Maths Solutions Chapter 24 The Plane Ex 24(f)

Question 1.
Find the equation of the plane which
(i) passes through P(3, -2, 4) and is perpendicular to a line whose direction ratios are 2, 2, -3;
(ii) passes through P(2, -3, 5) and has the line joining A(1, -3, -5) and B(2, 2, 3) as a normal;
(iii) bisects the line joining (5, -2, 6) and (7, 2, 0) at right angles;
(iv) passes through P(1, -2, -4) and is parallel to the plane 7 x – 4 y + 6 z + 2 = 0;
(v) passes through the points (-8, 6, 0),(0, 12, 0), and (-10, 0, -9);
(vi) passes through the points (6, 2, 3),(3, 3, -2),(2, -2, -1);
(vii) passes through the y-axis and the point (4, 2, -3).
Answer:
(i) Given direction ratios of normal to plane are ∴ < 2, 2, 3 >.
Thus, eqn. of plane through the point (3, -2, 4) and having direction ratios of normal to plane are
∴ < 2, 2, -3 > is given by 2(x – 3) + 2(y + 2) – 3(z – 4) = 0
⇒ 2 x + 2 y – 3 z + 10 = 0 be the required eqn. of plane.

(ii) ∴ directon ratios of normal to required plane are
∴< 2 – 1, 2 + 3, 3 + 5 >
i.e. ∴ < 1, 5, 8 > Thus required eqn. of plane through the point P(2, -3, 5) is given by
1(x – 2) + 5(y + 3) + 8(z – 5) = 0
⇒ x + 5 y + 8 z – 27 = 0

(iii) D ratios of line AB are
< 7 – 5, 2 + 2, 0 – 6 > i.e.
< 2, 4, -6 > Thus A B be normal to required plane. Also the required plane passes through the mid point of AB
i.e. \(\frac{5+7}{2}\), \(\frac{-2+2}{2}\), \(\frac{6+0}{2}\) i.e. (6, 0, 3).
∴ eqn. of plane through the point (6, 0, 3) is given by
1(x – 6) + 2(y – 0) – 3(z – 3) = 0
⇒ x + 2 y – 3 z + 3 = 0

(iv) eqn. of plane parallel to given plane
7 x – 4 y + 6 z + 2 = 0 be given by 7 x – 4 y + 6 z + k = 0
Since eqn. (1) passes through the point P(1,-2,-4).
∴ 7 × 1 – 4 × (-2) + 6(-4) + k = 0
⇒ 7 + 8 – 24 + k = 0 ⇒ k = 9
putting the value of k in eqn. (1) ; we have
7 x – 4 y + 6 z + 9 = 0 be the reqd. eqn. of plane.

(v) Let the eqn. of plane through the point (-8, 6, 0) be given by
a(x + 8) + b(y – 6) + c(z – 0) = 0
Now eqn. (1) passes through the point (0, 12, 0).
∴ 8 a + 6 b + 0 c = 0
⇒ 4 a + 3 b + 0 c = 0
Also, plane (1) passes through the point (-10, 0, -9)
∴ a(-10 + 8) + b(0 – 6) + c(-9 – 0) = 0
⇒ -2 a – 6 b – 9 c = 0
on solving eqn. (2) and eqn. (3); we have
\(\frac{a}{-27-0}\) = \(\frac{b}{0+36}\)
= \(\frac{c}{-24+6}\)
⇒ \(\frac{a}{-27}\)
= \(\frac{b}{36}\) = \(\frac{c}{-18}\)
⇒ \(\frac{a}{3}\) = \(\frac{b}{-4}\)
= \(\frac{c}{2}\) = k (say) where k ≠ 0
∴ a = 3 k ; b = -4 k and c = 2 k
putting the values of a, b and c in eqn. (1); we have
3 k(x + 8) – 4 k(y – 6) + 2 k(z) = 0
⇒ 3 x – 4 y + 2 z + 48 = 0

(vi) Let the eqn. of plane through the point (6, 2, 3) is given by
a(x – 6) + b(y – 2) + c(z – 3) = 0
where ∴ < a, b, c > are the direction rates of normal to plane (1). eqn. (1) passes through the point (3, 3, -2).
a(3 – 6) + b(3 – 2) + c(-2 – 3) = 0
⇒ -3 a + b – 5 c = 0
Also, plane (1) passes through the point (2, -2, -1).
a(2 – 6) + b(-2 – 2) + c(-1 – 3) = 0
a + b + c = 0
⇒ -4 a – 4 b – 4 c = 0
By cross-multiplication method; we have
\(\frac{a}{1+5}\) = \(\frac{b}{-5+3}\)
= \(\frac{c}{-3-1}\)
⇒ \(\frac{a}{6}\) = \(\frac{b}{-2}\)
= \(\frac{c}{-4}\)
⇒ \(\frac{a}{3}\) = \(\frac{b}{-1}\) = \(\frac{c}{-2}\) = k (say); k ≠ 0
∴ a = 3 k ; b = -k ; c = -2 k
putting the values of a, b and c in eqn. (1); we have
3 k(x – 6) – k(y – 2) – 2 k(z – 3) = 0
⇒ 3 x – y – 2 z – 10 = 0 be the required eqn. of plane.

(vii) since y-axis be the line of intersection of x o y plane (i.e. z = 0 ) and y o z plane ( i.e. x = 0 ) is given by
z + k x = 0
Since plane (1) passes through the point (4, 2, -3)
∴ -3 + 4 k = 0
⇒ k = \(\frac{3}{4}\)
∴ from (1); z + \(\frac{3 x}{4}\) = 0
⇒ 3 x + 4 z = 0 be the reqd. plane.

Question 2.
Find the equation of the plane
(i) parallel to the plane 4 x – 4 y + 7 z – 3 = 0 and distant 4 units from the point (4, 1, -2);
(ii) which passes through the point (3, -2, 4) and is perpendicular to each of the planes 7 x – 3 y + z – 5 = 0 and 4 x – y – z + 9 = 0.
(iii) perpendicular to each of the planes 3 x – y + z = 0 and x + 5 y + 3 z = 0 and is at a distance of \(\sqrt{6}\) from the origin ;
(iv) through (2, 2, 2) and (0, -2, 0) and perpendicular to the plane x – 2 y + 3 z – 7 = 0.
Answer:
(i) eqn. of given plane be
4 x – 4 y + 7 z – 3 = 0 …………………….. (1)
Thus eqn. of plane parallel to plane (1) be given by
4 x – 4 y + 7 z + k = 0 …………………….. (2)
also it is given that ⊥ distance of point (4, 1, -2) from given plane (2) = 4 units
\(\frac{|4 \times 4-4 \times 1+7 \times(-2)+k|}{\sqrt{4^2+(-4)^2+7^2}}\) = 4
⇒ \(\frac{|16-4-14+k|}{\sqrt{16+16+49}}\) = 4
⇒ \(\frac{|k-2|}{9}\) = 4
⇒ |k – 2| = 36 ⇒ k – 2 = ± 36
⇒ k = ± 36 + 2
⇒ k = 38, – 34
putting the values of k in eqn. (2); we have
4 x – 4 y + 7 z + 38 = 0 and 4 x – 4 y + 7 z – 34 = 0 be the required eqns. of planes.

(ii) The equations of given planes are ;
7 x – 3 y + z – 5 = 0 …………………….. (1)
4 x – y – z + 9 = 0 …………………….. (2)
Thus the eqn. of plane through the point (3, -2, 4) is given by
a(x – 3) + b(y + 2) + c(z – 4) = 0 …………………….. (3)
where < a, b, c > are the direction ratios of normal to plane (3).
Since the required plane (3) is ⊥ to plane (1) and (2).
∴ 7 a – 3 b + c = 0 …………………….. (4)
4 a – b – c = 0 …………………….. (5)
on solving eqn. (4) and (5) simultaneously using cross multiplication method, we have
\(\frac{a}{3+1}\) = \(\frac{b}{4+7}\)
= \(\frac{c}{-7+12}\)
i.e. \(\frac{a}{4}\) = \(\frac{b}{11}\) = \(\frac{c}{5}\) = k (say)
⇒ a = 4 k ; b = 11 k and c = 5 k
putting the values of a, b, c in eqn. (3); we have
4 k(x – 3) + 11 k(y + 2) + 5 k(z – 4) = 0
⇒ 4 x + 11 y + 5 z – 10 = 0 be the reqd. plane.

(iii) Let the eqn. of required plane be
a x + b y + c z + d = 0 …………………….. (1)
where < a, b, c > be the direction ratios of normal to plane (1).
The eqns. of given planes are
3 x – y + z = 0 …………………….. (2)
x + 5 y + 3 z = 0 …………………….. (3)
and
hiven planes.
Since the plane (1) is ⊥ to both given planes.
∴ 3 a – b + c = 0 …………………….. (4)
a + 5 b + 3 c = 0 …………………….. (5)
on solving (4) and (5) simultaneously
∴ \(\frac{a}{-8}\) = \(\frac{b}{1-9}\) = \(\frac{c}{15+1}\)
⇒ \(\frac{a}{1}\) = \(\frac{b}{1}\)
= \(\frac{c}{-2}\) = k (say) ; where k ≠ 0
∴ a = k ; b = k ; c = -2 k
putting the values of a, b, c in eqn. (1); we have
x + y – 2 z + \(\frac{d}{k}\) = 0 …………………….. (6)
⇒ x + y – 2 z + d = 0
Also it is given that ⊥ distance from (0, 0, 0) to given plane (6) = \(\sqrt{6}\)
\(\frac{\left|0+0-2 \times 0+d^{\prime}\right|}{\sqrt{1^2+1^2+(-2)^2}}\)
= \(\sqrt{6}\)
⇒ d = ± 6 .
∴ from eqn. (6); we have x + y – 2 z ± 6 = 0 be the reqd. equations of planes.

(iv) Let the eqn. of plane through the point (2, 2, 2) is given by
a(x – 2) + b(y – 2) + c(z – 2) = 0 …………………….. (1)
plane (1) passes through the point (0, -2, 0).
∴ a(0 – 2) + b(-2 – 2) + c(0 – 2) = 0
⇒ a + 2 b + c = 0 …………………….. (2)
⇒ -2 a – 4 b – 2 c = 0
Since eqn. (1) is ⊥ to plane
x – 2 y + 3 z – 7 = 0
a – 2 b + 3 c = 0 …………………….. (3)
\(\frac{a}{6+2}\) = \(\frac{b}{1-3}\)
= \(\frac{c}{-2-2}\)
i.e. \(\frac{a}{8}\) = \(\frac{b}{-2}\) = \(\frac{c}{-4}\)
i.e. \(\frac{a}{4}\) = \(\frac{b}{-1}\) = \(\frac{c}{-2}\) = k
∴ a = 4 k ; b = -k ; c = -2 k ; k ≠ 0
putting the values of a, b and c in eqn. (1); we have
4 k(x – 2) – k(y – 2) – 2 k(z – 2) = 0
⇒ 4 x – y – 2 z – 2 = 0 be the required eqn. of plane.

Question 3.
Find the equation of the plane which contains the line of intersection of the planes x + 2 y + 3 z – 4 = 0 and 2 x + y – z + 5 = 0 and is perpendicular to the plane 5 x + 3 y – 6 z + 8 = 0.
Answer:
Given eqns. of planes are ; and
x + 2 y + 3 z – 4 = 0 …………………….. (1)
2 x + y – z + 5 = 0 …………………….. (2)
Thus the eqn. of any plane through the line of intersection of given planes be
(x + 2 y + 3 z – 4) + k(2 x + y – z + 5) = 0
(1 + 2 k) x + (2 + k) y + (3 – k) z – 4 + 5 k = 0 …………………….. (3)
Now plane (3) is ⊥ to given plane
5 x + 3 y – 6 z + 8 = 0
∴ (1 + 2 k) 5 + ( 2 + k) 3 + (3 – k)(-6) = 0 …………………….. (4)
⇒ 5 + 10 k + 6 + 3 k – 18 + 6 k = 0
⇒ 19 k – 7 = 0
⇒ k = \(\frac{7}{19}\)
putting the value of k = \(\frac{7}{19}\) in eqn. (3) ; we get
(1 + \(\frac{14}{19}\)) x + (2 + \(\frac{7}{19}\)) y + (3 – \(\frac{7}{19}\)) z – 4 + \(\frac{35}{19}\) = 0
⇒ 33 x + 45 y + 50 z – 41 = 0 be the reqd. plane.

Question 4.
Find the equation of the plane through the intersection of the planes x + y + z = 1 and 2 x + 3 y – z + 4 = 0 and parallel to the x-axis.
Answer:
The eqn. of any palen through the line if intersection of two given planes x + y + z – 1 = 0 and 2 x + 3 y – z + 4 = 0 is given by
(x + y + z – 1) + k(2 x + 3 y – z + 4) = 0
⇒ (1 + 2 k) x + (1 + 3 k) y + (1 – k) z – 1 + 4 k = 0 …………………….. (1)
Now plane (1) is parallel to x-axis whose direction ratios are < 1, 0, 0 >
∴ Normal to plane (1) is ⊥ to x-axis.
Thus, (1 + 2 k) 1 + (1 + 3 k) 0 + (1 – k) 0 = 0
⇒ 2 k = -1
⇒ k = \(\frac{-1}{2}\)
∴ from (1); we have
\(\frac{-1}{2}\) y + \(\frac{3}{2}\) z – 1 – 2 = 0
⇒ -y + 3 z – 6 = 0
⇒ y – 3 z + 6 = 0 be the required eqn. of plane.

Question 5.
(i) Find the equation of the plane through (2, 3, -4) and (1, -1, 3) and parallel to the x-axis.
(ii) Find the equation of the plane passing through the points (2, 3, 1) and (4, -5, 3) and parallel to the x-axis.
Answer:
(i) The eqn. of any plane through the point (2, 3, -4) is given by
a(x – 2) + b(y – 3) + c(z + 4) = 0 …………………….. (1)
where < a, b, c > are the direction ratios of normal to plane (1).
Also paine (1) passes through the point (1, -1, 3).
∴ a(1 – 2) + b(-1 – 3) + c(3 + 4) = 0
⇒ -a – 4 b + 7 c = 0 ⇒ a + 4 b – 7 c = 0 …………………….. (2)
Also plane (1) is parallel to x-axis.
∴ Normal to plane (1) is ⊥ to x-axis.
∴ a + 0 b + 0 c = 0 …………………….. (3)
on solving eqn. (2) and eqn. (3) simultaneously by cross-multiplication method, we have
\(\frac{a}{0-0}\) = \(\frac{b}{-7-0}\) = \(\frac{c}{0-4}\)
⇒ \(\frac{a}{0}\) = \(\frac{b}{-7}\) = \(\frac{c}{-4}\) = k (say)
∴ a = 0 ; b = -7 k ; c = -4 k ; k ≠ 0
∴ from (1); we have
0(x – 2) – 7 k (y – 3) – 4 k(z + 4) = 0
⇒ -7 y + 21 – 4 z – 16 = 0
⇒ -7 y – 4 z + 5 = 0
⇒ 7 y + 4 z – 5 = 0 which is the required plane.

(ii) eqn. of any plane through the point (2, 3, 1) is given by
a(x – 2) + b(y – 3) + c(z – 1) = 0 …………………….. (1)
where < a, b, c > be the d ratios of normal to plane (1).
Now plane (1) passes through the point (4, -5, 3).
a(4 – 2) + b(-5 – 3) + c(3 – 1) = 0
⇒ 2 a – 8 b + 2 c = 0
⇒ a – 4 b + c = 0 …………………….. (2)
Since plane (1) is parallel to x-axis.
∴ Normal to plane (1) is ⊥ to x-axis whose direction ratios are < 1, 0, 0 > i.e. a + 0 b + 0 c = 0 …………………….. (3)
on solving eqn. (2) and eqn. (3) simultaneously
\(\frac{a}{0}\) = \(\frac{b}{1-0}\) = \(\frac{c}{0+4}\) = k (say); k ≠ 0
⇒ a = 0 k = 0 ; b = k ; c = 4 k
putting the values of a, b and c in eqn. (1); we get
0(x – 2) + k(y – 3) + 4 k(z – 1) = 0
⇒ y + 4 z – 7 = 0
which is the required eqn. of plane.

Question 6.
Find the equation of a plane which is perpendicular to the plane 2 x – 3 y + 6 z + 8 = 0 and passes through the intersection of the planes x + 2 y + 3 z – 4 = 0 and 2 x – y – z + 5 = 0.
Answer:
eqns. of given planes are ;
x + 2 y + 3 z – 4 = 0 …………………….. (1)
2 x – y – z + 5 = 0 …………………….. (2)
and
Thus the eqn. of any plane through the line of intersection of given planes is given by
x + 2 y + 3 z – 4 + k(2 x – y – z + 5) = 0
⇒ (1 + 2 k) x + (2 – k) y + (3 – k) z – 4 + 5 k = 0 …………………….. (1)
Now plane (1) is normal to plane 2 x – 3 y + 6 z + 8 = 0
∴ 2(1 + 2 k) – 3(2 – k) + 6(3 – k) = 0
⇒ 2 + 4 k – 6 + 3 k + 18 – 6 k = 0
⇒ k + 14 = 0
⇒ k = -14
putting the value of k in eqn. (1); we have
-27 x + 16 y + 17 z – 74
⇒ 27 x – 16 y – 17 z + 74 = 0
⇒ 27, which is the reqd. plane.

Question 7.
(i) Find the equation of the plane passing through A(-1, 1, 1) and B(1, 1, 1) and perpendicular to the plane x – 2 y + 2 z = 3.
(ii) Also, find the distance of the point A from the plane x – 2 y + 2 z = 3.
Answer:
(i) Any plane through the point A(-1, 1, 1) be given by
a(x + 1) + b(y – 1) + c(z – 1) = 0 …………………….. (1)
where < a, b, c > are the direction ratios of normal to plane (1).
The point B(1, -1, 1) lies an eqn. (1); we have
a(1 + 1) + b(-1 – 1) + c(1 – 1) = 0
⇒ 2 a – 2 b + 0 c = 0
⇒ a – b + 0 c = 0 …………………….. (2)
Now plane (1) ⊥ to given plane
x – 2 y + 2 z = 3
a – 2 b + 2 c = 0 …………………….. (3)
∴ on solving eqn. (2) and eqn. (3) simultaneously by cross multiplication method, we have
\(\frac{a}{-2-0}\) = \(\frac{b}{0-2}\) = \(\frac{c}{-2+1}\)
i.e. \(\frac{a}{-2}\) = \(\frac{b}{-2}\) = \(\frac{c}{-1}\)
i.e. \(\frac{a}{2}\) = \(\frac{b}{2}\) = \(\frac{c}{1}\) = k (say) ; k ≠ 0
∴ a = 2 k ; b = 2 k and c = k
putting the value of a, b and c in eqn. (1); we have
2 k(x + 1) + 2 k(y – 1) + k(z – 1) = 0
⇒ 2 x + 2 y + z – 1 = 0 be the reqd. plane

(ii) Required distance of A(-1, 1, 1) from x – 2 y + 2 z – 3 = 0
= \(\frac{|-1-2 \times 1+2 \times 1-3|}{\sqrt{1^2+(-2)^2+2^2}}\)
= \(\frac{|-1-2+2-3|}{3}\) = \(\frac{4}{3}\) units

Question 8.
A plane meets the plane x = 0, where x = 0, 2 y – 3 z = 5, and the plane z = 0 where z = 0, 7 x + 4 y = 10. Find the equation to the plane.
Answer:
The eqn. of any plane through the line of intersection of planes x = 0 and 2 y – 3 z – 5 = 0 be given by
2 y – 3 z – 5 + k x = 0 …………………….. (1)
Now plane (1) meets the planes z = 0 and 7 x + 4 y = 10
∴ from (1); 2 y – 5 + k (\(\frac{10-4 y}{7}\)) = 0
i.e. 14 y – 35 + 10 k – 4 k y = 0
i.e. (14 – 4 k) y + 10 k – 35 = 0
∴ 14 – 4 k = 0 and 10 k – 35 = 0
i.e. k = \(\frac{7}{2}\) and k = \(\frac{7}{2}\)
∴ from (1); 2 y – 3 z – 5 + \(\frac{7}{2}\) x = 0
⇒ 4 y – 6 z – 10 + 7 x = 0
which is the required eqn. of plane.

Question 9.
Prove that the plane 2 x + y – 3 z + 5 = 0, 5 x – 7 y + 2 z + 3 = 0, 5 and x + 10 y – 11 z + 12 = 0 have a line in common.
Answer:
The eqn. of plane through the line of intersection of first two planes is given by
2 x + y – 3 z + 5 + k(5 x – 7 y + 2 z + 3) = 0
⇒ (2 + 5 k) x + (1 – 7 k) y + (-3 + 2 k) z + 5 + 3 k = 0 …………………….. (1)
Now plane (1) is identical to given plane
if
x + 10 y – 11 z + 12 = 0
\(\frac{2-15 k}{1}\) = \(\frac{1-7 k}{10}\)
= \(\frac{-3-12 k}{-11}\) = \(\frac{5+3 k}{12}\) …………………….. (2)
From first two fractions ; 20 + 50 k = 1 – 7 k
⇒ 57 k = -19 ⇒ k = \(\frac{-1}{3}\)
putting k = \(\frac{-1}{3}\) in last two fractions ; we have
\(\frac{-3-\frac{2}{3}}{-11}\) = \(\frac{5-1}{12}\)
⇒ \(\frac{1}{3}\) = \(\frac{1}{3}\) which is true.
Thus the given three planes have same line of intersection.

Question 10.
Find the equation of the plane passing through the intersection of the plane.
\(\vec{r}\) (\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)) = 1 and \(\vec{r}\) (2 \(\hat{i}\) + 3 \(\hat{j}\) – \(\hat{k}\)) + 4 = 0 and parallel to x-axis.
Answer:
The eqn. of plane passing through the line of intersection of given planes
\(\vec{r}\) (\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)) = 1 and \(\vec{r}\) (2 \(\hat{i}\) + 3 \(\hat{j}\) – \(\hat{k}\)) + 4 = 0
be given by \(\vec{r}\) (\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)) – 1] + λ[\(\vec{r}\) (2 \(\hat{i}\) + 3 \(\hat{j}\) – \(\hat{k}\)) + 4] = 0
⇒ \(\vec{r}\) [(1 + 2λ) \(\hat{i}\) + (1 + 3λ) \(\hat{j}\) + (1 – λ) \(\hat{k}\)] + 4λ – 1 = 0 …………………….. (1)
Since eqn. (1) is parallel to x-axis.
∴ Normal to plane is ⊥ to x-axis.
∴(1 + 2 λ) 1 + (1 + 3λ) 0 + (1 – λ) 0 = 0
⇒ 1 + 2λ = 0
⇒ λ = \(\frac{-1}{2}\)
∴ from (1) ; we have
\(\vec{r}\) [\(\frac{-\hat{j}}{2}\) + \(\frac{3}{2}\) \(\hat{k}\)] – 3 = 0

Examples:

Question 1.
Show that the equation b y + c z + d = 0 represents a plane parallel to the axis OX. Find the equation to a plane through the points (2, 3, 1), (4, -5, 3) and parallel to OX.
Answer:
The eqn. of given plane be b y + c z + d = 0
∴ direction no’s of normal to plane (1) are < 0, b, c >
Now plane (1) parallel to x-axis whose direction ratios are < 1, 0, 0>
if normal to plane (1) is ⊥ to x-axis i.e. if 0 1 + b 0 + c 0 = 0 if 0 = 0, which is true.
Thus given plane (1) is parallel to x-axis.
Eqn. of any plane through the point (2, 3, 1) is given by
a(x – 2) + b(y – 3) + c(z – 1) = 0 …………………….. (1)
where < a, b, c > be the d ratios of normal to plane (1).
Now plane (1) passes through the point (4, -5, 3).
a(4 – 2) + b(-5 – 3) + c(3 – 1) = 0
⇒ 2 a – 8 b + 2 c = 0
⇒ a – 4 b + c = 0 …………………….. (2)
Since plane (1) is parallel to x-axis.
∴ Normal to plane (1) is ⊥ to x-axis whose direction ratios are < 1, 0, 0> i.e.
a + 0 b + 0 c = 0 …………………….. (3)
on solving eqn. (2) and eqn. (3) simultaneously
\(\frac{a}{0}\) = \(\frac{b}{1-0}\) = \(\frac{c}{0+4}\) = k (say) ; k ≠ 0
⇒ a = 0 k = 0 ; b = k ; c = 4 k
putting the values of a, b and c in eqn. (1); we get
0(x – 2) + k(y – 3) + 4 k(z – 1) = 0
⇒ y + 4 z – 7 = 0 which is the required eqn. of plane.

Question 2.
Find the equations of the planes parallel to the plane 3 x – 6 y + 2 z = 12 and 6 units away from it.
Answer:
The eqn. of plane || to plane 3 x – 6 y + 2 z – 12 = 0 is given by
3 x – 6 y + 2 z + k = 0 …………………….. (1)
Let P(x₁, y₂, z₁) be any point on given plane
3 x – 6 y + 2 z = 12 …………………….. (2)
3 x₁ – 6 y₁ + 2 z₁ = 12 …………………….. (3)
i.e. also it is given that distance between planes (1) and (2) = 6
⇒ \(\frac{|3 x_1-6 y_1+2 z_1+k|}{\sqrt{3^2+(-6)^2+2^2}}\) = 6
⇒ \(\frac{|12+k|}{7}\) = 6
⇒ 12 + k = ± 42
⇒ k = ± 42 – 12
∴ k = 30, – 54 [using (3)]
∴ from (1); 3 x – 6 y + 2 z + 30 = 0 and 3 x – 6 y + 2 z – 54 = 0 be the required eqns. of planes.

Question 3.
Find the equation of the plane passing through the point (2, 3, 4) and making equal intercepts on the axis.
Answer:
The eqn. of any plane making equal intercepts on coordiantes axes be given by
\(\frac{x}{a}\) + \(\frac{y}{a}\) + \(\frac{z}{a}\) = 1 …………………….. (1)
where a be the length of intercept made by plane on coordiante axes.
Since plane (1) passes through the point (2, 3, 4).
2 + 3 + 4 = a
⇒ a = 9
∴ from (1); x + y + z = 9 be the required eqn. of plane.

Question 4.
Show that the four points (0, 4, 3), (-1, -5, -3), (-2, -2, 1) and (1, 1, -1) are coplanar and find the equation of the common plane.
Answer:
The eqn. of any plane through the point (0,4,3) is given by
a(x – 0) + b(y – 4) + c(z – 3) = 0 …………………….. (1)
where < a, b, c > are the direction ratios of normal to plane.
The point (-1, -5, -3) lies on eqn. (1).
∴ a(-1) + b(-5 – 4) + c(-3 – 3) = 0
⇒ a + 9 b + 6 c = 0 …………………….. (2)
The plane (1) passes through the point (-2, -2, 1).
-2 a + b(-2 – 4) + c(1 – 3) = 0
a + 3 b + c = 0
⇒ -2 a – 6 b – 2 c = 0 …………………….. (3)
on solving eqn. (2) and eqn. (3) simultaneously by cross multiplication method, we have
\(\frac{a}{9-18}\) = \(\frac{b}{6-1}\) = \(\frac{c}{3-9}\)
⇒ \(\frac{a}{-9}\) = \(\frac{b}{5}\) = \(\frac{c}{-6}\) = k (say)
∴ a = -9 k ; b = 5 k ; c = -6 k ; k ≠ 0
putting the value of a, b and c in eqn. (1); we have
-9 k x + 5 k(y – 4) – 6 k(z – 3) = 0
9 x – 5 y + 6 z + 2 = 0
⇒ -9 x + 5 y – 6 z – 2 = 0 …………………….. (4)
Also the point (1, 1, -1) lies on plane (4).
if 9 – 5 – 6 + 2 = 0
⇒ 0 = 0, which is true.
Hence the given four points are coplanar.

Question 5.
Find the equation of the plane which is parallel to x-axis and passes through the points (2, 3, 1) and (4, -5, 3).
Answer:
Eqn. of any plane through the point (2,3,1) is given by
a(x – 2) + b(y – 3) + c(z – 1) = 0 …………………….. (1)
where < a, b, c > be the d ratios of normal to plane (1).
Now plane (1) passes through the point (4, -5, 3).
a(4 – 2) + b(-5 – 3) + c(3 – 1) = 0
⇒ 2 a – 8 b + 2 c = 0
⇒ a – 4 b + c = 0 …………………….. (2)
Since plane (1) is parallel to x-axis.
∴ Normal to plane (1) is ⊥ to x-axis whose direction ratios are < 1, 0, 0 >
i.e. a + 0 b + 0 c = 0 …………………….. (3)
on solving eqn. (2) and eqn. (3) simultaneously
\(\frac{a}{0}\) = \(\frac{b}{1-0}\) = \(\frac{c}{0+4}\) = k (say); k ≠ 0
⇒ a = 0 k = 0 ; b = k ; c = 4 k
putting the values of a, b and c in eqn. (1); we get
0(x – 2) + k(y – 3) + 4 k(z – 1) = 0
⇒ y + 4 z – 7 = 0
which is the required eqn. of plane.

Question 6.
Find the equation of the plane passing through to the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2 x + 6 y + 6 z = 9.
Answer:
The eqn. of any plane through the point (2, 2, 1) be given by
a(x – 2) + b(y – 2) + c(z – 1) = 0 …………………….. (1)
where < a, b, c > are the direction ratios of normal to plane (1).
Now plane (1) passes through the point (9, 3, 6).
∴ a(9 – 2) + b(3 – 2) + c(6 – 1) = 0
i.e. 7 a + b + 5 c = 0 …………………….. (2)
Also plane (1) is ⊥ to given plane
2 x + 6 y + 6 z – 9 = 0
2 a + 6 b + 6 c = 0 …………………….. (3)
on solving eqn. (2) and (3) simultaneously
By cross-multiplication method, we have
\(\frac{a}{6-30}\) = \(\frac{b}{10-42}\) = \(\frac{c}{42-2}\)
i.e. \(\frac{a}{-24}\) = \(\frac{b}{-32}\) = \(\frac{c}{40}\)
i.e. \(\frac{a}{3}\) = \(\frac{b}{4}\) = \(\frac{c}{-5}\) = k (say); where k ≠ 0
∴ a = 3 k ; b = 4 k and c = -5 k
∴ from (1); we have
3 k(x – 2) + 4 k(y – 2) – 5 k(z – 1) = 0
⇒ 3 x + 4 y – 5 z – 9 = 0 be the required eqn. of plane.

Question 7.
A plane is passing through the point (2, -3, 1) and perpendicular to the straight line joining the points (3, 4, -1) and (2, -1, 5). Find the equation of the plane.
Answer:
Here D ratios of normal to plane are < 2 – 3, -1 – 4, 5 + 1 > i.e. < -1, -5, 6 > The eqn. of any plane through the point (2, -3, 1) be given by
-1(x – 2) – 5(y + 3) + 6(z – 1) = 0
⇒ – x – 5 y + 6 z – 19 = 0
⇒ x + 5 y – 6 z + 19 = 0

Question 8.
Find the equation of the plane passing through (1, 2, 3) and perpendicular to the straight line \(\frac{x}{-2}\) = \(\frac{y}{4}\) = \(\frac{z}{3}\)
Answer:
The eqn. of given line be \(\frac{x}{-2}\) = \(\frac{y}{4}\) = \(\frac{z}{3}\)
∴ direction ratios of line (1) are < -2, 4, 3 > …………………….. (1)
Since line (1) is ⊥ to required plane.
∴ a ratios of normal to reqd. plane are < -2, 4, 3 >
Hence the eqn. of plane through the point (1, 2, 3) be given by
-2(x – 1) + 4(y – 2) + 3(z – 3) = 0
2 x – 4 y – 3 z + 15 = 0
⇒ -2 x + 4 y + 3 z – 15 = 0

Question 9.
Find the cosine of the angle between the planes x + 2 y – 2 z + 6 = 0 and 2 x + 2 y + z + 6 = 0.
Answer:
Given eqns. of planes are
x + 2 y – 2 z = -6 …………………….. (1)
2 x + 2 y + z = -6 …………………….. (2)
and
2 x + 2 y + z = -6
direction ratios of normal to plane (1) are < 1, 2, -2 > and d ratios of normal to plane (2) are < 2, 2, 1 >
Let θ be the angle between given planes
∴ cosθ = \(\frac{(1 \times 2+2 \times 2-2 \times 1)}{\sqrt{1^2+2^2+(-2)^2} \sqrt{2^2+2^2+1^2}}\) = \(\frac{4}{3 \times 3}\) = \(\frac{4}{9}\)
∴ θ = cos^-1 \(\frac{4}{9}\)

Question 10.
Find the angle between the line
\(\frac{x-6}{3}\) = \(\frac{y-7}{2}\) = \(\frac{z-7}{-2}\) and the plane x + y + 2 z = 0 .
Answer:
eqn. of given line be \(\frac{x-6}{3}\) = \(\frac{y-7}{2}\) = \(\frac{z-7}{-2}\)
∴ direction ratios of given line are < 3, 2, -2 >
and eqn. of given plane be x + y + 2 z = 0
∴ direction ratios of normal to given plane are < 1, 1, 2 >
Let θ be angle between line and plane then 90° – θ be the angle between normal to plane and given line.
OP Malhotra Class 12 Maths Solutions Chapter 24 The Plane Ex 24(f) 1
∴ cos(90° – θ ) = \(\frac{3 \times 1+2 \times 1-2 \times 2}{\sqrt{3^2+2^2+(-2)^2} \sqrt{1^2+1^2+2^2}}\)
⇒ sinθ = \(\frac{1}{\sqrt{17} \sqrt{6}}\) = \(\frac{1}{\sqrt{102}}\)
∴ θ = sin^-1 (\(\frac{1}{\sqrt{102}}\))

Question 11.
Find the equation of a plane through the point (-1, -1, 2) and perpendicular to the planes 3 x + 2 y – 3 z = 1 and 5 x – 4 y + z = 5.
Answer:
The eqn. of plane through a the point (-1 ,-1, 2) be given by
a(x + 1) + b(y + 1) + c(z – 2) = 0 …………………….. (1)
where < a, b, c > are the direction ratios of normal to plane (1).
Since plane (1) is ⊥ to planes 3 x + 2 y – 3 z = 1 and 5 x – 4 y + z = 5
∴ 3 a + 2 b – 3 c = 0 …………………….. (2)
5 a – 4 b + c = 0
and
5 a – 4 b + c = 0 …………………….. (3)
on solving eqn. (2) and (3) simultaneously by cross multiplication method, we have
\(\frac{a}{2-12}\) = \(\frac{b}{-15-3}\) = \(\frac{c}{-12-10}\)
i.e. \(\frac{a}{-10}\) = \(\frac{b}{-18}\) = \(\frac{c}{-22}\)
i.e. \(\frac{a}{5}\) = \(\frac{b}{9}\) = \(\frac{c}{11}\) = k (say) where k ≠ 0
∴ a = 5 k ; b = 9 k and c = 11 k
putting the vaues of a, b and c in eqn. (1); we have
5 k(x + 1) + 9 k(y + 1) + 11 k(z – 2) = 0
⇒ 5 x + 9 y + 11 z – 8 = 0 be the reqd. plane.

Question 12.
Find the equations to the planes passing through the points (0, 4, – 3) and (6, -4, 3) if the sum of their intercepts on the three axes is zero.
Answer:
Let the eqn. of plane (using intercept form) be given by
\(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1 …………………….. (1)
It passes through the point (0, 4, – 3) and (6, -4, 3).
∴ \(\frac{0}{a}\) + \(\frac{4}{b}\) – \(\frac{3}{c}\) = 1
⇒ \(\frac{4}{b}\) – \(\frac{3}{c}\) = 1 …………………….. (2)
\(\frac{6}{a}\) – \(\frac{4}{b}\) + \(\frac{3}{c}\) = 1 …………………….. (3)
a + b + c = 0
also
a + b + c = 0 …………………….. (4)
on adding (2) and (3); we get
\(\frac{6}{a}\) = 2 ⇒ a = 3
∴ from (4); b + c = -3 ⇒ c = -3 – b …………………….. (5)
∴ from (2); 4 c – 3 b = b c ⇒ 4(-3 – b) – 3 b = b(-3 – b)
⇒ 2 – 7 b = -3 b – b²
⇒ b² – 4 b – 12 = 0
⇒ (b + 2)(b – 6) = 0
⇒ b = -2, 6
when b = -2 ∴ from (5); c = -3 + 2 = -1
when b = 6 ∴ from (5); c = -3 – 6 = -9
putting the values of a, b and c in eqn. (1); we have
\(\frac{x}{3}\) + \(\frac{y}{-2}\) + \(\frac{z}{-1}\) = 1 and
\(\frac{x}{3}\) + \(\frac{y}{6}\) + \(\frac{z}{-9}\) = 1 be the required eqn’s of planes.

Question 13.
Find the equation of the plane through the point (1, 2, 3) and perpendicular to the planes
x + y + 2 z = 3 and 3 x + 2 y + z = 4
Answer:
The eqn. of any plane through the point (1, 2, 3) be given by
a(x – 1) + b(y – 2) + c(z – 3) = 0 …………………….. (1)
where < a, b, c > are the direction ratios of normal to plane (1).
Now plane (2) is ⊥ to given planes x + y + 2 z = 3 and 3 x + 2 y + z = 4
a + b + 2 c = 0 …………………….. (2)
3 a + 2 b + c = 0 …………………….. (3)
and on solving eqn. (2) and eqn. (3) simultaneusly by cross multiplication method, we have
\(\frac{a}{1-4}\) = \(\frac{b}{6-1}\) = \(\frac{c}{2-3}\)
i.e. \(\frac{a}{-3}\) = \(\frac{b}{5}\) = \(\frac{c}{-1}\) = k (say) where k ≠ 0
∴ a = -3 k ; b = 5 k ; c = -k
putting the values of a, b and c in eqn. (1); we get
-3 k(x – 1) + 5 k(y – 2) – k(z – 3) = 0
⇒ -3 x + 5 y – z – 4 = 0
⇒ 3 x – 5 y + z + 4 = 0 which is the required plane.

Question 14.
Find the coordinates of the point where the line joining the points (1, -2, 3) and (2, -1, 5) cuts the plane x – 2 y + 3 z = 19. Hence, find the distance of this point from the point (5, 4, 1).
Answer:
The eqn. of line through the points (1, -2, 3) and (2, -1, 5) is given by
\(\frac{x-1}{2-1}\) = \(\frac{y+2}{-1+2}\) = \(\frac{z-3}{5-3}\) …………………….. (1)
i.e. \(\frac{x-1}{1}\) = \(\frac{y+2}{1}\) = \(\frac{z-3}{2}\) = t (say)
Any point on line (1) be given by (t + 1, t – 2, 2 t + 3). It is given that this point lies on given plane
x – 2 y + 3 z = 19
⇒ t + 1 – z(t – 2) + 3(2 t + 3) = 19
⇒ 5 t = 5 ⇒ t = 1
∴ required point of intersection be (1 + 1, 1 – 2, 2 + 3) i.e. (2, -1, 5)
Thus required distance of point P(2, -1, 5) from given point (5, 4, 1)
= \(\sqrt{(5-2)^2+(4+1)^2+(1-5)^2}\)
= \(\sqrt{9+25+16}\)
= \(\sqrt{50}\) = 5 \(\sqrt{2}\) units

Question 15.
Find the equation of the plane which contains the line \(\frac{x-1}{2}\) = \(\frac{y+1}{-1}\) = \(\frac{z-3}{4}\) and is perpendicular to the plane x + 2 y + z = 12.
Answer:
Given eqn. of line be \(\frac{x-1}{2}\) = \(\frac{y+1}{-1}\) = \(\frac{z-3}{4}\) …………………….. (1)
The eqn. of any plane through the line (1) be given by
a(x – 1) + b(y + 1) + c(z – 3) = 0 …………………….. (2)
where
2 a – b + 4 c = 0 …………………….. (3)
plane (1) is ⊥ to given plane x + 2 y + z – 12 = 0
∴ a + 2 b + c = 0 …………………….. (4)
on solving eqn. (3) and enq. (4) simultaneously by cross-multiplication method, we have
\(\frac{a}{-1-8}\) = \(\frac{b}{4-2}\) = \(\frac{c}{4+1}\)
i.e. \(\frac{a}{-9}\) = \(\frac{b}{2}\) = \(\frac{c}{5}\) = k (say)
∴ a = -9 k ; b = 2 k ; c = 5 k ; where k ≠ 0
putting the value of a, b, c in eqn. (2); we have
-9 k(x – 1) + 2 k(y + 1) + 5 k(z – 3) = 0
⇒ -9 x + 2 y + 5 z – 4 = 0
9 x – 2 y – 5 z + 4 = 0 which is the reqd. plane of plane.

Question 16.
Find the shortest distance between the lines whose vector equations are
\(\vec{r}\) = (4 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\)) + λ(\(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\)) and \(\vec{r}\) = (2 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)) + μ(3 \(\hat{i}\) + 2 \(\hat{j}\) – 4 \(\hat{k}\))
Answer:
Given eqns. of lines are ;
\(\vec{r}\) = (4 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\)) + λ(\(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\)) and \(\vec{r}\)
= (2 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)) + λ(3 \(\hat{i}\) + 2 \(\hat{j}\) – 4 \(\hat{k}\))
on comparing with \(\vec{r}\)
= \(\vec{a}_1\) + λ \(\overrightarrow{b_1}\) and \(\vec{r}\)
= \(\vec{a}_2\) + λ \(\overrightarrow{b_2}\) we have
\(\overrightarrow{a_1}\) = 4 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\) ;
\(\overrightarrow{a_2}\) = 2 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)
\(\overrightarrow{b_1}\) = \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\) ;
\(\overrightarrow{b_2}\)
= 3 \(\hat{i}\) + 2 \(\hat{j}\) – 4 \(\hat{k}\)
∴ \(\overrightarrow{b_1}\) × \(\vec{b}_2\)
= \(|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k}
1 & 2 & -3
3 & 2 & -4
\end{array}|\)
= \(\hat{i}\)(-8 + 6) – \(\hat{j}\)(-4 + 9) + \(\hat{k}\)(2 – 6)
= -2 \(\hat{i}\) – 5 \(\hat{j}\) – 4 \(\hat{k}\)
∴ \(|\vec{b}_1 \times \vec{b}_2|\)
= \(\sqrt{(-2)^2+(-5)^2+(-4)^2}\) = \(\sqrt{4+25+16}\) = \(\sqrt{45}\)
and \(\vec{a}_2\) – \(\vec{a}_1\)
= (2 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)) – (4 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\))
= -2 \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\)
Thus (\(\vec{a}_2\) – \(\vec{a}_1\)) \(\overrightarrow{b_1}\) × \(\overrightarrow{b_2}\)
= (-2 \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\)) (-2 \(\hat{i}\) – 5 \(\hat{j}\) – 4 \(\hat{k}\))
= (-2)(-2) + 2(-5) – 3(-4) = 4 – 10 + 12 = 6
Thus, required S.D between given lines = \(\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1}) \cdot(\overrightarrow{b_1} \times \vec{b}_2)|}{|\overrightarrow{b_1} \times \overrightarrow{b_2}|}\)
= \(\frac{6}{\sqrt{45}}\) = \(\frac{6}{3 \sqrt{5}}\)
= \(\frac{2}{\sqrt{5}}\) units.

Question 17.
Find the equation of the plane passing through the line of intersection of the planes x + 2 y + 3 z – 4 = 0 and 3 z – y = 0 and perpendicular to the plane 3 x + 4 y – 2 z + 6 = 0.
Answer:
The given planes are
x + 2 y + 3 z – 4 = 0 …………………….. (1)
3 z – y = 0 …………………….. (2)
3 x + 4 y – 2 z + 6 = 0 …………………….. (3)
and
The eqn. of any plane through the intersection of plane (1) and plane (2) be given by
x + 2 y + 3 z – 4 + λ(3 z – y) = 0
⇒ x + (2 – λ) y + (3 + 3λ) z – 4 = 0 …………………….. (4)
Also d ratios of normal to plane (4) are < 1, 2, – λ, 3 + 3 λ >
D ratios of normal to plane (3) are given by < 3, 4, -2 >
Since pane (4) is ⊥ plane (3).
∴ 3.1 + 4(2 – λ) – 2(3 + 3λ) = 0
⇒ 3 + 8 – 4 λ – 6 – 6 λ = 0
⇒ 5 – 10λ = 0
⇒ λ = \(\frac{1}{2}\)
putting the value of λ in eqn. (4); we have
x + 2 y + 3 z – 4 + \(\frac{1}{2}\)(3 z – y) = 0
⇒ 2 x + 3 y + 9 z – 8 = 0 be the required plane.

Question 18.
Find the vector equation of the line passing through the point (-1, 2, 1) and parallel to the line \(\vec{r}\) = 2 \(\hat{i}\) + 3 \( \hat{j}\) – \(\hat{k}\) + λ(\(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\)). Also, find the distance between them.
Answer:
Given vector eqn. of line be \(\vec{r}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) – \(\hat{k}\) + λ(\(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\)) …………………….. (1)
So line (1) has direction ratios < 1, -2, 1 >
Since the required line is || to line (1)
∴ d ratios of required line are < 1, -2, 1 >
Hence required vector eqn. of line passing through the point where P.V
\(\overrightarrow{a_2}\) = \(-\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\) and || to vector \(\vec{b}\) = \(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\) be given by
\(\vec{r}\) = \(\overrightarrow{a_2}\) + λ \(\vec{b}\)
= (\(-\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\)) + λ(\(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\))

…………………….. (2)
where λ be the parameter
Here \(\vec{a}_1\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) – \(\hat{k}\) ;
\(\vec{a}_2\) = \(-\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\)
and \(\vec{b}\) = \(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\)
Thus required distance between || lines =
S.D between lines = \(\frac{|\vec{b} \times(\overrightarrow{a_2}-\overrightarrow{a_1})|}{|\vec{b}|}\) …………………….. (3)
Now, \(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\) = -3 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\)
∴ \(\overrightarrow{b_1}\) × \(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\)
= \(|\hat{i} \hat{j} \hat{k}
1 -2 1
-3 -1 2
|\)
= \(\hat{i}\)(-4 + 1) – \(\hat{j}\)(2 + 3) + \(\hat{k}\)(-1 – 6)
= -1 \(\hat{i}\) – 5 \(\hat{j}\) – 7 \(\hat{k}\)
∴ from (3); required distance = \(\frac{|-3 \hat{i}-5 \hat{j}-7 \hat{k}|}{\sqrt{1^2+(-2)^2+1^2}}\) = \(\frac{\sqrt{(-3)^2+(-5)^2+(-7)^2}}{\sqrt{6}}\) = \(\sqrt{\frac{83}{6}}\) units

Question 19.
Find the equation of the plane passing through the points A(2, 1, -3), B(-3, -2, 1) and C(2, 4, -1).
Answer:
The eqn. of any plane through the point (2, 1, -3) be given by
a(x – 2) + b(y – 1) + c(z + 3) = 0 …………………….. (1)
where < a, b, c > are the direction ratios of normal to plane (1).
The point B(-3, -2, 1) lies on eqn. (1); we have
a(-3 – 2) + b(-2 – 1) + c(1 + 3) = 0
⇒ -5 a – 3 b + 4 c = 0 …………………….. (2)
Also the plane (1) passes through the point C(2, 4, -1).
i.e. a(2 – 2) + b(4 – 1) + c(-1 + 3) = 0
i.e. 0 a + 3 b + 2 c = 0 …………………….. (3)
on solving eqn. (2) and eqn. (3) simultaneously using cross multiplication method, we have
\(\frac{a}{-6-12}\) = \(\frac{b}{0+10}\)
= \(\frac{c}{-15-0}\)
i.e. \(\frac{a}{-18}\) = \(\frac{b}{10}\) = \(\frac{c}{-15}\) = k (say);
where k ≠ 0
∴ a = -18 k ; b = 10 k ; c = -15 k
putting the values of a, b and c in eqn. (1) we have
-18 k(x – 2) + 10 k(y – 1) – 15 k(z + 3) = 0
⇒ -18 x + 10 y – 15 z – 19 = 0
18 x – 10 y + 15 z + 19 = 0 which is the required eqn. of plane.

Question 20.
Find the shortest distance between the line
\(\frac{x-8}{3}\) = \(\frac{y+9}{-16}\) = \(\frac{z-10}{7}\) and \(\frac{x-15}{3}\)
= \(\frac{y-29}{8}\) = \(\frac{5-z}{5}\)
Answer:
Equations of given lines are
and
\(\frac{x-8}{3}\) = \(\frac{y+9}{-16}\) = \(\frac{z-10}{7}\) …………………………… (1)
\(\frac{x-15}{3}\) = \(\frac{y-29}{8}\) = \(\frac{z-5}{-5}\) ……………………………. (2)
OP Malhotra Class 12 Maths Solutions Chapter 24 The Plane Ex 24(f) 2
Let < l, m, n > be the d cosines of line of S.D PQ
Since line P Q is \perp to both given lines (1) and (2).
3 l – 16 m + 7 n = 0 …………………………… (3)
3 l + 8 m – 5 n = 0 …………………………… (4)
on solving eqn. (3) and (4) simultaneously using cross-multiplication methods, we have
\(\frac{l}{80-56}\) = \(\frac{m}{21+15}\)
= \(\frac{n}{24+48}\)
i.e. \(\frac{l}{24}\) = \(\frac{m}{36}\) = \(\frac{n}{72}\)
i.e. \(\frac{l}{2}\) = \(\frac{m}{3}\) = \(\frac{n}{6}\) = k say
∴ l = 2 k ; m = 3 k and n = 6 k; where k ≠ 0
Since l² + m² + n² = 1
⇒ 4 k² + 9 k² + 36 k² = 1
⇒ 49 k² = 1
⇒ k = ± \(\frac{1}{7}\)
∴ l = ± \(\frac{2}{7}\) ; m = ± \(\frac{3}{7}\) and n = ± \(\frac{6}{7}\)
Thus D cosines of line of S.D are < \(\frac{2}{7}\), \(\frac{3}{7}\), \(\frac{6}{7}\) >
∴ SD = |PQ| = projection of line of AB and PQ
= |l(x₂ – x₁) + m(y₂ – y₁) + n(z₂ – z₁)|
= |\(\frac{2}{7}\)(15 – 8) + \(\frac{3}{7}\) (29 + 9) + \(\frac{6}{7}\)(5 – 10)|
= \(\frac{2}{7}\) × 7 + \(\frac{3}{7}\) × 38 + \(\frac{6}{7}\) × (-5)
= \(\frac{14+114-30}{7}\) = 14 units
Alter : In vector form;
\(\vec{a}_1\) = 8 \(\hat{i}\) – 9 \(\hat{j}\) + 10 \(\hat{k}\) ;
\(\vec{a}_2\) = 15 \(\hat{i}\) + 29 \(\hat{j}\) + 5 \(\hat{k}\)
\(\vec{b}_1\) = 3 \(\hat{i}\) – 16 \(\hat{j}\) + 7 \(\hat{k}\) ;
\(\vec{b}_2\) = 3 \(\hat{i}\) + 8 \(\hat{j}\) – 5 \(\hat{k}\)
∴ \(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\) = 7 \(\hat{i}\) + 38 \(\hat{j}\) – 5 \(\hat{k}\)
\(\overrightarrow{b_1}\) × \(\overrightarrow{b_2}\)
= \(|\hat{i} \hat{j} \hat{k}
3 -16 7
3 8 -5
|\)
= \(\hat{i}\) (80 – 56) – \(\hat{j}\)(-15 – 21) + \(\hat{k}\)(24 + 48)
= 24 \(\hat{i}\) + 36 \(\hat{j}\) + 72 \(\hat{k}\)
∴ |\(\overrightarrow{b_1}\) × \(\overrightarrow{b_2}\)|
= 12(2 \(\hat{i}\) + 3 \(\hat{j}\) + 6 \(\hat{k}\)
= 12 \(\sqrt{2^2+3^2+6^2}\) = 12 × 7 = 84
Thus, (\(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\)) (\(\overrightarrow{b_1}\) × \(\overrightarrow{b_2}\))
= (7 \(\hat{i}\) + 38 \(\hat{j}\) – 5 \(\hat{k}\)) (24 \(\hat{i}\) + 36 \(\hat{j}\) + 72 \(\hat{k}\))
= 7 × 24 + 38 × 36 – 5 × 72
= 12[14 + 114 – 30] = 12 × 98
∴ required S.D between lines
= \(\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1}) (\overrightarrow{b_1} \times \overrightarrow{b_2})|}{|(\overrightarrow{b_1} \times \overrightarrow{b_2})|}\)
= \(\frac{12 \times 98}{84}\)
= 14 units

Question 21.
Find the equation of the plane passing through the line of intersection of the planes x + 2 y + 3 z – 5 = 0 and 3 x – 2 y – z + 1 = 0 and cutting off equal intercepts on the x and z axes.
Answer:
The eqns. of given planes are ;
x + 2 y + 3 z – 5 = 0 …………………………… (1)
3 x – 2 y – z + 1 = 0 …………………………… (2)
and
Thus the eqn. of any plane through the line of intersection of two given planes be given by
(x + 2 y + 3 z – 5) + λ (3 x – 2 y – z + 1) = 0
⇒ (1+3λ) x + (2 – 2λ) y + (3 – λ) z – 5 + λ = 0
⇒ \(\frac{x}{\frac{5-\lambda}{1+3 \lambda}}\) + \(\frac{y}{\frac{5-\lambda}{2-2 \lambda}}\) + \(\frac{z}{\frac{5-\lambda}{3-\lambda}}\) = 1 …………………………… (3)
Here intercepts made by plane (3) on x and z-axis are ; \(\frac{5-\lambda}{1+3 \lambda}\) and \(\frac{5-\lambda}{3-\lambda}\)
According to given condition, we have
\(\frac{5-\lambda}{1+3 \lambda}\) = \(\frac{5-\lambda}{3-\lambda}\)
⇒ (5 – λ)[3 – λ – 1 – 3λ] = 0
⇒ (5 – λ)(2 – 4 λ) = 0
⇒ λ = 5, \(\frac{1}{2}\)
but 5 ≠ λ
∴ λ = \(\frac{1}{2}\)
putting the value of λ in eqn. (3); we have
\(\frac{5}{x}\) + y + \(\frac{5}{2}\) z – \(\frac{9}{2}\) = 0
⇒ 5 x + 2 y + 5 z – 9 = 0
be the required plane.

Question 22.
Find the equation of a line passing through the point (-1, 3, -2) and perpendicular to the lines: \(\frac{x}{1}\) = \(\frac{y}{2}\) = \(\frac{z}{3}\) and \(\frac{x+2}{-3}\) = \(\frac{y-1}{2}\) = \(\frac{z+1}{5}\)
Answer:
The eqns. of given lines are ;
\(\frac{x}{1}\) = \(\frac{y}{2}\) = \(\frac{z}{3}\) …………………………… (1)
and
\(\frac{x+2}{-3}\) = \(\frac{y-1}{2}\) = \(\frac{z+1}{5}\) …………………………… (2)
∴ D ratios of given lines (1) and (2) are < 1, 2, 3 > and < -3, 2, 5 >
Let the direction ratios of required line are < a, b, c >
Since the required line is ⊥ to line (1) and (2)
∴ a + 2 b + 3 c = 0 …………………………… (3)
-3 a + 2 b + 5 c = 0 …………………………… (4)
on solving eqn. (3) and eqn. (4) simultaneously
\(\frac{a}{10-6}\) = \(\frac{b}{-9-5}\) = \(\frac{c}{2+6}\)
i.e. \(\frac{a}{4}\) = \(\frac{b}{-14}\) = \(\frac{c}{8}\)
i.e. \(\frac{a}{2}\) = \(\frac{b}{-7}\) = \(\frac{c}{4}\)
Thus the required eqn. of line throught he point (-1, 3, -2) having d ratios < 2, -7, 4 > be given by \(\frac{x+1}{2}\) = \(\frac{y-3}{-7}\) = \(\frac{z+2}{4}\)

Question 23.
Find the equations of planes parallel to the plane 2 x – 4 y + 4 z = 7 and which are at a distance of five units from the point (3, -1, 2).
Answer:
eqn. of given plane be 2 x – 4 y + 4 z – 7 = 0 …………………………… (1)
∴ eqn. of any plane parallel to plane (1) be given by
2 x – 4 y + 4 z + k = 0 …………………………… (2)
Also given ⊥ distance from P(3, -1, 2) to plane (2) = 5 units
\(\frac{|2 \times 3-4 \times(-1)+4 \times 2+k|}{\sqrt{2^2+(-4)^2+4^2}}\) = 5
⇒ \(\frac{|18+k|}{6}\) = 5
⇒ |18 + k| = 30
⇒ 18 + k = ± 30
∴ k = ± 30 – 18 = 12, -48
∴ from (2); we have
2 x – 4 y + 4 z + 12 = 0 and 2 x – 4 y + 4 z – 48 = 0
which are the required eqns. of planes.

Question 24.
Find the equation of a line passing through the points P(-1, 3, 2) and Q(-4, 2, -2). Also, Q the point R(5, 5, λ) is collinear w,ith the points. P and Q, then find the values of λ
Answer:
Now, direction ratios of the line joining the points P(-1, 3, 2) and Q(-4, 2, -2) are
< -4 + 1, 2 – 3, -2 – 2 >
i.e. < -3, -1, -4 >
i.e. < 3, 1, 4 >
Thus the required eqn. of line through the point (-1, 3, 2) and is having direction ratios < 3, 1, 4 > be given by
\(\frac{x+1}{3}\) = \(\frac{y-3}{1}\) = \(\frac{z-2}{4}\) …………………………… (1)
Since the point R(5, 5, λ) is collinear with P and Q.
∴ P, Q and R lies on same straight line.
∴ R(5, 5, λ) lies on (1); we get
\(\frac{5+1}{3}\) = \(\frac{5-3}{1}\) = \(\frac{\lambda-2}{4}\)
⇒ 2 = 2 = \(\frac{\lambda-2}{4}\)
λ – 2 = 8
⇒ λ = 10

Question 25.
Find the equation of the plane passing through the points (2, -3, 1) and (-1, 1, -7) and perpendicular to the plane x – 2 y + 5 z + 1 = 0.
Answer:
The eqn. of any plane through the point (2, -3, 1) be given by
a(x – 2) + b(y + 3) + c(z – 1) = 0 …………………………… (1)
where < a, b, c > are the direction ratios of normal to plane (1).
Now plane (1) passes through the point (-1, 1, -7).
a(-1 – 2) + b(1 + 3) + c(-7 – 1) = 0
-3 a + 4 b – 8 c = 0 …………………………… (2)
i.e. x – 2 y + 5 z + 1 = 0
a – 2 b + 5 c = 0 …………………………… (3)
on solving eqn. (2) and eqn. (3) simultaneously by cross-multiplication method, we have
\(\frac{a}{20-16}\) = \(\frac{b}{-8+15}\)
= \(\frac{c}{6-4}\)
i.e. \(\frac{a}{4}\) = \(\frac{b}{7}\)
= \(\frac{c}{2}\) = k (say ); where k ≠ 0
∴ a = 4 k ; b = 7 k ; c = 2 k
putting the value of a, b and c in eqn. (1); we have
4 k(x – 2) + 7 k(y + 3) + 2 k(z – 1) = 0
⇒ 4 x + 7 y + 2 z + 11 = 0 be the reqd. eqn. of plane.

Question 26.
Find the equation of the plane passing through the intersection of the planes:
x + y + 1 = 0 and 2 x – 3 y + 5 z – 2 = 0 and the point (-1, 2, 1).
Answer:
The eqn. of any plane through the line of intersection of given planes be given by
x + y + z – 1 + k(2 x – 3 y + 5 z – 2) = 0 …………………………… (1)
Now plane (1) passes through the point (-1, 2, 1).
∴ -1 + 2 + 1 – 1 + k(-2 – 6 + 5 – 2) = 0
⇒ 1 – 5 k = 0
⇒ k = 1 / 5
putting the value of k in eqn. (1); we have
x + y + z – 1 + \(\frac{1}{5}\)(2 x – 3 y + 5 z – 2) = 0
⇒ 7 x + 2 y + 10 z – 7 = 0
which is the reqd. eqn. of plane.

Question 27.
Find the shortest distance between the lines
\(\vec{r}\) = \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\) + λ(2
\(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\)) and \(\vec{r}\) = 2 \(\hat{i}\) + 4 \(\hat{j}\) + 5 \(\hat{k}\) + µ(4 \(\hat{i}\) + 6 \(\hat{j}\) + 8 \(\hat{k}\))
Answer:
Given lines are ;
and
\(\vec{r}\) = \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\) + λ(2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\)) …………………………… (1)
\(\vec{r}\) = 2 \(\hat{i}\) + 4 \(\hat{j}\) + 5 \(\hat{k}\) + µ(4 \(\hat{i}\) + 6 \(\hat{j}\) + 8 \(\hat{k}\))
⇒ \(\vec{r}\) = 2 \(\hat{i}\) + 4 \(\hat{j}\) + 5 \(\hat{k}\) + λ(2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\)) …………………………… (2)
where λ = 2µ
on comparing eqn. (1) and eqn. (2) with
\(\vec{r}\) = \(\overrightarrow{a_1}\) + λ\(\vec{b}\) and \(\vec{r}_2\) = \(\overrightarrow{a_2}\) + λ \(\vec{b}\)
∴ \(\vec{a}_1\) = \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\) ;
\(\vec{a}_2\) = 2 \(\hat{i}\) + 4 \(\hat{j}\) + 5 \(\hat{k}\)
and \(\vec{b}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\)
⇒ \(\vec{a}_2\) – \(\vec{a}_1\) = \(\hat{i}\) + 2 \(\hat{j}\) + 2 \(\hat{k}\)
∴ \(\vec{b}\) × (\(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\))
= \(|\hat{i} \hat{j} \hat{k}
2 3 4
1 2 2
|\)
= \(\hat{i}\)(6 – 8) – \(\hat{j}\)(4 – 4) + \(\hat{k}\)(4 – 3)
= -2 \(\hat{i}\) + 0 \(\hat{j}\) + \(\hat{k}\)
Thus |\(\vec{b}\) × (\(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\))|
= \(\sqrt{(-2)^2+0^2+1^2}\) = \(\sqrt{5}\) and
\(\vec{b}\) = \(\sqrt{2^2+3^2+4^2}\) = \(\sqrt{29}\)
∴ required S.D between parallel lines = \(\frac{|\vec{b} \times (\overrightarrow{a_2}-\overrightarrow{a_1})|}{|\vec{b}|}\)
= \(\sqrt{\frac{5}{29}}\) units

Question 28.
Find the image of the point (2, -1, 5) in the line \(\frac{x-11}{10}\) = \(\frac{y+2}{-4}\) = \(\frac{z+8}{-11}\). Also, find the length of the perpendicular from the point (2, -1, 5) to the line.
Answer:
Let P(2, -1, 5) be the given point and eqn. of given line AB be
\(\frac{x-11}{10}\) = \(\frac{y+2}{-4}\) = \(\frac{z+8}{-11}\) = t (say)
So any point on given line AB be M(10 t + 11, – 4 t – 2, -11 t – 8)
Let this point M} be the foot of ⊥ drawn from P on AB.
OP Malhotra Class 12 Maths Solutions Chapter 24 The Plane Ex 24(f) 3
Produce PM to P s.t PM = MP.
Then P be the image of P in AB and let (α, β, γ) be its coordinates.
∴ d ratios of line PM are
< 10 t + 11 – 2, -4 t – 2 + 1, -11 t – 8 – 5 >
i.e. < 10 t + 9, -4 t – 1, -11 t – 13 >
and d ratios of given line AB are < 10, -4, -11 >
Since line PM is ⊥ to line AB.
∴ 10(10 t + 9) – 4(-4 t – 1) – 11(-11 t – 13) = 0
⇒ 100 t + 90 + 16 t + 4 + 121 t + 143 = 0
⇒ 237 t + 237 = 0
⇒ t = -1
∴ coordinates of M are (-10 + 11, 4 – 2, 11 – 8) i.e. (1, 2 , 3)
∴ coordinates of M are (-10 + 11, 4 – 2, 11 – 8) i.e. (1, 2, 3)
Since M be the mid point of PP.
∴ \(\frac{α+2}{2}\) = 1
⇒ α + 2 = 2
⇒ α = 0
\(\frac{β-1}{2}\) = 2
⇒ β – 1 = 4
\(\frac{γ+5}{2}\) = 3
⇒ γ + 5 = 6
⇒ β = 5
γ = 1
Thus, the required image of P be P(0, 5, 1).

Question 29.
Find the cartesian equation of the plane passing through the line of intersection of the planes:
\(\vec{r}\) (2 \(\hat{i}\) + 3 \(\hat{j}\) – 4 \(\hat{k}\)) + 5 = 0 and
\(\vec{r}\)(\(\hat{i}\) – 5 \(\hat{j}\) + 7 \(\hat{k}\)) + 2 = 0 and intersecting y-axis at (0, \(\mathbf{3}\), \(\mathbf{0}\)).
Answer:
Cartesian eqns. of given planes are
\(\vec{r}\) (2 \(\hat{i}\) + 3 \(\hat{j}\) – 4 \(\hat{k}\)) + 5 = 0
⇒ (x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)) (2 \(\hat{i}\) + 3 \(\hat{j}\) – 4 \(\hat{k}\)) + 5 = 0
⇒ 2 x + 3 y – 4 z + 5 = 0 …………………………… (1)
and \(\vec{r}\)(\(\hat{i}\) – 5 \(\hat{j}\) + 7 \(\hat{k}\)) = -2
⇒ (x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)) (\(\hat{i}\) – 5 \(\hat{j}\) + 7 \(\hat{k}\)) + 2 = 0
⇒ x – 5 y + 7 z + 2 = 0 …………………………… (2)
Thus, the eqn. of any plane through the line of intersection of eqn. (1) and (2) be given by
(2 x + 3 y – 4 z + 5) + k(x – 5 y + 7 z + 2) = 0 …………………………… (3)
Since plane (3) passes through the point (0, 3, 0).
∴(0 + 9 – 0 + 5) + k(0 – 15 + 0 + 2) = 0
⇒ 14 – 13 k = 0
⇒ k = \(\frac{14}{13}\)
putting the value of k in eqn. (3); we have
(2 x + 3 y – 4 z + 5) + \(\frac{14}{13}\)(x – 5 y + 7 z + 2) = 0
⇒ 40 x – 31 y + 46 z + 93 = 0
which is the required eqn. of plane.

Question 30.
A line making angle 45° and 60° with the positive directions of the axes of x and y makes with the positive direction of z-axis, an angle of
(a) 60°
(b) 120°
(c) 60° and 120°
(d) None of these
Answer:
Let θ be the acute angle made by line with z-axis.
Then direction cosines of required line are < cos \(\frac{\pi}{4}\), cos \(\frac{\pi}{3}\), cosθ >
i.e. < \(\frac{1}{\sqrt{2}}\), \(\frac{1}{2}\), cosθ >
Here l = \(\frac{1}{\sqrt{2}}\), m = \(\frac{1}{2}\) ; n = cosθ
Now l² + m² + n² = 1
⇒ (\(\frac{1}{\sqrt{2}}\))² + (\(\frac{1}{2}\))² + cos²θ = 1
⇒ \(\frac{1}{2}\) + \(\frac{1}{4}\) + cos²θ = 1
⇒ cos²θ = \(\frac{1}{4}\)
⇒ cosθ = \(\frac{1}{2}\)
⇒ θ = \(\frac{\pi}{3}\)
[ ∵ cosθ > 0 since θ be the acute angle]

Question 31.
A line makes equal angles with the coordinate axes. Its direction cosines are
(a) < f{0 , 0 , 0} >
(b) < ± 1, ± 1, ± 1 >
(c) < ± \(\frac{1}{\sqrt{3}}\), ± \(\frac{1}{\sqrt{3}}\), ± \(\frac{1}{\sqrt{3}}\) >
(d) < ± \(\frac{1}{3}\), ± \(\frac{1}{3}\), ± \(\frac{1}{3}\) >
Answer:
Let α be the equal angle made by line with the coordinate axes. Then direction cosines of line are < cosα, cosα, cosα > i.e. cos² α + cos² α + cos²α = 1
⇒ cosα = ± \(\frac{1}{\sqrt{3}}\)
∴ required direction cosines of line are < ± \(\frac{1}{\sqrt{3}}\), ± \(\frac{1}{\sqrt{3}}\), ± \(\frac{1}{\sqrt{3}}\) >
∴ Ans. (c)

Question 32.
The direction cosines of the line \(\frac{x+2}{2}\) = \(\frac{2 y-5}{3}\), z = -1 are
(a) (\(\frac{4}{5}\), \(\frac{3}{5}\), 0)
(b) (\(\frac{3}{5}\), \(\frac{4}{5}\), \(\frac{1}{5}\))
(c) (\(-\frac{3}{5}\), \(\frac{4}{5}\), 0)
(d) (\(\frac{4}{5}\), – \(\frac{1}{5}\), \(\frac{1}{5}\))
Answer:
Given eqn. of line be \(\frac{x+2}{2}\) = \(\frac{2 y-5}{3}\), z = -1
i.e. \(\frac{x+2}{2}\) = \(\frac{y-5 / 2}{3 / 2}\) = \(\frac{z+1}{0}\)
i.e. \(\frac{x+2}{4}\) = \(\frac{y-5 / 2}{3}\) = \(\frac{z+1}{0}\)
The direction ratios of given line are proportional to < 4, 3, 0 >.
∴ Direction cosines of given line are ;
< \(\frac{4}{\sqrt{16+9+0}}\), \(\frac{3}{\sqrt{16+9+0}}\), \(\frac{0}{\sqrt{16+9+0}}\)
i.e. < \(\frac{4}{5}\), \(\frac{3}{5}\), 0 >
∴ Ans. (a)

Question 33.
Find the length of the perpendicular to the line \(\frac{x}{1}\) = \(\frac{y-1}{2}\) = \(\frac{z-2}{3}\) from the point
(1, 6, 3).
(a) \(\sqrt{13}\)
(b) \(\sqrt{10}\)
(c) 3
(d) 2
Answer:
eqn. of given line be
\(\frac{x}{1}\) = \(\frac{y-1}{2}\) = \(\frac{z-2}{3}\) = t (say) ……………………… (1)

OP Malhotra Class 12 Maths Solutions Chapter 24 The Plane Ex 24(f) 4
Any point on given line be Q(t, 2 t + 1, 3 t + 2) and coordinates of given point P are (1, 6, 3)
∴ D’ratios of line PQ are < t – 1, 2 t + 1 – 6, 3 t + 2 – 3 >
i.e. < t – 1, 2 t – 5, 3 t – 1 >
Since PQ is ⊥ to given line (1).
∴(t – 1) 1 + (2 t – 5) 2 + (2 t – 1) 3 = 0
⇒ 14 t – 14 = 0 ⇒ t = 1
Thus coordinates of Q are (1, 3, 5)
∴|P Q| = \(\sqrt{(1-1)^2+(6-3)^2+(3-5)^2}\)
= \(\sqrt{0+9+4}\) = \(\sqrt{13}\)
∴ Ans. (a)

Question 34.
(i) The angle between the lines
\(\frac{x-5}{-3}\) = \(\frac{y+3}{-4}\) = \(\frac{z-7}{0}\)
\(\frac{x}{1}\) = \(\frac{y-1}{-2}\) = \(\frac{z-6}{2}\) is
(a) \(\frac{\pi}{3}\)
(b) tan^-1 (\(\frac{1}{5}\))
(c) cos^-1 (\(\frac{1}{3}\))
(d) \(\frac{\pi}{2}\)
(ii) The angle between the lines \(\vec{r}\)
= (2 \(\hat{i}\) – 3 \(\hat{j}\) + \(\hat{k}\)) + λ(\(\hat{i}\) + 4 \(\hat{j}\) + 3 \(\hat{k}\)) and \(\vec{r}\)
= (\(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\)) + µ(\(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\)) is
(a) cos^-1 (\(\frac{9}{\sqrt{91}}\))
(b) cos ^-1 (\(\frac{7}{\sqrt{84}}\))
(c) \(\frac{\pi}{3}\)
(d) \(\frac{\pi}{2}\)
Answer:
(i) D’ ratio’s of given lines are < -3, -4, 0 > and < 1, -2, 2 >
Let θ be the angle between given lines.
Then cos θ = \(\frac{-3 \times 1-4 \times(-2)+0 \times 2}{\sqrt{9+16+0} \sqrt{1+4+4}}\)
= \(\frac{5}{5 \times 3}\) = \(\frac{1}{3}\)
⇒ θ = cos^-1 (\(\frac{1}{3}\))
∴ Ans. (c)

(ii) Comparing given lines with
\(\vec{r}\) = \(\overrightarrow{a_1}\) + λ \(\overrightarrow{b_1}\) and \(\vec{r}\)
= \(\overrightarrow{a_2}\) + λ \(\overrightarrow{b_2}\)
∴ \(\overrightarrow{b_1}\) = \(\hat{i}\) + 4 \(\hat{j}\) + 3 \(\hat{k}\);
\(\overrightarrow{b_2}\) = \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\)
⇒ \(\overrightarrow{b_1}\) \(\overrightarrow{b_2}\)
= 1 × 1 + 4 × 2 + 3 × (-3) = 0
Let θ be the angle between given lines.
Then cos θ
= \(\frac{\overrightarrow{b_1}\overrightarrow{b_2}}{|\overrightarrow{b_1}||\overrightarrow{b_2}|}\)
= 0 ⇒ θ = \(\frac{\pi}{2}\)
∴ Ans. (d)

Question 35.
The value of λ for which the lines are \(\frac{1-x}{3}\) = \(\frac{y-2}{2 \lambda}\)
= \(\frac{z-3}{2}\) and \(\frac{x-1}{3 \lambda}\) = \(\frac{y-1}{1}\)
= \(\frac{6-z}{7}\) are perpendicular to each other is
(a) -1
(b) -2
(c) 1
(d) 2
Answer:
Given eqns. of lines are
\(\frac{1-x}{3}\) = \(\frac{y-2}{2 \lambda}\) = \(\frac{z-3}{2}\)
Here a₁ = -3 ; b₁ = 2λ ; c₁ = 2
and \(\frac{x-1}{3 \lambda}\) = \(\frac{y-1}{1}\) = \(\frac{6-z}{7}\)
i.e. a₂ = 3λ ; b₂ = 1 ; c₂ = -7
Since given linnes are perpendicular to each other.
∴ a₁ a₂ + b₁ b₂ + c₁ c₂
= 0 – 3(3λ) + 2 λ(1) + 2(-7) = 0
⇒ -7λ – 14 = 0
⇒ λ = -2

Question 36.
A straight line joining the points (1, 1, 1) and (0, 0, 0) intersects the plane 2 x + 2 y + z = 10 at
(a) (1, 2, 5)
(b) (2, 2, 2)
(c) (2, 1, 5)
(d) (1, 1, 6)
Answer:
eqn. of any line joining the points (1, 1, 1) and (0, 0, 0) be given by
\(\frac{x-1}{1}\) = \(\frac{y-1}{1}\) = \(\frac{z-1}{1}\) = t (say)
So any point on line (1) be given by
P(t + 1, t + 1, t + 1)
Now given line intersects given plane
2 x + 2 y + z = 10
⇒ 2(t + 1) + 2(t + 1) + t + 1 = 10
⇒ 5 t + 5 = 10
⇒ t = 1
Thus the required point of intersection of given line and plane be (2, 2, 2).

Question 37.
Lines \(\frac{x-2}{1}\) = \(\frac{y-3}{1}\) = \(\frac{z-4}{-k}\) and \(\frac{x-1}{k}\)
= \(\frac{y-4}{2}\) = \(\frac{z-5}{1}\) are coplanar if
(a) k = 2
(b) k = 0
(c) k = 3
(d) k = -1
Answer:
We know that, the lines \(\frac{x-x_1}{a_1}\) = \(\frac{y-y_1}{b_1}\) = \(\frac{z-z_1}{c_1}\) and
\(\frac{x-x_2}{a_2}\) = \(\frac{y-y_2}{b_2}\) = \(\frac{z-z_2}{c_2}\)
are coplanar if
\(|\begin{array}{ccc}
x_2-x_1 y_2-y_1 z_2-z_1
a_1 b_1 c_1
a_2 b_2 c_2
\end{array}|\) = 0
For given lines ;
x₁ = 2 ; y₁ = 3 ; z₁ = 4
x₂ = 1 ; y₂ = 4 ; z₂ = 5
a₁ = 1 ; b₁ = 1 ; c₁ = -k
a₂ = k ; b₂ = 2 ; c₂ = 1
Thus,
\(|\begin{array}{ccc}
1-2 4-3 5-4
1 1 -k
k 2 1
\end{array}|\) = 0
⇒ \( |\begin{array}{rrr}
-1 1 1
1 1 -k
k 2 1
\end{array}|\) = 0
Expanding along R₃
-1(1 + 2 k) – 1(1 + k² ) + 1(2 – k) = 0
⇒ -1 – 2 k – 1 – k² + 2 – k = 0
⇒ -3 k – k² = 0
⇒ -k(3 + k) = 0
⇒ k = 0,-3

Question 38.
Angle between the planes x + y + 2 z = 6 and 2 x – y + z = 9 is
(a) \(\frac{\pi}{4}\)
(b) \(\frac{\pi}{6}\)
(c) \(\frac{\pi}{3}\)
(d) \(\frac{\pi}{2}\)
Answer:
D’ratios of normal to given planes are < 1, 1, 2 > and < 2, -1, 1 > Let θ be the angle between given planes.
Then cos θ = \(\frac{1 \times 2-1 \times 1+2 \times 1}{\sqrt{4+1+1} \sqrt{1+1+4}}\)
= \(\frac{3}{\sqrt{6} \sqrt{6}}\)
= \(\frac{3}{6}\)
= \(\frac{1}{2}\)
⇒ θ = \(\frac{\pi}{3}\)

Question 39.
If the planes \(\vec{r}\) (2 \(\hat{i}\) – λ\(\hat{j}\) + 3 \(\hat{k}\)) = 0 and
\(\vec{r}\) (λ\(\hat{i}\) + 5 \(\hat{j}\) – \(\hat{k}\)) = 5 are perpendicular to each other then the value of λ² + λ is
(a) 0
(b) -2
(c) -1
(d) 2
Answer:
Cartesian eqns. of given planes are ;
2 x – λ y + 3 z = 0
and λx + 5 y – z = 5
On comparing with a₁ x + b₁y + c₁z = d₁ and a₂ x + b₂y + c₂z
= d₂
∴ a₁ = 2 ; b₁ = -λ ; c₁ = 3
a₂ = λ ; b₂ = 5 ; c₂ = -1
planes (1) and (2) are ⊥ to each other.
∴ a₁ a₂ + b₁ b₂ + c₁ c₂ = 0
2 × λ – λ × 5 + 3 × (-1) = 0
⇒ -3λ – 3 = 0
⇒ λ = -1
∴ λ² + λ = (-1)² – 1 = 0

Question 40.
If α, β and γ are the direction cosine of a line in space, then the value of sin²α + sin²β + sin²γ.
(a) 0
(b) 1
(c) -1
(d) 2
Answer:
Let \(\vec{a}\) be the vector that makes α, β, γ with OX, OY and OZ respectively.
Let < l, m, n> be the direction cosines of \(\vec{a}\)
∴ l = cos α ; m = cos β ; n = cos γ
Also l² + m² + n² = 1
cos² α + cos² β + cos²γ = 1
⇒ (1 – sin²α) + (1 – sin²β) + (1 – sin²γ) = 1
⇒ sin²α + sin²β + sin²γ = 3 – 1 = 2

Question 41.
The length of the perpendicular drawn from (1, 2, 3) to the line \(\frac{x-6}{3}\) = \(\frac{y-7}{2}\) = \(\frac{z-7}{-2}\) is
(a) 4
(b) 5
(c) 6
(d) 7
Answer:
Let L be the foot of ⊥ drawn from P(1, 2, 3) on given line. any point on given line be,
\(\frac{x-6}{3}\) = \(\frac{y-7}{2}\) = \(\frac{z-7}{-2}\) = t (say)

OP Malhotra Class 12 Maths Solutions Chapter 24 The Plane Ex 24(f) 5
i.e. x = 3 t + 6 ; y = 2 t + 7 ; z = -2 t + 7
Thus the coordinates of L are (3 t + 6, 2 t + 7, -2 t + 7)
∴ D’ ratios of line PL are proportional to < 3 t + 6 – 1, 2 t + 7 – 2, -2 t + 7 – 3 >
i.e. < 3 t + 5, 2 t + 5, -2 t + 4 >
also D’ ratios of given line be proportional to < 3, 2, -2 >.
Since P L is ⊥ to given line.
∴ (3 t + 5) 3 + (2 t + 5) 2 + (-2 t + 4)(-2) = 0
⇒ 9 t + 15 + 4 t + 10 + 4 t – 8 = 0
⇒ 17 t + 17 = 0
⇒ t = -1
∴ Coordinates of point L are (3, 5, 9).
∴ required ⊥ distance = |PL|
= \(\sqrt{(3-1)^2+(5-2)^2+(9-3)^2}\)
= \(\sqrt{4+9+36}\) = 7

Question 42.
The lines \(\frac{x-2}{1}\) = \(\frac{y-3}{1}\) = \(\frac{4-z}{k}\) and \(\frac{x-1}{k}\) = \(\frac{y-4}{2}\) = \(\frac{z-5}{-2}\)
are mutually perpendicular, if the value of k is
(a) \(-\frac{2}{3}\)
(b) \(\frac{2}{3}\)
(c) -2
(d) 2
Answer:
eqns. of given lines are ; \(\frac{x-2}{1}\) = \(\frac{y-3}{1}\) = \(\frac{4-z}{k}\)
and \(\frac{x-1}{k}\) = \(\frac{y-4}{2}\) = \(\frac{z-5}{-2}\)
Since lines (1) and (2) are mutually ⊥ to each other
∴ 1(k) + 1(2) + (-k)(-2) = 0
⇒ k + 2 + 2 k = 0
[∵ a₁ a₂ + b₁ b₂ + c₁ c₂ = 0]
⇒ 3 k = -2
⇒ k = \(-\frac{2}{3}\)

Question 43.
Similar question. The two lines x = a y + b, z = c y + d; and x = a y + b, z = c y + d are perpendicular to each other, if
(a) \(\frac{a}{a^{\prime}}\) + \(\frac{c}{c^{\prime}}\) = 1
(b) \(\frac{a}{a^{\prime}}\) + \(\frac{c}{c^{\prime}}\) = -1
(c) a a + c c = 1
(d) a a + c c = -1
Answer:
eqns. of given lines in cartesian form, can be written as ;
\(\frac{x-b}{a}\) = \(\frac{y}{1}\) = \(\frac{z-d}{c}\)
and \(\frac{x-b^{\prime}}{a^{\prime}}\) = \(\frac{y}{1}\)
= \(\frac{z-d^{\prime}}{c^{\prime}}\)
Direction ratios of given lines (1) and (2) are < a, 1, c > and < a, 1, c >
Now lines (1) and (2) are \perp to each other
if a × a + 1 × 1 + c × c = 0
if a a + 1 + c c = 0

Question 44.
The distance of the origin from the plane -2 x + 6 y – 3 z = -7 is
(a) 1 unit
(b) \(\sqrt{2}\) units
(c) 2 \(\sqrt{2}\) units
(d) 3 units
Answer:
∴ required ⊥ distance of O(0, 0, 0) from given plane = \(\frac{|-2 \times 0+6 \times 0-3 \times 0+7|}{\sqrt{4+36+9}}\)
OP Malhotra Class 12 Maths Solutions Chapter 24 The Plane Ex 24(f) 6
= \(\frac{7}{7}\) = 1

Question 45.
The two planes x – 2 y + 4 z = 10 and 18 x + 17 y – k z = 50 are perpendicular, if k is equal to
(a) -4
(b) 4
(c) 2
(d) -2
Answer:
eqns. of given planes are ;
x – 2 y + 4 z = 10
18 x + 17 y – k z = 50
On comparing eqns. (1) and (2) with
a₁ x + b₁y + c₁z = d₁
and a₂x + b₂y + c₂z = d₂
Here a₁ = 1 ; b₁ = -2 ; c₁ = 4 ;
a₂ = 18 ; b₂ = 17 ; c₂ = -k
We know that planes (1) and (2) are \perp to each other.
Then a₁ a₂ + b₁ b₂ + c₁ c₂ = 0
⇒ 1 × 18 – 2 × 17 + 4 × (-k) = 0
⇒ 18 – 34 – 4 k = 0
⇒ 4 k = -16
⇒ k = -4

Question 46.
The line \(\frac{x-2}{3}\) = \(\frac{y-3}{4}\) = \(\frac{z-4}{5}\) is parallel to the plane
(a) 2 x + 3 y + 4 z = 0
(b) 3 x + 4 y – 5 z = 7
(c) 2 x + y – 2 z = 0
(d) x – y + z = 2
Answer:
eqn. of given line be,
\(\frac{x-2}{3}\) = \(\frac{y-3}{4}\) = \(\frac{z-4}{5}\)
∴ D’ratios of given line (1) are < 3, 4, 5 > Let us take the option (b) ; the eqn. of given plane be
3 x + 4 y – 5 z = 7
∴ D’ratios of normal to plane (2) are < 3, 4, -5 >
Now line (1) is parallel to plane (2) iff normal to plane (2) is ⊥ to line (1)
Here, 3 × 3 + 4 × 4 + 5(-5) = 9 + 16 – 25 = 0

Question 47.
What is the distance (in units) between the two planes 3 x + 5 y + 7 z = 3 and 9 x + 15 y + 21 z = 9 ?
(a) 0
(b) 3
(c) \(\frac{6}{\sqrt{83}}\)
(d) 6
Answer:
eqns. of given planes are
3 x + 5 y + 7 z = 3
and 9 x + 15 y + 21 z = 9
Here \(\frac{3}{9}\) = \(\frac{5}{15}\) = \(\frac{7}{21}\) = \(\frac{1}{3}\)
Thus given planes (1) and (2) are coincident.
Let P(x, y, z) be any point on plane (1).
∴ distance between given planes = ⊥ distance of P(x, y, z) from plane (2)
= \(\frac{|3(3 x+5 y+7 z)-9|}{\sqrt{9^2+15^2+21^2}}\)
= \(\frac{|3 \times 3-9|}{\sqrt{81+225+441}}\) = 0

Question 48.
The equation of the line in vector form passing through the point (-1, 3, 5) and parallel to the line \(\frac{x-3}{2}\) = \(\frac{y-4}{3}\), z = 2 is
(a) \(\vec{r}\) = (\(-\hat{i}\) + 3 \(\hat{j}\) + 5 \(\hat{k}\)) + λ(2 \(\hat{i}\) + 3 \(\hat{j}\) + \(\hat{k}\))
(b) \(\vec{r}\) = (\(-\hat{i}\) + 3 \(\hat{j}\) + 5 \(\hat{k}\)) + λ(-2 \(\hat{i}\) + 3 \(\hat{j}\))
(c) \(\vec{r}\) = (2 \(\hat{i}\) + 3 \(\hat{j}\) – 2 \(\hat{k}\)) + λ(\(-\hat{i}\) + 3 \(\hat{j}\) + 5 \(\hat{k}\))
(d) \(\vec{r}\) = (2 \(\hat{i}\) + 3 \(\hat{j}\)) + λ(\(-\hat{i}\) + 3 \(\hat{j}\) + 5 \(\hat{k}\))
Answer:
eqn. of given line can be written as ;
\(\frac{x-3}{2}\) = \(\frac{y-4}{3}\) = \(\frac{z-2}{0}\)
∴ direction ratios of required line which is parallel to line (1) are < 2, 3, 0 >.
Hence vector eqn. of line through the point (-1, 3, 5) whose position vector \(\vec{a}\)
= \(-\hat{i}\) + 3 \(\hat{j}\) + 5 \(\hat{k}\) and having direction ratios < 2, 3, 0 >
i.e. || to vector
\(\vec{b}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) be given by \(\vec{r}\)
= \(\vec{a}\) + λ\(\vec{b}\)
where λ be the parameter \(\vec{r}\) = (\(-\hat{i}\) + 3 \(\hat{j}\) + 5 \(\hat{k}\)) + λ(2 \(\hat{i}\) + 3 \(\hat{j}\))

Question 49.
(i) The sine of the angle between the straight line \(\frac{x-2}{3}\) = \(\frac{y-3}{4}\) = \(\frac{z-4}{5}\) and the plane 2 x – 2 y + z = 5 is
(a) \(\frac{10}{6 \sqrt{5}}\)
(b) \(\frac{4}{5 \sqrt{2}}\)
(c) \(\frac{2 \sqrt{3}}{5}\)
(d) \(\frac{\sqrt{2}}{10}\)
Answer:
Equation of given line be \(\frac{x-2}{3}\) = \(\frac{y-3}{4}\) = \(\frac{z-4}{5}\)
Thus the line passing through the point (2, 3, 4) and having direction ratios < 3, 4, 5 >.
Thus vector equation of line passing through the point whose P.V be 2
\(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\) and parallel to \(\vec{b}\)
= 3 \(\hat{i}\) + 4 \(\hat{j}\) + 5 \(\hat{k}\) is given by \(\vec{r}\)
= 2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\) + λ(3 \(\hat{i}\) + 4 \(\hat{j}\) + 5 \(\hat{k}\)) and
given equation of plane be 2 x – 2 y + z = 5
D’ ratios of normal to plane are < 2, -2, 1 >
∴ \(\vec{n}\) = 2 \(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\)
Let θ be the angle between given plane and given line.
Then sinθ = \(\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\)
= \(\frac{(3 \hat{i}+4 \hat{j}+5 \hat{k}) \cdot(2 \hat{i}-2 \hat{j}+\hat{k})}{\sqrt{3^2+4^2+5^2} \sqrt{2^2+(-2)^2+1^2}}\)
= \(\frac{3(2)+4(-2)+5(1)}{\sqrt{9+16+25} \sqrt{4+4+1}}\)
= \(\frac{3}{5 \sqrt{2} \times 3}\) = \(\frac{1}{5 \sqrt{2}}\)
= \(\frac{\sqrt{2}}{10}\)

(ii) Similar questions. The plane 2 x – 3 y + 6 z – 11 = 0 makes an angle sin^-1 (α) with x-axis. The value of α is equal to
(a) \(\frac{\sqrt{3}}{2}\)
(b) \(\frac{\sqrt{2}}{3}\)
(c) \(\frac{2}{7}\)
(d) \(\frac{3}{7}\)
Answer:
The given plane be 2 x – 3 y + 6 z – 11 = 0
∴ \(\vec{n}\) = 2 \(\hat{i}\) – 3 \(\hat{j}\) + 6 \(\hat{k}\)
equation of given line be x-axis and direction ratios of x-axis are
< 1,0,0 > i.e., \(\vec{b}\) = \(\hat{i}\) + 0 \(\hat{j}\) + 0 \(\hat{k}\)
Since θ be the angle between given plane and given line.
∴ sinθ = \(\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\)
= \(\frac{(2 \hat{i}-3 \hat{j}+6 \hat{k}) \cdot(\hat{i}+0 \hat{j}+0 \hat{k})}{\sqrt{2^2+(-3)^2+6^2} \sqrt{1^2+0^2+0^2}}\)
= \(\frac{2(1)-3(0)+6(0)}{7 \times 1}\) = \(\frac{2}{7}\)
⇒ θ = sin^-1 \(\frac{2}{7}\)
Also given plane makes an angle sin^-1 α with given line.
⇒ α = \(\frac{2}{7}\)

Question 50.
(i) The point of intersection of the straight line \(\frac{x-2}{2}\) = \(\frac{y-1}{-3}\) = \(\frac{z+2}{1}\) with the plane x + 3 y – z + 1 = 0 is
(a) (4, -2, -1)
(b) (-5, 1, -1)
(c) (2, 0, 3)
(d) (5, -1, 3)
(ii) Similar questions. The value of λ for which the straight line \(\frac{x-\lambda}{3}\) = \(\frac{y-1}{2+\lambda}\)
= \(\frac{z-3}{-1}\) may line on the plane x – 2 y = 0 is
(a) 2
(b) 0
(c) -1 / 2
(d) no such value of λ exists.
Answer:
(i) eqn. of given line be, \(\frac{x-2}{2}\) = \(\frac{y-1}{-3}\) = \(\frac{z+2}{1}\) = t (say)
and eqn. of given plane be, x + 3 y – z + 1 = 0
Any point on line (1) be P(2 t + 2, -3 t + 1, t – 2)
For point of intersection of line (1) and plane (2) then any point P lies on plane (2).
2 t + 2 + 3(-3 t + 1) – (t – 2) + 1 = 0
⇒ 2 t + 2 – 9 t + 3 – t + 2 + 1 = 0
⇒ -8 t + 8 = 0 ⇒ t = 1
Hence, the coordinates of required point of intersection of line (1) and plane (2) are (4, -2, -1)

(ii) eqn. of given line be \(\frac{x-\lambda}{3}\) = \(\frac{y-1}{2+\lambda}\) = \(\frac{z-3}{-1}\)
and eqn. of given plane be x – 2 y = 0
D’ ratios of normal to plane (2) are < 1, -2, 0 >
and D’ratios of line (1) are < 3, 2 + λ, -1 >
Now line (1) lies on plane (2) if normal to plane (2) is ⊥ to line (1).
∴ 1 × 3 – 2(2 + λ) + 0 × (-1) = 0
⇒ 3 – 4 – 2λ = 0
⇒ -1 – 2λ = 0
⇒ λ = \(-\frac{1}{2}\)

Question 51.
The distance of the point (1, 0, 2) from the point of intersection of the line \(\frac{x-2}{3}\) = \(\frac{y+1}{4}\) = \(\frac{z-2}{12}\) and the plane x – y + z = 16, is
(a) 3 \(\sqrt{21}\)
(b) 13
(c) 2 \(\sqrt{14}\)
(d) 8
Answer:
eqn. of given line be, \(\frac{x-2}{3}\) = \(\frac{y+1}{4}\) = \(\frac{z-2}{12}\) = t (say)
and eqn. of given plane be, x – y + z – 16 = 0
Any point on line (1) be P(3 t + 2, 4 t – 1, 12 t + 2)
Now line (1) intersects plane (2). Then point P lies on plane (2).
∴ 3 t + 2 – (4 t – 1) + 12 t + 2 = 16 ⇒ 11 t = 11 ⇒ t = 1
Thus coordinates of point of intersection of line (1) and plane (2) are (3 + 2, 4 – 1, 12 + 2) i.e. (5, 3, 14)
∴ Required distance of Q(1, 0, 2) from P(5, 3, 14) = |P Q|
= \(\sqrt{(5-1)^2+(3-0)^2+(14-2)^2}\)
= \(\sqrt{16+9+144}\) = 13

Question 52.
Let P(-7, 1, -5) be a point on a plane and let O be the origin. If O P is a normal to the plane, then the equation of the plane is
(a) 7 x-y+5 z+75=0
(b) 7 x-y+5 z+80=0
(c) 7 x+y+5 z+80=0
(d) 7 x-y-5 z-75=0
Answer:
Let the eqn. of plane through P(-7, 1, -5) be a(x + 7) + b(y – 1) + c(z + 5) = 0 where < a, b, c > are the D’ratios of normal to plane (1).
given O P be also the normal to plane (1) and its direction ratios are <-7,1,-5 >
OP Malhotra Class 12 Maths Solutions Chapter 24 The Plane Ex 24(f) 8
Thus eqn. (1) becomes ;
-7(x + 7) + 1(y – 1) – 5(z + 5) = 0
⇒ – 7 x + y – 5 z – 75 = 0
⇒ 7 x – y + 5 z + 75 = 0

Question 53.
If the straight lines \(\frac{x-2}{1}\) = \(\frac{y-3}{1}\) = \(\frac{z-4}{0}\) and \(\frac{x-1}{k}\) = \(\frac{y-4}{2}\) = \(\frac{z-5}{1}\) are coplanar, then the value of k is
(a) -3
(b) 0
(c) 1
(d) 6
Answer:
We know that, \(\frac{x-x_1}{a_1}\) = \(\frac{y-y_1}{b_1}\) = \(\frac{z-z_1}{c_1}\) and \(\frac{x-x_2}{a_2}\) = \(\frac{y-y_2}{b_2}\) = \(\frac{z-z_2}{c_2}\) are coplanar
if \(|\begin{array}{ccc}x_2-x_1 y_2-y_1 z_2-z_1 a_1 b_1 c_1 a_2 b_2 c_2\end{array}|\) = 0
For given lines ;
x₁ = 2 ; y₁ = 3 ; z₁ = 4 ; x₂ = 1 ; y₂ = 4 ; z₂ = 5
a₁ = 1 = b₁ ; c₁ = 0 ; a₂ = k ; b₂ = 2 ; c₂ = 1
Here, \(|\begin{array}{ccc}-1 1 1 1 1 0 k 2 1\end{array}|\) = 0;
Expanding along R 1
-1(1 – 0) – 1(1 – 0) + 1(2 – k) = 0
⇒ -1 – 1 + 2 – k = 0
⇒ k = 0

Question 54.
The distance of the point (2, 1, 0) from the plane 2 x + y + 2 z + 5 = 0 is
(a) 10
(b) \(\frac{10}{3}\)
(c) \(\frac{10}{9}\)
(d) 5
Answer:
Required ⊥ distance of P(2, 1, 0) from given plane
= \(\frac{|2 \times 2+1 \times 1+2 \times 0+5|}{\sqrt{2^2+1^2+2^2}}\)
= \(\frac{10}{3}\)

Question 55.
If the distance of point 2 \(\hat{i}\) + 3 \(\hat{j}\) + λ \(\hat{k}\) from the plane \(\vec{r}\) (3 \(\hat{i}\) + 2 \(\hat{j}\) + 6 \(\hat{k}\)) = 13 is 5 units, then
(a) 6,- \(\frac{17}{3}\)
(b) -6, \(\frac{17}{3}\)
(c) -6, \(-\frac{17}{3}\)
(d) -6, \(\frac{17}{3}\)
Answer:
Cartesian eqn. of given plane be,
3 x + 2 y + 6 z = 13
given ⊥ distance of point (2, 3λ) from plane (1) = 5
⇒ \(\frac{|3 \times 2+2 \times 3+6 \times \lambda-13|}{\sqrt{3^2+2^2+6^2}}\) = 5
⇒ \(\frac{|12+6 \lambda-13|}{7}\) = 5
⇒ |6λ + 12 – 13| = 35
⇒ (6λ – 1)= ± 35
⇒ 6 λ = ± 35 + 1
⇒ λ = 6, – \(\frac{17}{3}\)

Question 56.
If the foot of the perpendicular from the origin to a plane is (1, 2, 3), then equation of the plane is
(a) 2 x – y + z = 3
(b) x + y + z = 6
(c) x – y – z = -4
(d) x + 2 y + 3 z = 14
Answer:
The eqn. of required plane through P(1, 2, 3) be given by a(x – 1) + b(y – 2) + c(z – 3) = 0 where < a, b, c > be the direction ratios of normal to plane (1).
Now OP is normal to plane (1).

OP Malhotra Class 12 Maths Solutions Chapter 24 The Plane Ex 24(f) 9
∴ its D’ratios are < 1 – 0, 2 – 0, 3 – 0 > i.e. < 1, 2, 3 >
Therefore eqn. (1) becomes ;
1(x – 1) + 2(y – 2) + 3(z – 3) = 0
⇒ x + 2 y + 3 z = 14

Question 57.
(i) If a line makes angles \(\frac{\pi}{2}\), \(\frac{3 \pi}{4}\) and \(\frac{\pi}{4}\) with x, y, z axes respectively, then its direction cosines are?
(ii) If a line has direction ratio 2, -1, 2, then what are the-direction cosives ?
(iii) Write the direction cosines of the line joining the points (1, 0, 0) and (0, 1, 1).
(iv) What are the direction cosines of a line which makes equal angles with the coordinate axes.
(v) Find the direction cosives of the vector joining the points A(1, 2, -3) and B(-1, -2, 1) directed from B to A.
Answer:
(i) We know that, direction cosines of a line are the cosines of the angles made by line with positive direction of coordinate axes.
Since the line makes an angle of \(\frac{\pi}{2}\), \(\frac{3 \pi}{4}\) and \(\frac{\pi}{4}\) with coordinate axes.
Then direction cosines of line are < cos \(\frac{\pi}{2}\), cos \(\frac{3 \pi}{4}\), cos \(\frac{\pi}{4}\) >
i.e. < 0, \(\frac{-1}{\sqrt{2}}\), \(\frac{1}{\sqrt{2}}\) >
[∵ cos \(\frac{3 \pi}{4}\) = cos (π –\(\frac{\pi}{4}\)) = -cos \(\frac{\pi}{4}\) = \(-\frac{1}{\sqrt{2}}\)

(ii) Given direction ratios of line are < 2, -1, -2 >
∴ Its direction cosines are
< \(\frac{2}{\sqrt{2^2+(-1)^2+2^2}}\), \(\frac{-1}{\sqrt{2^2+(-1)^2+2^2}}\), \(\frac{2}{\sqrt{2^2+(-1)^2+2^2}}\) >
i.e. < \(\frac{2}{3}\), \(-\frac{1}{3}\), \(\frac{2}{3}\) >

(iii) D’ratios of line joining the points (1, 0, 0) and (0, 1, 1) are < 0 – 1, 1 – 0, 1 – 0 >
i.e. < -1, 1, 1 >
∴ D’ cosines of line are < \(\frac{-1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\) >

(iv) Let α be the equal angle made by line with the coordinate axes. Then direction cosines of line are < cos α, cos α, cos α >.
i.e. cos² α + cos² α + cos² α = 1
⇒ cos α = ± \(\frac{1}{\sqrt{3}}\)
∴ required direction cosines of line are < ± \(\frac{1}{\sqrt{3}}\), ± \(\frac{1}{\sqrt{3}}\), ± \(\frac{1}{\sqrt{3}}\) >

(v) D’ratios of \(\overrightarrow{\mathrm{BA}}\) are < 1-(-1), 2-(-2),-3 – 1 >
i.e. < 2, 4, -4 >
∴ D cosines of \(\overrightarrow{B_A}\) are < \(\frac{2}{\sqrt{4+16+16}}\), \(\frac{4}{\sqrt{4+16+16}}\), \(\frac{-4}{\sqrt{4+16+16}}\) >
< \(\frac{2}{6}\), \(\frac{4}{6}\), \(\frac{-4}{6}\) >
i.e. < \(\frac{1}{3}\), \(\frac{2}{3}\), \(\frac{-2}{3}\) >

Question 58.
Write the distance of the point (2, 3, 4) from the x-axis.
Answer:.
Required distance of P(2, 3, 4) from x-axis = distance of P(2, 3, 4) from (2, 0, 0)
= (2 – 2)² + (3 – 0)² + (4 – 0)²
= \(\sqrt{9+16}\) = 5 units

Question 59.
The equations of a line are 5 x – 3 = 15 y + 7 = 3 – 10 z. Write the distance cosines of the line.
Answer:
Given line can be written as ;
5(x – \(\frac{3}{5}\)) = 15(y + \(\frac{7}{15}\)) = -10(z – \(\frac{3}{10}\))
⇒ \(\frac{x-\frac{3}{5}}{6}\) = \(\frac{y+\frac{7}{15}}{2}\) = \(\frac{z-\frac{3}{10}}{-3}\)
∴ D cosines of given line are
< \(\frac{6}{\sqrt{6^2+2^2+(-3)^2}}\), \(\frac{2}{\sqrt{6^2+2^2+(-3)^2}}\), \(\frac{-3}{\sqrt{6^2+2^2+(-3)^2}}\) >
i.e. < \(\frac{6}{7}\), \(\frac{2}{7}\), \(\frac{-3}{7}\) >

Question 60.
Find the direction cosines of the line \(\frac{4-x}{2}\) = \(\frac{y}{6}\) = \(\frac{1-z}{3}\)
Answer:
eqn. of given line can be written as ; \(\frac{x-4}{-2}\) = \(\frac{y}{6}\) = \(\frac{z-1}{-3}\)
∴ direction cosines of given line (1) be
< \(\frac{-2}{\sqrt{4+36+9}}\), \(\frac{6}{\sqrt{4+36+9}}\), \(\frac{-3}{\sqrt{4+36+9}}\) >
i.e. < \(\frac{-2}{7}\), \(\frac{6}{7}\), \(\frac{-3}{7}\) >

Question 61.
The equation of a line are given by \(\frac{3-x}{-3}\) = \(\frac{y+2}{-2}\) = \(\frac{z+2}{6}\). Write the direction cosines of a line parallel to the above line.
Answer:
eqn. of given line can be written as ;
\(\frac{x-3}{3}\) = \(\frac{y+2}{-2}\) = \(\frac{z+2}{6}\)
∴ D’ratios of line parallel to line (1) are < 3, -2, 6 >
∴ D’ratios of line parallel to given line (1) are ;
< \(\frac{3}{\sqrt{9+4+36}}\), \(\frac{-2}{\sqrt{9+4+36}}\), \(\frac{6}{\sqrt{9+4+36}}\) >
i.e. < \(\frac{3}{7}\), \(-\frac{2}{7}\), \(\frac{8}{7}\) >

Question 62.
Write the equation of a line parallel to the line \(\frac{x-2}{-3}\) = \(\frac{y+3}{2}\) = \(\frac{z+5}{6}\) and passing through the point (1, 2, 3).
Answer:
eqn. of given line be, \(\frac{x-2}{-3}\) = \(\frac{y+3}{2}\) = \(\frac{z+5}{6}\)
∴ D’ ratios of line || to line (1) are < -3, 2, 6 >
Hence eqn. of line passing through the point (1, 2, 3)
and having direction ratios are < -3, 2, 6 > be \(\frac{x-1}{-3}\) = \(\frac{y-2}{2}\) = \(\frac{z-3}{6}\)

Question 63.
Find the vector equation of a line which passes through the points (3, 4, -7) and (1, -1, 6).
Answer:
We know that, vector equation of line passing through the points with position vectors \(\vec{a}\) and \(\vec{b}\) is \(\vec{r}\) = \(\vec{a}\) + λ(\(\vec{b}\) – \(\vec{a}\))
Here \(\vec{a}\) = 3 \(\hat{i}\) + 4 \(\hat{j}\) – 7 \(\hat{k}\) ;
\(\vec{b}\) = \(\hat{i}\) – \(\hat{j}\) + 6 \(\hat{k}\)
∴ Required vector eqn. of line be
\(\vec{r}\) = 3 \(\hat{i}\) + 4 \(\hat{j}\) – 7 \(\hat{k}\) + λ\(\hat{i}\) – \(\hat{j}\) + 6 \(\hat{k}\) – 3 \(\hat{i}\) – 4 \(\hat{j}\) + 7 \(\hat{k}\)
⇒ \(\vec{r}\) = 3 \(\hat{i}\) + 4 \(\hat{j}\) – 7 \(\hat{k}\) + λ(-2 \(\hat{i}\) – 5 \(\hat{j}\) + 13 \(\hat{k}\))

Question 64.
Write the vector equation of the line given by \(\frac{x-5}{3}\) = \(\frac{y+4}{7}\) = \(\frac{z-6}{2}\)
Answer:
Given equation of line in cartesian form be given by \(\frac{x-5}{3}\) = \(\frac{y+4}{7}\) = \(\frac{z-6}{2}\)
So the given line pass through the point (5, -4, 6) whose P.V be \(\vec{a}\) = 5 \(\hat{i}\) – 4 \(\hat{j}\) + 6 \(\hat{k}\) and having direction ratios < 3, 7, 2 > i.e. || to vector \(\vec{b}\) = 3 \(\hat{i}\) + 7 \(\hat{j}\) + 2 \(\hat{k}\)
Thus the required vector equation of line be \(\vec{r}\) = \(\vec{a}\) + λ \(\vec{b}\) i.e. \(\vec{r}\) = (5 \(\hat{i}\) – 4 \(\hat{j}\) + 6 \(\hat{k}\)) + λ(3 \(\hat{i}\) + 7 \(\hat{j}\) + 2 \(\hat{k}\)) where λ be any scalar.

Question 65.
Find the vector equation of the line which passes through the point (3, 4, 5) and is parallel to the vector 2 \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\)
Answer:
We know that, vector eqn. of line passing through the point with P.V \(\vec{a}\) and || to \(\vec{b}\) be \(\vec{r}\) = \(\vec{a}\) + λ \(\vec{b}\)
Here, \(\vec{a}\) = 3 \(\hat{i}\) + 4 \(\hat{j}\) + 5 \(\hat{k}\) and \(\vec{b}\) = 2 \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\)
∴ required vector eqn. of line be \(\vec{r}\) = (3 \(\hat{i}\) + 4 \(\hat{j}\) + 5 \(\hat{k}\)) +
λ(2 \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\))

Question 66.
Find a Cartesian form of the equation of a line which passes through a point with position vector 2 \(\hat{i}\) – 3 \(\hat{j}\) + 4 \(\hat{k}\) and makes angles 60° , 120° and 45° with x, y, z-axis respectively.
Answer:
Since the required line pass through the point with P.V. 2 \(\hat{i}[latex] – 3 [latex]\hat{j}\) + 4 \(\hat{k}\)
∴ line must pass through the point (2, -3, 4).
D’ cosines of required line are < cos 60°, cos 120°, cos 45° >
i.e. < \(\frac{1}{2}\), \(-\frac{1}{2}\), \(\frac{1}{\sqrt{2}}\) >.
∴ D’ ratios of required line are < 1, -1, \(\sqrt{2}\) >
Hence eqn. of required line through (2, -3, 4) and having direction ratios < 1, -1, \(\sqrt{2}\) > be \(\frac{x-2}{1}\) = \(\frac{y+3}{-1}\) = \(\frac{z-4}{\sqrt{2}}\)

Question 67.
Find the acute angle between the lines \(\frac{x-4}{3}\) = \(\frac{y+3}{4}\) = \(\frac{z+1}{5}\) and \(\frac{x-1}{4}\) = \(\frac{y+1}{-3}\) = \(\frac{z+10}{5}\)
Answer:
We know that, The acute angle between the given lines \(\frac{x-x_1}{a_1}\) = \(\frac{y-y_1}{b_1}\) = \(\frac{z-z_1}{c_1}\)
and \(\frac{x-x_2}{a_2}\) = \(\frac{y-y_2}{b_2}\) = \(\frac{z-z_1}{c_2}\) be
cosθ = \(\frac{\left|a_1 a_2+b_1 b_2+c_1 c_2\right|}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\)
Here a₁ = 3 ; b₁ = 3 ; c₁ = 5 and a₂ = 4 ; b₂ = -3 ; c₂ = 5
∴ cosθ = \(\frac{|3 \times 4+4 \times(-3)+5 \times 5|}{\sqrt{9+16+25} \sqrt{9+16+25}}\)
= \(\frac{25}{\sqrt{50} \sqrt{50}}\) = \(\frac{1}{2}\)
⇒ θ = \(\frac{\pi}{3}\)

Question 68.
Find the Cartesian equation of the plane \(\vec{r}\) ( \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)) = 1.
Answer:
Let P(x, y, z) be any point on given plane so its P. V be \(\vec{r}\) = x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)
Thus eqn. of plane in cartesian form becomes ;
(x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\))(\(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)) = 1
⇒ x + y – z = 1

Question 69.
Find the equation of the plane which passes through the points (2, 0, 0)(0, 3, 0) and (0, 0, 4).
Answer:
Since the required plane passes through the points A(2, 0, 0), B(0, 3, 0) and C(0, 0, 4). Thus the plane made intercepts on x-axis, y-axis and z-axis are 2, 3 , and 4.
We know that equation of plane having intercepts a, b, c on coordinate axes is given by
\(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1
Here a = 2 ; b = 3 and c = 4
Thus equation (1) reduces to \(\frac{x}{2}\) + \(\frac{y}{3}\) + \(\frac{z}{4}\) = 1.

Question 70.
Find the sum of the intercepts made by the plane 2 x + y – z = 5 on the coordinate axes.
Answer:
Given eqn. of plane be 2 x + y – z = 5
⇒ \(\frac{x}{\frac{5}{2}}\) + \(\frac{y}{5}\) + \(\frac{z}{-5}\) = 1
We know that, lengths of intercepts made by plane \(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1
on coordinate axes are a, b and c respectively.
∴ a = \(\frac{5}{2}\) ; b = 5 and c = -5
∴ Required sum = \(\frac{5}{2}\) + 5 – 5 = \(\frac{5}{2}\)

Question 71.
Find the acute angle between the planes \(\vec{r}\) (\(\hat{i}\) – 2 \(\hat{j}\) – 2 \(\hat{k}\)) = 1 and \(\vec{r}\) (3 \(\hat{i}\) – 6 \(\hat{j}\) + 2 \(\hat{k}\)) = 0.
Answer:
Given eqn. of planes are \(\vec{r}\) (\(\hat{i}\) – 2 \(\hat{j}\) – 2 \(\hat{k}\)) = 1
and \(\vec{r}\)(3 \(\hat{i}\) – 6 \(\hat{j}\) + 2 \(\hat{k}\)) = 0
We know that, the angle between the planes \(\vec{r}\) \(\overrightarrow{n_1}\) = d₁ and \(\vec{r}\) \(\overrightarrow{n_2}\) = d₂be given by
cosθ = \(\frac{|\overrightarrow{n_1} \cdot \overrightarrow{n_2}|}{|\overrightarrow{n_1}||\overrightarrow{n_2}|}\)
Here \(\overrightarrow{n_1}\) = \(\hat{i}\) – 2 \(\hat{j}\) – 2 \(\hat{k}\);
\(\overrightarrow{n_2}\) = 3 \(\hat{i}\) – 6 \(\hat{j}\) + 2 \(\hat{k}\)
∴ \(\overrightarrow{n_1}\) \(\overrightarrow{n_2}\) = (\(\hat{i}\) – 2 \(\hat{j}\) – 2 \(\hat{k}\)) (3 \(\hat{i}\) – 6 \(\hat{j}\) + 2 \(\hat{k}\))
= 1(3) – 2(-6) – 2(2) = 3 + 12 – 4 = 11
and |\(\vec{n}_1\)| = \(\sqrt{1^2+(-2)^2+(-2)^2}\)
= 3 and |\(\overrightarrow{n_2}\)| = \(\sqrt{3^2+(-5)^2+2^2}\) = 7
∴ cosθ = \(\frac{11}{3 \times 7}\) = \(\frac{11}{21}\)
⇒ θ = cos^-1 \(\frac{11}{21}\)

Question 72.
Find the angle between the line \(\vec{r}\) = (5 \(\hat{i}\) – \(\hat{j}\) – 4 \(\hat{k}\))
+ λ(2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)) and the plane
\(\vec{r}\)(3 \(\hat{i}\) – 4 \(\hat{j}\) – \(\hat{k}\)) + 5 = 0 .
Answer:
Equation of given line be \(\vec{r}\) = (5 \(\hat{i}\) – \(\hat{j}\) – 4 \(\hat{k}\)) + λ(2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\))
on comparing with \(\vec{r}\) = \(\vec{a}\) + λ \(\vec{b}\)
Here \(\vec{b}\) = 2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)
and equation of given plane be \(\vec{r}\) (3 \(\hat{i}\) – 4 \(\hat{j}\) – \(\hat{k}\)) + 5 = 0
Here \(\vec{n}\) = 3 \(\hat{i}\) – 4 \(\hat{j}\)
\(-\hat{k}\)
Let θ be the angle between given line and given plane
Then
sinθ = \(\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\)
= \(\frac{(2 \hat{i}-\hat{j}+\hat{k}) (3 \hat{i}-4 \hat{j}-\hat{k})}{\sqrt{2^2+(-1)^2+1^2} \sqrt{3^2+(-4)^2+(-1)^2}}\)
= \(\frac{2(3)-1(-4)+1(-1)}{\sqrt{4+1+1} \sqrt{9+16+1}}\)
= \(\frac{9}{\sqrt{6} \sqrt{26}}\)

ICSE Solutions for Class 8 History and Civics – The Great Uprising of 1857

September 9, 2024

ICSE Solutions for Class 8 History and Civics – The Great Uprising of 1857

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for ICSE Solutions for Class 8 History and Civics. You can download the History and Civics ICSE Solutions for Class 8 with Free PDF download option. History and Civics for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

ICSE Solutions Class 8 History & Civics Geography Biology Chemistry Physics Maths

Time To Learn
I. Fill in the blanks:

By his policy of Doctrine of Lapse Lord Dalhousie annexed Nagpur and Jhansi.
Rani Laxmibai captured Gwalior with Tantya Tope’s help.
In Lucknow Hazrat Mahal led the revolt.
The rebels proclaimed Bahadur Shah Zafar as the emperor of India.

II. Match the contents of Column A and Column B:

Answer:
Column A Column B

III. State whether the following statements are True or False:

Awadh was annexed on the ground of malad-ministration.
True
The British supported Sati.
False.
The Indian soldiers were given lesser salary as compared to their British counterparts.
True.
The Revolt of 1857 did not involve the masses.
True.

IV. Answer the following questions:

Question 1.
How were the economic policies of the British responsible for the Revolt of 1857?
Answer:
The British had changed the entire economic structure of India.

Introduction of the new land revenue system which allowed the government to confiscate land and accept revenue in cash caused and suffering to the Indian peasants.
Due to Industrial Revolution markets were flooded with machine made goods. Indian Industries could not compete with them and thus declined.
When an Indian state was annexed, the administration was replaced by the company. This created unemployment and caused resentment.

Question 2.
Discuss the social and religious causes responsible for the Revolt of 1857.
Answer:

British advocated many reforms such as abolition of Sati, female infanticide, child marriage, widow remarriage. This caused discontent all over. These reforms were interpreted by Indians as interference in their religious and social customs.
Introduction of Railways and telegraph was viewed with fear and suspicion.
Introduction of western Education was viewed as an attack on Indian social and religious customs.
Policy of racial discrimination practiced by British led to anger and resentment.

Question 3.
Why do you think the Revolt of 1857 failed? Discuss four, causes for its failure.
Answer:

Though it was wide spread it failed to involve all sections of the population.
The revolt did not involve the masses. Only those people joined revolt who had been adversely affected by the British. Scindia, Holkar, Raja of Jodhpur did not join it. Instead they supported the British. Merchants, western educated middle class did not join it.
The British soldiers were better equipped. Indians had outdated weapons. The British had experienced and competent Generals. Indian leaders lacked this expertise.
The revolutionaries did not have enough resources. The British were stronger financially and militarily.
The revolt was not planned and organised.
The British had the advantage of post and telegraph. This helped them to communicate and exchange messages, plan strategies and act immediately.

Question 4.
Discuss the nature of the Uprising/Revolt of 1857.
Answer:
There are divergent view regarding the nature of the outbreak of 1857. British historians have called it a Mutiny – confined to the army which did not command the support of the people at large.
Same view was held by .contemporary Indians like Munshi Jiwan Lai, Moinuddin, (Both eye witnesses at Delhi) and Durgadas Bandopadhyaya. Others have described it as racial struggle for supremacy between the Black and White. Some describe it as a struggle between oriental and occidental civilisation and culture.
A few have described it as Hindu-Muslim conspiracy to overthrow the British rule. Some Indian have called it a “well planned struggle” and as “the first war of Indian Independence”.

V. Give reasons why:

1. State of Awadh was annexed by the British.
Ans. The state of Awadh was annexed on account of maladministration.

2. Indian artisans, weavers and craftsmen lost their means of livelihood.
Ans. With the coming of Industrial Revolution markets were flooded with machine made goods. Indian Industries could not compete

3.The Indian sepoys refused to bite the cartridges of the Enfield rifle.
Ans. The Indian sepoys refused to bite the cartridges of the Enfield rifle as it was rumoured that these cartridges were greased with the fat of cows and pigs. This enraged the Hindus and Muslims as the cow is a holy animal of Hindus and the pig is considered taboo for the Muslims.

VI. Picture Study –
The picture shows the revolt of 1857
ICSE Solutions for Class 8 History and Civics - The Great Uprising of 1857 3

Question 1.
How far were the greased catridges responsible for the Revolt?
Answer:
It was an immediate cause which supplied the spark to ignite diy firewood awaiting to be ignited. Other causes like oppressive economic policy, aggressive annexation policy, religious social interference by the British were already there waiting to be triggered into Revolt. This was provided by the rumour that cartridges were greased with the fat of cow and pig and hence this enraged the Hindus and Muslims.

Question 2.
Give your opinion as to how the Revolt would have been successful.
Answer:
It would have been successful if it had been well organised and well planned or if it had involved the masses. It would have been successful if merchants, educated middle class and zamindars had supported it and the Indian soldiers had better weapons and goods generals.

Additional Questions

EXERCISES
A. Fill in the blanks:

Dalhousie annexed the states of Satara, Nagpur, and Jhansi on the basis of the Doctrine of Lapse.
Prior to the outbreak of the Revolt, Bahadur Shah Zafar lived in Delhi as a Pensioner of the British.
The Revolt of 1857 started as a Mutiny of the sepoys.
The immediate cause of the Revolt was the issue of the greased cartridges.
The Revolt ended the rule of the English East India Company.

B. Match the following:

Answer:

ICSE Solutions for Class 8 History and Civics - The Great Uprising of 1857 6

C. Choose the correct answer:

1. When Awadh/Nagpur/Jhansi was annexed, the estates of the zamindars and talukdars were confiscated by the British.
Ans. When Awadh was annexed, the estates of the zamindars and talukdars were confiscated by the British.

2. The rumour regarding greased cartidges started in Madras/ Calcutta/Delhi.
Ans. The rumour regarding greased cartidges started in Calcutta.

3. Mangal Pandey was a sepoy at Barrackpore/Nagpur/Satara, who refused to use the greased cartidges in 1857.
Ans. Mangal Pandey was a sepoy at Barrackpore, who refused to use the greased cartidges in 1857.

4.The Revolt in Lucknow/Meerut/Kanpur was led by Nana Saheb.
Ans. The Revolt in Kanpur was led by Nana Saheb.

5. Nana Saheb/Hazrat Mahal/Rani Lakshmibai led the Revolt in Lucknow.
Ans. Hazrat Mahal led the Revolt in Lucknow.

D. State whether the following are true or false:

Indian rulers were quite satisfied with Lord Dalhousie’s expansionist policies.
False.
Correct: Indian rulers were dissatisfied with Lord Dalhousies expansionist policies.
Peasants benefitted from the land revenue system of the British.
False.
Correct : Peasants were not benefitted from the land revenue system of the British.
Social reforms such as the abolition of sati and female infanticide, and the Widow Remarriage Act caused deep resentment among the orthodox sections of society.
True.
The Doctrine of Lapse was abolished after the Revolt.
True
The Revolt of 1857 had come as a shock to the British.
True.

E. Answer the following questions in one or two words/ sentences:
Question 1.
Why did Rani of Jhansi become a staunch enemy of the British?
Answer:
Dalhousie annexed the states of Satara, Nagpur and Jhansi by applying the Doctrine of Lapse. The annexation transformed the courageous Rani of Jhansi into a staunch enemy of the British.

Question 2.
Why did Nana Saheb fight against the British during the Revolt of 1857?
Answer:
Nana Saheb, the adopted son of Peshwa Baji Rao II (a pensioner of the British), was denied a pension after his father’s death. Nana Saheb became one of the leaders of the Revolt.

Question 3.
Why did the zamindars and talukdars became sworn enemies of the British?
Answer:
When Awadh was annexed, the estates of the zamindars and talukdars were confiscated by the British. They became sworn enemies of British rule.

Question 4.
What happened to the nawab’s army when Awadh was annexed?
Answer:
After the annexation of Awadh, the nawab’s army was disbanded. The soldiers lost their means of livelihood and their bitterness against the British increased.

Question 5.
When and where did the Revolt of 1857 begin?
Answer:
May 10, 1857 was a Sunday. The British officers at the Meerut cantonment in North India were preparing to attend church, while many other British soldiers were off duty.

Question 6.
Name any two main centres of Revolt.
Answer:
The important centres of the Revolt were Meerut, Delhi, Kanpur, Lucknow and Jhansi.

Question 7.
What step was taken by the British to reorganize the army after the Revolt?
Answer:
The army was reorganized and strengthened. The number of British soldiers was increased and the artillery placed exclusively under their control.

Question 8.
Mention any two important results of the Revolt.
Answer:
Results of the Revolt 1857

The rule of the English East India Company came to an end.
Treaties with Indian states would be honoured.
The British government would not interfere in the social and religious customs of the people.

F. Answer the following questions briefly:

Question 1.
In the context of the Revolt of 1857, answer the following questions:

Mention any three political causes of the Revolt.
Mention the three economic factors that led to the outbreak of the great Revolt.
Explain briefly any four social and religious causes that led to the Revolt of 1857.

Answer:
(a) Political Causes:

Dalhousie annexed the states of Satara, Nagpur and Jhansi by applying the Doctrine of Lapse. The annexation transformed the courageous Rani into a staunch enemy of the British.
Nana Saheb, the adopted son of Peshwa Baji Rao II (pensioner of the British), was denied a pension after his father’s death. Nana Saheb became one of the leaders of the Revolt.
Bahadur Shah Zafar, the Mughal Emperor, lived in Delhi as a pensioner of the British. Dalhousie announced that Bahadur’s successor would not be allowed to stay on in the historic Red Fort. He would have to move to a place near the Qutb Minar, on the outskirts of Delhi. This was a great blow to the dignity of the Mughal emperor and deeply hurt the sentiments of the Muslims.
(The annexation of Awadh, on grounds of maladministration, outraged the people of India, in general, and Awadh, in particular. Awadh had always been a triendly. faithful and subordinate ally. The Nawab of Awadh was exiled to Calcutta.
The British showed no respect for the treaties they had signed with the Indians. Treaties were broken whenever it suited them to do so. This created a sense of fear and insecurity among the rulers of subordinate states. The axe could fall on them anywhere, at any time.

(b) Economic Causes:

The land revenue system, introduced by the British, caused great hardship and misery among the peasants. Under the zamindari system, for instance, the peasants were oppressed by the Zamindars and exploited by the moneylenders. If the cultivators failed to pay the land revenue to the Zamindars or return the loans to the moneylenders on time, they were often flogged, tortured or jailed. The impoverishment of the peasantry led to numerous famines.
Landlords also suffered from a sense of insecurity. Thousands of jagirs were confiscated by Bentinck and Dalhousie when they were unable to produce written title deeds of ownership.
The interests of the Indian economy was sacrificed for the interests of British trade and industry. This led to the utter collapse of traditional handicraft industries. Indian artisans and craftspersons were ruined.
The annexation of Indian states was followed by large- scale unemployment and economic distress. When Awadh was annexed, the administration was replaced by Company Administration. As such, hundreds of court officials and their subordinates lost their means of livelihood.

(c) Social and Religious Causes:

Social reforms such as the abolition of sati, and female infanticide, the Widow Remarriage Act and the introduction of women ‘s education caused deep resentment among the orthodox sections of society.
The efforts of the missionaries to convert people to Christianity caused great alarm. Some of the missionaries ridiculed the religious beliefs and practices of the Hindus and Muslims in their effort to convert people to their faith. This hurt the religious sentiments of the people.
The introduction of Western education undermined the position and importance of the Pundits and Maulvis and was seen as an attack on ancient traditions and values. The office of the Inspector of Schools in Patna was referred to as the ‘shaitane dafitar’.
The introduction of the railways and posts and telegraphs aroused grave doubts and fears, especially among the simple, backward villagers. They thought that the telegraph system was a form of Western magic. They grew fearful of the intentions of the British
The British judicial system introduced the principle of equality. This was regarded as a threat to the existing caste norms and privileges of the upper classes.
The British looked down upon the Indians and followed a policy of racial discrimination. They made no effort to interact socially with the Indians. They were convinced of the superiority of the European race and treated the Indians with great contempt.

Question 2.
In the context of the military causes of the Revolt of 1857, answer the following questions:

Mention any three grievances that the sepoys had against their British masters.
How did the Act passed in 1856 by the British hurt the sentiments of the Hindu sepoys?
Explain the immediate cause of the great Revolt?

Answer:
(a)
The sepoys had numerous grievances against the British-masters:

The sepoys had helped the British to establish their empire in India but they were neither appreciated nor rewarded for their efforts. On the contrary, they were treated with great contempt by the British officers.
There was grave discrimination between the Indian sepoy and his British counterpart. A capable and dedicated sepoy could not rise above the post of subedar.
In 1856, an Act was passed which made it compulsory for all new recruits to serve overseas if required. This hurt the sentiments of the Hindus because they belived that overseas travel would lead to a loss of caste. The sepoys interpreted the regulation as another attack on their caste and religion.
After the annexation of Awadh, the Nawab’s army was disbanded. The soldiers lost their means of livelihood and their bitterness against the British increased.
The Indian soldiers greatly outnumbered the British soldiers. In 1856, the number of sepoys in the British army was more than five times that of the British soldiers. This emboldened the sepoys to take up arms against their foreign masters.

(b)
In 1856, an Act was passed which made it compulsory for all new recruits to serve overseas if required. This hurt the sentiments of the Hindus because they believed that overseas travel would lead to a loss of caste. The sepoys interpreted the regulation as another attack on their caste and religion.
(c)
In January 1857, a rumour started at the Dum Dum cantonment (in Calcutta) that the cartridges, which the British had introduced, were greased with cow fat and pig lard. The rumour spread like wildfire among the Hindu and Muslim sepoys. They were convinced that the government was deliberately trying to defile their religion. A wave of indignation and anger swept through all the military stations. On 29 march 1857 Mangal Pandey, a sepoy at Barrackpore refused to use the cartridge and attacked his senior officers. He was hanged to death.

Question 3.
In the context of the Revolt of 1857, briefly discuss:

The decline of the Mughal dynasty
Any four results of the Revolt of 1857
Nature of the Revolt of 1857

Answer:
(a)
On 11 May 1857, Bahadur Shah Zafar was persuaded to accept the leadership of the Revolt. He was proclaimed the emperor of Hindustan. There was jubilation all around. The restoration of the Mughal empire was proclaimed with the booming of guns.
The success of the Revolt proved to be shortlived. British reinforcements arrived from Punjab, and Delhi was recovered in September 1857.
Bahadur Shah Zafar was taken prisoner, tried and exiled to Rangoon. The royal princes (two sons and one grandson of Bahadur Shah Zafar) were shot and and their bodies displayed on the streets. The once great dynasty of the Mughals finally came to an end.
(b)
Results of the Revolt 1857

The rule of the English East India Company came to an end.
India came under the direct rule of the British Parliament and the Queen of England.
Queen Victoria issued a Proclamation promising to look after the welfare of the Indian people.
Treaties with Indian states would be honoured.
A general pardon was granted to all the rebels, except those who had killed British subjects.
The British government would not interfere in the social and religious customs of the people.
Indians would be given opportunities to be associated with the administration. High posts in government services would be given on the basis of merit, not race.
The army was reorganized and strengthened. The number of British soldiers was increased and the artillery placed exclusively under their control.

The Revolt was the outcome of the accumulated grievances of different sections of people and not the sepoys alone.
Those who joined the Revolt had different reasons and different motives but they were all united in their hatred of British rule and their determination to overthrow it.
It had wide popular support of various sections of society.
The struggle created a strong bonding and a sense of unity between the Hindus and the Muslims. They fought shoulder to shoulder, as single brethren against a common enemy.
Of the estimated 1,50,000 people killed in the Revolt, 1,00,000 were civilians.

G Picture study:

This is the picture of the queen who led the Revolt in Central India
ICSE Solutions for Class 8 History and Civics - The Great Uprising of 1857 7

Identify the queen.
Ans. Rani Laxmi Bai
Name the city where she led the Revolt.
Ans. Jhansi
What were the economic causes of the Revolt of 1857?
Ans. Refer Answer F-l (b) above
What was the major outcome of the Revolt of 1857?
Ans. Refer Answer F-3 (b) above.

ICSE Solutions for Class 7 Geography Voyage – Industries : Their Need and Classification

September 9, 2024

ICSE Solutions for Class 7 Geography Voyage – Industries : Their Need and Classification

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for ICSE Solutions for Class 7 Geography Voyage. You can download the Voyage Geography ICSE Solutions for Class 7 with Free PDF download option. Geography Voyage for Class 7 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

ICSE Solutions Class 7 Geography History & Civics Maths Physics Chemistry Biology

Discuss

Discuss the importance of cottage industry for a country which has a large population with little education.
Answer:
A major benefit of cottage industries is that they allow people to work from their homes. Women have benefited the most because they can work from home while still tending to their families. Many cottage businesses include the entire family in their operations, however, including husbands and children. Some businesses start as cottage industries and then become too large, necessitating a move out of the home into a business environment, but at that point they likely have the financial means to make the move.

Discuss

What, according to you, would be the future of industries if we destroyed our forests and wildlife ?
Answer:
Forest-based Industries are based on forest products. If we destroys forest the industries such packaging industry, furniture industry, sports good industry, paper industry are shutdown because of unavailability of raw materials.

THINK AND ANSWER

Why do you think multinational companies are attracted to establish ventures in foreign countries ?
Answer:
Multinational companies are attracted to establish ventures in foreign countries as cheap labour and cheap raw material inputs, transport and power are easily available in developing countries.

VALUES & LIFE SKILLS

A lot of children below the age of 14 years work in various industries.
Do you think these children should work in these industries ?
Why do you think they have to work there ?
Answer:
Child labour is the employment of children at regular and sustained labour.
No, the children should not work in any industry because of following reasons:

Child labour does more than deprive children of their education and mental and physical development – their childhood is stolen.
Immature and inexperienced child labourers may be completely unaware of the short and long term risks involved in their work.
Working long hours, child labourers are often denied a basic school education, normal social interaction, personal development and emotional support from their family.
Poverty is undoubtedly a dominant factor in the use of child labour; families on or below the poverty line force their children into work to supplement their household’s meager income. Eradicating poverty, however, is only the first step on the road to eliminating child labour.

EXERCISES

A. Fill in the blanks.

1. People who work together to make cars work in the large- scale industry.
2. The USA and China are strong because they have thousands of industries.
3. Nepal has few industries.
4. The size of an industry depends on the number of people employed, the capital invested and the area it occupies.
5. A cottage industry is a household unit run by a family.

B. Match the following.

ICSE Solutions for Class 7 Geography Voyage Chapter 9 Industries Their Need and Classification 1
Answer:

ICSE Solutions for Class 7 Geography Voyage Chapter 9 Industries Their Need and Classification 2

C. Choose the correct answer.

Question 1.
China/Nepal has many industries.
Answer:
China has many industries.

Question 2.
Cottage industries produce handicrafts/cement.
Answer:
Cottage industries produce handicrafts.

Question 3.
Small-scale industries are smaller/bigger than cottage industries.
Answer:
Small-scale industries are bigger than cottage industries.

Question 4.
Sports equipment are produced by small-scale/large- scale industry.
Answer:
Sports equipment are produced by small-scale industry.

Question 5.
Chota Nagpur Plateau is well known for steel/textile production.
Answer:
Chota Nagpur Plateau is well known for steel production.

D. State whether the following is true or false.

1. The USA has many industries.
Answer. True.

2. Industries do not contribute much to national income.
Answer. False.
Correct : Industries contribute much to national income.

3. The number of labour employed has no relation to the size of an industry.
Answer. False.
Correct : The number of labour employed has relation to the size of an industry.

4. Cottage industry employ labour from outside.
Answer. False.
Correct : Cottage industry employ only family members.

5. Small-scale industries employ only family members.
Answer. False.
Correct : Small-scale industries employ labour from outside.

E. Answer the following questions in brief.

Question 1.
What is an industry?
Answer:
An Industry is a group of people or companies engaged in a particular kind of business enterprise. There are different groups of people that work together to form an industry. For example, people who work together to make cars form the car industry or people who work together to fashion clothes form the fashion industry.

Question 2.
Name the three major types of industries.
Answer:
Three types of industries are :

Cottage industry
Small-scale industry
Large-scale industry.
Agro-based industry.

Question 3.
What goods are produced by a cottage industry?
Answer:
Handicrafts, handloom products, jewellery, pottery, leather products, etc. are produced by cottage industry.

Question 4.
Name a few goods produced by small-scale industries.
Answer:
Electronic goods, sports equipment, brassware, bicycles, toys,
etc. are produced by small-scale industries.

Question 5.
Which type of industry produces cement and petrochemicals?
Answer:
Cement and petrochemicals are produced by large-scale industry.

F. Answer the following questions in detail.

Question 1.
Why do we need industries?
Answer:
Industries are required because of the following reasons :

Industries are needed to make a country strong financially: The more number of industries in a country the stronger the country becomes. Countries such as the USA and China are strong because of the hundreds and thousands of industries they have. India too has many industries but not as many as USA or China. Some countries like Nepal have very few industries.
Industries are needed for products of daily use : We will see items that we use everyday such as toothpaste, soap, bicycle, exercise books, pencils, medicines, cheese, and jams and so on. All these items are manufactured by different industries.
Industries are needed so there is no shortage of necessary goods :
Apart from fulfilling our daily needs, industries must produce
sufficiently so that all the things people need are available to them at reasonable prices.

Question 2.
What is a cottage industry?
Answer:
Cottage Industry : It is generally a household unit run by a family.
The craftsmen and their family members carry on the traditional work started by their forefathers generations ago. Cottage industries generally produce all kinds of handicrafts, handloom products, jewellery, pottery, leather products, etc. These industries are encouraged by the government as they generate employment and improve living conditions, especially in rural areas.

Question 3.
What are the differences between a cottage industry and a small-scale industry?
Answer:
The differences between a cottage industry and a smale- scale industry are:

The location of cottage industries is restricted in villages whereas the small-scale industries are mostly located in urban and semi- urban areas.
Cottage industry being a household industry is mostly run by the members of the family and therefore do not maintain hired labourers. But the small-scale industries are mostly run by hired labourers.
Cottage industries are producing goods for meeting local requirements whereas small-scale industries are producing goods to meet the demand for the people living in a wider area.
Cottage industries are investing a very little amount of capital and are working with simple tools. But the small industries are investing a comparatively higher amount of capital (presently the limit has been raised from Rs. 60 lakh to Rs. 3 crore) and are working with machines run by power.

Question 4.
What are large-scale industries?
Answer:
Large-scale industries involve huge investments, professional management and a large force of skilled and unskilled labour.
Big power-driven machines are used with considerable automation. These industries are generally established at places where raw materials, cheap transportation, sources of power, plenty of labour and market are easily available. Industries producing iron and steel, petrochemicals, cement, railway engines and coaches, automobiles, textiles, etc. are examples of large-scale industries.

Question 5.
Give an account of the factors that determine the establishment of an industry.
Answer:
Factors that determines the establishment of an Industry are:

Availability of Raw Material — All industries that require heavy and bulky raw material are generally set up near the sources of raw materials in order to save heavy transportation cost. For example, iron and steel plants in India and other countries have been set up near iron ore mines and coalfields as both iron ore and coal are heavy and difficult to transport.
Availability of Power — For any industry power is a very critical factor. Cheap, abundant and uninterrupted power supply is an essential need for any modem industry using large machines.
Availability of Transport —An important factor for setting up industries is the availability of efficient means of transportation as movement of raw material to the factory and finished goods to the market depend on it.
Availability of Labour — Though labour-skilled and unskilled-can be transported from different regions, it is advantageous to have an assured labour supply locally for setting up an industry.
Market — The ultimate aim of any industry is to sell its product easily, i.e., to find a market for its product without much difficulty. Industries of certain types are set up in specific regions where their goods can easily be sold. Woollen industries will do well in areas of cold climate while industries making cotton goods will

LET’S DO SOMETHING

With the help of the Internet, find out the names of three companies from the following industries :

Iron and steel
Textile
Cement

Answer:

Tata Iron and Steel Corp., Bhilai Steel Plant, Durgapur Steel Plant.
Acrow India ltd., Al chemist corp. ltd. Acil Cotton Industries ltd.
Ultratech,ACC,Ambuja Cement, Ramco Cements. thrive in areas that have warm and dry climate.
Other Factors — Factors like easy access to financial and banking facilities, climate, state-government policies, etc. also influence either directly or indirectly the location of an industry.

G. Ask your parents to take you to a cottage or small-scale industry in your city so that you can see how goods are produced there ?
Answer:
Do yourself with the help of parents.

H. Picture Study
This is a picture of an industry.

What type of industry is this ?
Can you give two features of this industry?

ICSE Solutions for Class 7 Geography Voyage Chapter 9 Industries Their Need and Classification 3

Answer:

This is a large-scale industry of automobiles sector.
Large-scale industries involve huge investments, professional management and a large force of skilled and unskilled labour. Big power-driven machines are used with considerable automation. These industries are generally established at places where raw materials, cheap transportation, sources of power, plenty of labour and market are easily available.

LET’S DO SOMETHING

With the help of the Internet, find out the names of three companies from the following industries:

Iron and steel
Textile
Cement

Answer:

Tata Iron and Steel Corp., Bhilai Steel Plant, Durgapur Steel Plant.
Acrow India ltd., A1 chemist corp. ltd. Acil Cotton Industries ltd.
Ultratech, ACC, Ambuja Cement, Ramco Cements.

ICSE Solutions for Class 7 History and Civics – Foundation of Mughal Empire

September 9, 2024

ICSE Solutions for Class 7 History and Civics – Foundation of Mughal Empire

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for ICSE Solutions for Class 7 History and Civics. You can download the History and Civics ICSE Solutions for Class 7 with Free PDF download option. History and Civics for Class 7 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Important Words

Sarkars were the provinces in Sher Shah’s empire. His empire was divided into 47 sarkars and each sarkar was divided into parganas.
Shiqdars were the officers responsible for law and order and the general administration of the parganas.
Mu ns if s were the officers who looked after the collection of revenue and civil cases.
Patwaris were responsible for the collection of land revenue in the villages.
Muqaddams were responsible for maintaining peace in the village.
Sarais were rest houses built by Sher Shah.
Dak chaukis were mail posts where two horsemen were kept ready to carry mail to the next post. The rest houses or sarais were used as dak chaukis.
Rupia was the standard silver coin introduced by Sher Shah.
Patta meant the title deed of land.
Qubuliat was the agreement of the land.

Time To Learn
I. Fill in the blanks:

Babur’s eldest son assumed the title Humayun which means fortunate.
Humayun reconquered the throne of Delhi in 1555 AD.
Humayun was in Persia after 1540.
Sher Shah ruled Delhi for five years.

II. Match Column A with Column B

Answer:

III. State whether the following statements are True or False:

The First Battle of Panipat decided once for all the fate of the Lodi dynasty.
True.
The Battle of Khanwa was fought between Rana Sanga and Babur.
True.
Babur introduced many new reforms for the administration of his new empire in India.
False.Sher Shah introduced many new reforms for the administration of his new empire in India.|
Humayun died in 1550 AD.
False.Humayun died in 1556 AD.
Sher Shah built the Grand Trunk Road.
True.

IV. Answer the following questions briefly:

Question 1.
Why was the First Battle of Panipat fought? What were its results?
Answer:
The First Battle of Panipat was fought:

The first battle of panipat was fought in 1526, between babur and Ibrahim lodi.
The battle was the result of policies of Ibrahim Lodhi which resulted in nobility turning against him and rebelling.
Ibrahim Lodhi’s attitude towards his officials and nobility was resented.
He also crushed and put to death nobility who rebelled against him.
Thus, Daulat Khan Lodi, and Alma Khan invited babur to fight against him. Hence, the first battle of panipat fought.It results in the babur declaring him as “Emperor of Hindustan” on April 27th, 1526.

Question 2.
What were the reasons for Babur’s victory in the First Battle of Panipat?
Answer:
There were several reasons for Babur’s victory at Panipat.

Babur had used artillery which was unknown to the Indians.
His army was better trained than the Sultan’s army and his cavalry was far superior to the Indian cavalry.
He also took over Agra, which was made the capital. On April 27th, 1526 he declared himself “Emperor of Hindustan”.

Question 3.
What difficulties did Humayun face in the beginning?When and how did Humayun reconquer India?
Answer:
When Humayun ascended the throne in 1530 A.D., he inherited many difficulties from his father Babur. They were:

He had to face political instability because the vast empire was not consolidated.
Empty treasure: The treasury was empty because collection of land revenue was not done in a systematic manner.
Troublesome relatives: Humayuns ambitious brothers rebelled against him.
1. Division of empire into jagirs was another difficulty he had to face.
2. The government was not well organised. So Humayun had to difficult task of strengthening his position when he ascended the throne.
Sher Shah died in AD 1545. Shershah’s successors were weak and inefficient. In 1555 AD, Humayun recovered Delhi and became emperor again.

Question 4.
Why is Sher Shah known as a-good administrator?
Answer:
Sher Shah known as a good administrator because of following reasons:

Sher Shah divided his empire into provinces (sarkars).
Each province had a governor with his team of officials to carry out the administrative duties.
A province was divided into a number of districts called parganas. Each pargana was made up of a number of villages. The village was the lowest administrative unit.
The two top officials in a pargana were shiqdar-i-shiqdaran and munsif-i-munsifan. The shiqdar handled law and order. The munsif took care of the revenue collection and other civil matters.
The panchayat looked after village administration.
The two main officials in a village were patwari (revenue collector) and muqaddam in charge of law and order.

V. Give reasons.

Question 1.
Sher Khan was able to strengthen his position in Bengal and Bihar.
Answer:
Sher Khan was able to strengthen his position in Bengal and Bihar because of following reasons:

Sher Shah personally supervised the recruitment and training of his soldiers who had to maintain strict discipline.
He revived Sultan Alauddin’s measures of branding horses (dagh) and maintaining descriptive rolls of soldiers (chehra).
He paid regular salary based on their skills.
He set up forts and garrisons in different parts of the empire for speedy deployment of soldiers.
Sher Shah kept a close watch over all his officials. He set up a spy system to know about the happenings in all parts of the empire.The officials were transferred every 2-3 years to prevent them from becoming corrupt or powerful.

Question 2.
Sher Shah was named so.
Answer:
Sher Shah was the son of a jagirdar of Sasaram (in Bihar). His actual name was Farid Khan. During his employment under the ruler of Bihar, he killed a tiger single-handedly, and came to be known as Sher Khan. Then he conquered many states and become Sher Shah.

Question 3.
We say that Sher Shah looked after the welfare of his people.
Answer:

Sher Shah constructed an excellent network of roads, mainly the Grand Trunk Road from Peshawar to Sonargaon (Bengal). He also built roads from Agra to Jodhpur and Chittor. He built another road from Lahore to Multan.
To make travel comfortable Sher Shah built sarais and wells at regular intervals.
Trees were also planted on both sides of the roads. In all, 1700 sarais (inns) were constructed for the travellers. These sarais gradually gained in importance and became the centres of trade.
Special officers known as Muqaddams were appointed to ensure the safety of travellers.
Roads helped Sher Shah to establish a good postal system and promoted inland trade.
His roads and sarais have been called ‘the arteries of the empire’.
Sher Shah divided his empire into 47 provinces (sarkars). Each sarkar was them divided into many districts (parganas). A pargana comprised several villages.

Additional Questions
(Foundation of The Mughal Empire)

A. Fill in the blanks:

Babur lost both Farghana and Samarkand and became a homeless wanderer. In 1504 ce, he became the ruler of Kabul.
Humayun lacked the qualities necessary to consolidate the vast empire he had inherited.
Sher Shah based his administration on the principle of a welfare state.
Sher Shah went on regular tours of inspection and set up an efficient spy system.
Sher Shah established a large standing army and introduced several reforms to make it disciplined, efficient and strong.

B. Match the following:

Answer:

C. Choose the correct answer:

1. In 1522 ce, Ibrahim Lodi/S her Khan Suri/Daulat Khan Lodi invited Babur to invade India.
Ans. In 1522 ce, Daulat Khan Lodi invited Babur to invade India.

2. With the help of the Shah of Persia/Afghanistan/Sind, Humayun returned to India in the year 1555
Ans. With the help of the Shah of Persia, Humayun returned to India in the year 1555 ce.

3. Sher Shah divided his empire into 47 sarkars/parganas/ villages.
Ans. Sher Shah divided his empire into 47 sarkars.

4. The rupia introduced by Sher Shah was made of gold/ silver/copper.
Ans. The rupia introduced by Sher Shah was made of silver.

5. Sher Shah’s greatest achievement was the construction of new roads/issuing of coins/his revenue system.
Ans. Sher Shah’s greatest achievement was his revenue system.

D. State whether the following are true or false:

Babur was a descendant of Timur and Chenghiz Khan.
True.
Babur’s war-weary and homesick soldiers wanted to return to Kabul after the Battle of Panipat.
True.
Babur won the Battle of Khanwa through treachery and cunning.
False. Correct: Babur won the Battle of Khanwa with his superior artillery and covering.
Sher Shah ruled for 15 years.
False. Correct: Sher Shah ruled for 5 years.
To prevent the officers from becoming corrupt, Sher Shah introduced the system of periodical transfers.
True.

E. Answer the following questions in one or two words/ sentences:

1.Why did Daulat Khan Lodi invite Babur to invade India?
Ans. In 1522 ce, Daulat Khan Lodi invited Babur to invade India and help him overthrow Ibrahim Lodi, the cruel and unpopular sultan of Delhi.

2.Between whom was the First Battle of Panipat fought?
Ans. Babur and Ibrahim Lodi came face-to-face at Panipat in 1526 ce.

3.What was the significance of the Battle of Panipat (1526 ce)?
Ans. After the First Battle of Panipat Delhi came under the rule of Babur and he was declared the emperor of Hindustan.

4.What is the name of Babur’s autobiography?
Ans.Tuzuk-i-Baburi it is written in flawless Turkish, and considered to be one of the best autobiographies in the world.

5.Who was Humayun’s most dangerous enemy?
Ans. Sher Khan was Humayun’s most dangerous enemy.

6.How did Humayun escape after his deafeat in the Battle of Chausa (1539 ce)?
Ans.Humayun jumped into the Ganga and floated down the river with the help of a water carrier’s inflated water bag.

7.How many years did Humayun spend in exile?
Ans.15 years (1540-55 ce), Humayun wandered about from place to place in search of shelter.

8.Who helped Humayan to recover Kabul and Kandahar?
Ans. Shah of Persia helped Humayun and he recovered Kabul and Kandahar from his brother Kamran.

9.What is the significance of the Battle of Kanauj (1540 CE)?
Ans. The large but fragile Mughal empire came to an end and was replaced by Afghan rule.

10.How were the affairs of the villages managed during Sher Shah’s reign?
Ans.The villages were looked after by the panchayats composed of village elders. Sher Shah himself kept in touch with the villages through village officials like patwaris and muqaddam.

11.Explain the Chehra system in Sher Shah’s military administration.
Ans.Sher Shah adopted a system of branding horses (dagh) and maintaining a descriptive roll of the soldiers (chehra) were revived.The army was divided into many units and each unit was placed under a commander.

12.Mention one reason why Sher Shah built an excellent network of good roads?
Ans.Sher Shah built an excellent network of good roads, to facilitate the quick movement of officials and troops throughout the empire and to promote trade and commerce and even to make travelling comfortable and easier.

F. Answer the following questions briefly:

Question 1.
Give a brief account of Babur’s early life.
Answer:
Babur lost his father, when he was eleven years old and at that tender age he became the ruler of a small principality in Central Asia called Farghana which he inherited from his father. He was also able to fulfill his dream of conquering Samarkand, when he was just 14 years old. But within a short time he lost both Farghana and Samarkand and became homeless but due to his courage and determination, after some time he became the ruler of Kabul.

Question 2.
Give an account of the causes, events and consequences of the First Battle of Panipat.
Answer:
Babur and Ibrahim Lodi came face-to-face at Panipat in 1526 ce. Ibrahim’s large army was no match for Babur’s small, disciplined and loyal army with its excellent cavalry and fine artillery. Babur was a bom leader and an experienced military general. Within a few hours, Ibrahim’s army was routed. By evening, Ibrahim and 15,000 of his soldiers lay dead on the battlefield. The following day, Babur triumphantly entered Delhi where he was proclaimed the emperor of Hindustan.

Question 3.
How many battles did Babur have to fight after the Battle of Panipat? Mention the significance of his victory in each of these battles.
Answer:
After the Battle of Panipat Babur fought three more battles.

The significance of each battle are:

Battle of Khanua: This battle was fought between Babur and Rana Sangha of Mewar at Khanua. Babur won this battle and this victory gave him supreme control over central India and it also removed all the obstacles in the way of Mughal rule.
Battle of Chanderi: This battle was fought between Babur and Medini Rai of Malwa. Babur won the battle and captured Chanderi. After this battle, no other Rajput chief ever dared to challenge Babur’s authority.
Battle of Ghagra: The Afghans had control over Bihar and Bengal and they posed a threat to Babur’s authority. So Babur had a war with them and after crushing them he became the master of Punjab, Delhi and the Ganga plains near Bihar.

Question 4.
Give a brief account of Sher Shah’s early life.
Answer:
Sher Shah was the son of Hasan Khan, a jagirdar of Sasaram in Bihar. His real name was Farid Khan. He was given the name Sher Khan after he single-handedly killed a tiger. He joined the Mughal army when Babur invaded India and learnt the techniques of Mughal warfare.

Question 5.
With reference to Sher Shah’s enlightened rule, explain:
(a)How was the central government organized? (b) The important features of provincial administration.
Answer:
(a)
The supreme head of the central government was Sher Shah himself. He divided the government into several departments. Each department was under the charge of a minister, who was assisted by other officials.
(b)
The whole empire was divided into forty seven provinces or sarkars. Each province was further subdivided into districts or parganas. Each paragana was made up of a number of villages. The villages were looked after by the panchayats composed’of village elders. Each Sarkar was placed under an Afghan chief. In every pargana there were several officials who were supervised by Shiqdar and Munsif. The Shiqdar was responsible for law, order and general administration and the Munsif looked after the collection of revenue and civil cases.

Question 6.
Sher Shah introduced several measures to make the army disciplined, efficient and strong. Explain?
Answer:
Sher Shah introduced many reforms to make his army efficient, disciplined and strong. He used to take personal interest in the appointment and training of soldiers. The salaries of the soldiers and the officers were fixed according to their skill and ability. The army was divided into many units and each unit was placed under a commander. The cavalry and infantry were highly trained, disciplined and were equipped with artillery.

Question 7.
Discuss the important measures taken by Sher Shah to boost the growth of trade and commerce.
Answer:
The growth of trade and commerce increased after the construction of roads and sarais. Sher Shah even abolished all duties, except two to boost it further more. He even instructed his officials to treat the merchants well and look after their interests. Proper safety was provided to traders and the village headmen was responsible for it in their respective regions. New weights and measures were introduced. Coins of gold, silver and copper of uniform standard were introduced.

Question 8.
What steps did Sher Shah take to maintain law and order in his kingdom?
Answer:
Sher Shah strongly believed injustice and he treated all the people equally. In his kingdom the criminal law was harsh and punishments were severe. The village headmen and the local chieftains were made responsible for any crime that took place in their area. They were severely punished if they failed to track down a robber or a murderer.

Question 9.
Sher Shah’s greatest achievement was his land revenue system. In this context explain: (a) Land assessments (b) Taxation policy and its impact
Answer:
(a) Land assessments: Sher Shah made proper arrangements that the land should be carefully surveyed and measured every year and the average produce to be calculated on the basis of its fertility.

(b) Taxation policy and its impact: One-third of the produce was paid as tax in cash or kind. The land was divided into three types – good, bad and medium on the basis of fertility and taxes were levied on this basis.

Question 10.
Sher Shah is one of the most striking personalities in medieval India. In this context, briefly state the important achievements of his 5-years reign.
Answer:

During his reign Sher Shah established a vast, powerful and prosperous empire. Some of his achievements are listed below:
In his empire there was a powerful central government and a sound administrative system.
He created a huge standing army that was disciplined, well organized and strong.
His revenue system was efficient progressive and humane and he also showed concern for the welfare of the peasants.
During his reign good roads and rest houses were constructed for travellers and merchants.
New weights and measures were introduced by him.
He also followed a policy of religious tolerance and appointed many Hindus to high posts of trust.

G Picture study:

This is a picture of the tomb of the successor of Babur
ICSE Solutions for Class 7 History and Civics - Foundation of Mughal Empire 6
Question 1.
Name the Mughal emperor who is buried in this tomb.
Answer:
Humayun.

Question 2.
Where is this monument located?
Answer:
Nizamuddin Delhi.

Question 3.
In which book could you find a detailed account of his life? Who wrote it?
Answer:
Humayunnama. Gulbadan wrote it.

Question 4.
Why did he spend 15 years in exile?
Answer:
After defeat from Sher Khan . Humayun wandered about 15 years from place to place in search of shelter. His brother refused to help him. In 1542 ce, in a small town in Amarkot in Sind, a son, Akbar was born to Humayun and his wife, Hamida Banu.

Question 5.
How did he reconquer his lost empire?
Answer:
With military help from the Shah of Persia, Humayun returned and recovered Kabul and Kandahar from his brother Kamran. Kamran was blinded and sent to Mecca. By 1554 ce, Humayun had consolidated his position in Afghanistan. Sher Shah had died in 1545 ce. His successors were weak and incompetent. Taking advantage of the situation, Humayun returned to India and recaptured Delhi and Agra in 1555 ce. Humayun died shortly afterwards in an accidental fall down the stairs of his library.

OP Malhotra Class 12 Maths Solutions Chapter 23 Three Dimensional Geometry Ex 23(a)

September 6, 2024

Students can cross-reference their work with ISC Class 12 Maths OP Malhotra Solutions Chapter 23 Three Dimensional Geometry Ex 23(a) to ensure accuracy.

S Chand Class 12 ICSE Maths Solutions Chapter 23 Three Dimensional Geometry Ex 23(a)

Question 1.
The direction ratios of a line are 1,-
2,-2. What are their direction cosines?
Answer:
The direction ratio of line are < 1, -2, -2 >
∴ direction cosines of line be
\(<\frac{1}{\sqrt{1^2+(-2)^2+(-2)^2}}\), \(\frac{-2}{\sqrt{1^2+(-2)^2+(-2)^2}}\), \(\frac{-2}{\sqrt{1^2+(-2)^2+(-2)^2}}\)
i.e., < \(\frac{1}{3}\), \(\frac{-2}{3}\), \(\frac{-2}{3}\) >

Question 2.
If α, β, γ are angles which a line makes with the axes, prove that
sin² α + sin² β + sin² γ = 2
Answer:
Given α, β and γ are the angles which a line makes with axes
∴ direction cosines of line are < cos α cosβ, cosγ >
∴ cos² α + cos² β + cos² γ = 1
⇒ 1 – sin² α + 1 – sin² β + 1 – sin² γ = 1
⇒ sin² α + sin² β + sin² γ = 3 – 1 = 2

Question 3.
Can a line have direction angles 45°, 60°, 120° ?
Answer:
Direction cosines of line be < cos 45°, cos 60°, cos 120° >
< \(\frac{1}{\sqrt{2}}\), \(\frac{1}{2}\), cos(180° – 60°) >
i.e., < \(\frac{1}{\sqrt{2}}\), \(\frac{1}{2}\), – \(\frac{1}{2}\) >
Here, l = \(\frac{1}{\sqrt{2}}\), m = \(\frac{1}{2}\) and n = \(-\frac{1}{2}\)
∴ l² + m² + n²
= (\(\frac{1}{\sqrt{2}}\))² + (\(\frac{1}{2}\))² + (\(-\frac{1}{2}\))²
= \(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) = 1
Yes, a line can hence direction angles 45°, 60° and 120°

Question 4.
Prove that 1,1,1 cannot be direction cosines of a straight line.
Answer:
Here, l = m = n = 1
∴l² + m² + n²
= 1 + 1 + 1 = 3 ≠ 1
Thus, < 1, 1, 1 > can’t be the direction cosines of straight line.

Question 5.
Find the direction cosines and direction ratios of the line joining the points
(i) A(0, 0, 0), B(4, 8, -8)
(ii) A(1, 3, 5), B(-1, 0, -1)
(iii) A(5, 6, -3), B(1, -6, 3)
(iv) A(4, 2, -6), B(-2, 1, 3).
Answer:
We know that, direction ratios of the line joining the points A(x₁, y₁, z₁) and B(x₂, y₂, z₂) are < x₂ – x₁, y₂ – y₁, z₂ – z₁ >
(i) Direction ratios of line AB are < 4 – 0, 8 – 0, -8 – 0 >
i.e., < 4, 8, -8 >
i.e., < 1, 2, -2 >
∴ Direction cosines of line AB are
< \(\frac{4}{\sqrt{4^2+8^2+(-8)^2}}\), \(\frac{8}{\sqrt{4^2+8^2+(-8)^2}}\), \(\frac{-8}{\sqrt{4^2+8^2+(-8)^2}}\) >
i.e., < \(\frac{4}{12}\), \(\frac{8}{12}\), \(\frac{-8}{12}\) >
i.e., < \(\frac{1}{3}\), \(\frac{2}{3}\), \(\frac{-2}{3}\) >

(ii) D ratios of line AB are < -1 – 1, 0 – 3, -1 – 5 >
i.e., < -2, -3, -6 >
i.e., < 2, 3, 6 >
∴ D cosines of line AB are
< \(\frac{2}{\sqrt{2^2+3^2+6^2}}\), \(\frac{3}{\sqrt{2^2+3^2+6^2}}\), \(\frac{6}{\sqrt{2^2+3^2+6^2}}\) >
i.e., < \(\frac{2}{7}\), \(\frac{3}{7}\), \(\frac{6}{7}\) > .

(iii) D ratios of line AB are < 1 – 5, -6 – 6, 3 + 3 >
i.e., < -4, -12, 6 >
i.e., < 2, 6, -3 >
∴ D cosines of line AB are
< \(\frac{2}{\sqrt{2^2+6^2+(-3)^2}}\), \(\frac{6}{\sqrt{2^2+6^2+(-3)^2}}\), \(\frac{-3}{\sqrt{2^2+6^2+(-3)^2}}\) >
i.e., < \(\frac{2}{7}\), \(\frac{6}{7}\), \(\frac{-3}{7}\) >

(iv) D ratios of line AB are < -2 – 4, 1 – 2, 3 + 6 >
i.e., < -6, -1, 9 >
i.e., < 6, 1, -9 >
∴ D cosines of line are
< \(\frac{6}{\sqrt{6^2+1^2+(-9)^2}}\), \(\frac{1}{\sqrt{6^2+1^2+(-9)^2}}\), \(\frac{-9}{\sqrt{6^2+1^2+(-9)^2}}\) >
i.e., < \(\frac{6}{\sqrt{118}}\), \(\frac{1}{\sqrt{118}}\), \(\frac{-9}{\sqrt{118}}\) >

Question 6.
By using direction ratios method, show that the following set of points are collinear:
(i) A(1, 2, 3), B(4, 0, 4) and C(-2, 4, 2)
(ii) (-2, 4, 7), (3, -6, -8), (1 -2, -2).
Answer:
(i) Direction ratios of line AB are < 4 – 1, 0 – 2, 4 – 3 > i.e., < 3, -2, 1 >
Direction ratios of line BC are < -2 – 4, 4 – 0, 2 – 4 >
i.e., < 3, -2, 1 >
Thus line AB is parallel to line BC and the point B is common in both lines
∴ points A, B and C line on same line
∴ points A, B and C are collinear.
(ii) Direction ratios of line AB are < 3 + 2, -6 – 4, -8 – 7 >
i.e., < 5, -10, -15 >
i.e., < 1, -2 – 3 > and Direction ratios of line BC are < 1 – 3, -2 + 6, -2 + 8 >
i.e., < -2, 4, 6 >
More \(\frac{1}{-2}\) = \(\frac{-2}{4}\) = \(\frac{-3}{6}\)
i.e., direction ratios of both lines are proportional and hence lines AB and BC are parallel and the point B in common to both lines
∴ A, B and C are collinear.

Question 7.
A line makes an angle of \(\frac{\pi}{4}\) with each of the x-axis and the y-axis. Find the angle made by it with the z-axis.
Answer:
Let θ the angle made by the line with z-axis
∴ direction cosines of given line be < cos \(\frac{\pi}{4}\), cos \(\frac{\pi}{4}\), cosθ >
i.e., < \(\frac{1}{\sqrt{2}}\), \(\frac{1}{\sqrt{2}}\), cosθ >
∴(\(\frac{1}{\sqrt{2}}\))² + (\(\frac{1}{\sqrt{2}}\))² + cos²θ = 1
[∵ l² + m² + n² = 1]
⇒ \(\frac{1}{2}\) + \(\frac{1}{2}\) + cos²
θ = 1 or cos²
θ = 0 or cosθ = 0
⇒ θ = \(\frac{\pi}{2}\)

Question 8.
If the line O P makes with the x-axis an angle of measure 120° and withy x-axis an angle of measure 60°. Find the angle made by the line with the z-axis.
Answer:
Let θ be the angle made by the line OP with z-axis
∴ direction casines of line OP are
< cos 120°, cos 60°, cos θ >
i.e., < \(-\frac{1}{2}\), \(\frac{1}{2}\), cos θ >
Since, l² + m² + n² = 1
⇒ (\(\frac{-1}{2}\))² + (\(\frac{1}{2}\))² + cos² θ = 1
⇒ \(\frac{1}{2}\) + cos² θ = 1
⇒ cos² θ = \(\frac{1}{2}\)
= (\(\frac{1}{\sqrt{2}}\))²
= cos² \(\frac{\pi}{4}\)
⇒ θ = n π ± \(\frac{\pi}{4}\)
Since, 0 < θ < π
∴ θ = \(\frac{\pi}{4}\), \(\frac{3 \pi}{4}\)

Question 9.
Find the angle between the vectors whose direction cosines are proportional to 2, 3, -6 and 3, -4. 5.
Answer:
Let θ be the angle between given vectors whose direction ratios are < 2, 3, -6 > and < 3, -4, 5 >
⇒ cos θ = \(\frac{-36}{7 \times 5 \sqrt{2}}\)
= \(-\frac{18 \sqrt{2}}{35}\)

Question 10.
If α, β, γ are the angles that a line makes with the axes, then find cos γ if
(i) cosα = \(\frac{14}{15}\), cos β = \(\frac{-1}{3}\)
(ii)α = 60°, β = 135°.
Answer:
(i) Sinceα, β, γ are the angles that a line makes with the axes.
∴ direction cosines of line are < cosα, cos β, cos γ >
∴ cos²α + cos² β + cos² γ = 1
⇒ (\(\frac{14}{15}\))² + (\(-\frac{1}{3}\))² + cos² γ = 1
⇒ \(\frac{196}{225}\) + \(\frac{1}{9}\) + cos² γ = 1
⇒ \(\frac{221}{225}\) + cos² γ = 1
⇒ cos² γ = 1 – \(\frac{221}{225}\)
= \(\frac{4}{225}\) = \(\frac{2}{15}\)²
⇒ cos² γ = ± \(\frac{2}{15}\)

(ii) Givenα = 60°, β = 135°
∴ cos² γ = 1 – cos² 60° – cos² 135°
= 1 – \(\frac{1}{2}\)² – \(\frac{1}{\sqrt{2}}\)²
= 1 – \(\frac{1}{4}\) – \(\frac{1}{2}\) = \(\frac{1}{4}\)
⇒ cos² γ = \(\frac{1}{2}\)²
⇒ cos γ = ± \(\frac{1}{2}\)

Question 11.
If the coordinates of A and B be (2, 3, 4) and (1, -2, 1) respectively, prove that O A is perpendicular to O B, where O is the origin.
Answer:
D ratios of line OA are < 2 – 0, 3 – 0, 4 – 0 > i.e., < 2, 3, 4 >
D ratios of line OB are < 1 – 0, -2 – 0, 1 – 0 > i.e., < 1, -2, 1 >
Here a₁ a₂ + b₁ b₂ + c₁ c₂
= 2(1) + 3(-2) + 4(1) = 0
∴ line O A be ⊥ to line O B.

Question 12.
Show that therein of the points (1, 2, 3), (4, 5, 7) is parallel to the join of the points (-4, 3, -6) and (2, 9, 2).
Answer:
Direction ratios of the line joining A(1, 2, 3) and B}(4, 5, 7) are
< 4 – 1, 5 – 2, 7 – 3 >
i.e., < 3, 3, 4 >
and D ratios of the loine joining C(-4, 3, -6) and D(2, 9, 2) are
< 2 + 4, 9 – 3, 2 + 6 > i.e. < 6, 6, 8 >
Here \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\)
= \(\frac{c_1}{c_2}\) since \(\frac{3}{6}\)
= \(\frac{3}{6}\) = \(\frac{4}{8}\) = \(\frac{1}{2}\)
i.e., D ratios of both lines are proportional.
∴ line AB is parallel to line CD.

Question 13.
Find the angles between the lines whose direction ratios are
(i) 5, -12, 13; -3, 4, 5;
(ii) 1, 1, 2 ; \(\sqrt{3}\) – 1, \(-\sqrt{3}\) – 1, 4.
Answer:
(i) Given direction ratios of lines are < 5, -12, 13 > and < -3, 4, 5 >
a₁ = 5 ; b₁ = -12 ; c₁ = 13
a₂ = -3 ; b₂ = 4 ; c₂ = 5
Let θ be the angle between the lines.
Then cosθ
= \(\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\)
= \(\frac{5(-3)-12(4)+13(5)}{\sqrt{5^2+(-12)^2+13^2} \sqrt{(-3)^2+4^2+5^2}}\)
= \(\frac{-15-48+65}{\sqrt{2 \times 169} \sqrt{25 \times 2}}\)
= \(\frac{2}{5 \times 13 \times 2}\) = \(\frac{1}{65}\)
∴ θ = cos^-1 (\(\frac{1}{65}\))

(ii) Here a₁ = 1 ; b₁ = 1 ; c₁ = 2
a₂ = \(\sqrt{3}\) – 1;
b₂ = \(-\sqrt{3}\) – 1;
c₂ = 4
∴ cosθ = \(\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\)
= \(\frac{1(\sqrt{3}-1)+1(-\sqrt{3}-1)+2(4)}{\sqrt{1^2+1^2+2^2} \sqrt{(\sqrt{3}-1)^2(-\sqrt{3}-1)^2+4^2}} \)
= \(\frac{\sqrt{3}-1-\sqrt{3}-1+8}{\sqrt{6} \sqrt{3+1+3+1+16}}\)
= \(\frac{6}{\sqrt{144}}\)
= \(\frac{6}{12}\)
= \(\frac{1}{2}\)
⇒ θ = \(\frac{\pi}{3}\)

Question 14.
If P, Q, R are respectively (2, 3, 5),(-1, 3, 2) and (3, 5, -2), find the direction cosines of the sides of the triangle P Q R.
Answer:
∴ D ratios of line PQ are < -1 – 2, 3 – 3, 2 – 5 >
i.e., < -3, 0, -3 > i.e., < 1, 0, 1 >
∴ D cosines of lines PQ are < \(\frac{1}{\sqrt{1+1}}\), \(\frac{0}{\sqrt{1+1}}\), \(\frac{1}{\sqrt{1+1}}\) >
i.e., < \(\frac{1}{\sqrt{2}}\), 0, \(\frac{1}{\sqrt{2}}\) >
D ratios of line QR are < 3 + 1, 5 – 3, -2 – 2 >
i.e., < 4, 2, -4 >
i.e., < 2, 1, -2 >
∴ D cosines of line QR are < \(\frac{2}{\sqrt{4+1+4}}\), \(\frac{1}{\sqrt{4+1+4}}\), \(\frac{-2}{\sqrt{4+1+4}}\) >
i.e., < \(\frac{2}{3}\), \(\frac{1}{3}\), \(\frac{-2}{3}\) >
D ratios of line PR are < 3 – 2, 5 – 3, -2 – 5 >
i.e., < 1, 2, -7 >
∴ D cosines of line PR are < \(\frac{1}{\sqrt{1+4+49}}\), \(\frac{2}{\sqrt{1+4+49}}\), \(\frac{-7}{\sqrt{1+4+49}}\) >
i.e., < \(\frac{1}{3 \sqrt{6}}\), \(\frac{2}{3 \sqrt{6}}\), \(\frac{-7}{3 \sqrt{6}}\) >

Question 15.
Prove that the three points P, Q, R, whose coordinates are respectively (3, 2, -4), (5, 4, -6) and (9, 8, -10) are collinear and find the ratio in which Q divides P R.
Answer:
Here D ratios of line PQ are < 5 – 3, 4 – 2, -6 + 4 >
i.e., < 2, 2, -2 > and D ratios of line QR are < 9 – 5, 8 – 4, -10 + 6 >
i.e., < 4, 4, -4 > More \(\frac{2}{4}\) = \(\frac{2}{4}\) = \(\frac{-2}{-4}\)
i.e., D ratios of both lines PQ and QR are proportional
∴ line PQ and QR are parallel and the point Q is common to both lines
∴ P, Q and R lies on same line
∴ the points P, Q and R are collinear let the point Q divides the time PR in the ratio K : 1.
∴ Coordinates of point Q are
(\(\frac{9{~K}+3}{{~K}+1}\), \(\frac{8{~K}+2}{{~K}+1}\), \(\frac{-10{~K}-4}{{~K}+1}\) )
Also coordinates of point Q be (5, 4, -6)
∴ \(\frac{9{~K}+3}{{~K}+1}\) = 5 ⇒ 9K + 3 = 5K + 5 ⇒ 4K = 2 ⇒ K = \(\frac{1}{2}\)
and \(\frac{8{~K}+2}{{~K}+1}\) = 4 ⇒ 8K + 2 = 4K + 4 ⇒ 4K = 2 ⇒ K = \(\frac{1}{2}\)
Also, \(\frac{-10{~K}-4}{{~K}+1}\) = -6 ⇒ -10K – 4 = -6K – 6 ⇒ 4K = 2 ⇒ K = \(\frac{1}{2}\)
Times required rastio be K}: 1, i.e., 1: 2.

Question 16.
Find the angle not greater than 90° between the lines joining the following pairs of points:
(i) (8, 2, 0), (4, 6, -7), and (-3, 1, 2), (-9, -2, 4);
(ii) (4, -2, 3), (6, 1, 7), and (4, -2, 3),(5, 4, -2);
(iii) (3, 1, -2), (4, 0, -4), and (4, -3, 3), (6, -2, 2).
Answer:
(i) D ratios of line joining the points A(8, 2, 0) and B(4, 6, -7) are < 4 – 8, 6 – 2, -7 – 0 > i.e., < -4, 4, – 7 >
and D ratios of line joining the points C(-3, 1, 2) and D(-9, -2, 4) and < -9 + 3, -2 – 1, 4 – 2 > i.e., < -6, -3, 2 >
Let θ be the acute angle between the lines A B and C D
∴ cosθ = \(\frac{|-4(-6)+4(-3)-7(2)|}{\sqrt{16+16+49} \sqrt{36+9+4}}\)
= \(\frac{2}{9 \times 7}\) = \(\frac{2}{63}\)
∴ θ = cos¹ \(\frac{2}{63}\)

(ii) D ratios of the line joining the points A(4, -2, 3) and B(6, 1, 7) are < 6 – 4, 1 + 2, 7 – 3 >
i.e., < 2, 3, 4 >
and D ratios of the line joining the points C(4, -2, 3) and D(5, 4, -2) are < 5 – 4, 4 + 2, -2 – 3 > i.e., < 1, 6, -5 >
Let θ be the angle between the lines A B and C D.
∴ cosθ = \(\frac{|2(1)+3(6)+4(-5)|}{\sqrt{4+9+16} \sqrt{1+36+25}}\)
= 0 ⇒ θ = \(\frac{\pi}{2}\)

(iii) D ratio of the line joining the points A(3, 1, -2) and B(4, 0, -4) are < 4 – 3, 0 – 1, -4 + 2 > i.e. < 1, -1, -2 >
and D ratios of the line joining the points C(4, -3, 3) and D(6, -2, 2) are < 6 – 4, -2 + 3, 2 – 3 > i.e., < 2, 1, -1 >
Let θ be the angle between the lines AB and CD.
Then cosθ = \(\frac{|1(2)-1(1)-2(-1)|}{\sqrt{1+1+4} \sqrt{4+1+1}}\)
= \(\frac{3}{\sqrt{6} \sqrt{6}}\) = \(\frac{3}{6}\)
= \(\frac{1}{2}\) ⇒ θ = \(\frac{\pi}{3}\)

Question 17.
Find the direction cosines of the line which is perpendicular to the lines with direction cosines proportional to 1, – 2, -2 ; 0, 2, 1.
Answer:
Let the required direction ratios of given line be < a, b, c > and given line is ⊥ to the lines having direction ratios are < 1, -2, -2 > and < 0, 2, 1 >.
∴ a – 2 b – 2 c = 0
0 a + 2 b + c = 0
using cross-multiplication method, we have
\(\frac{a}{-2+4}\) = \(\frac{b}{0-1}\) = \(\frac{c}{2-0}\)
i.e., \(\frac{a}{2}\) = \(\frac{b}{-1}\) = \(\frac{c}{2}\)
∴ D ratios of line are < 2, -1, 2 >
Time required direction cosines of line are
< \(\frac{2}{\sqrt{2^2+(-1)^2+2^2}}\), \(\frac{-1}{\sqrt{2^2+(-1)^2+2^2}}\), \(\frac{2}{\sqrt{2^2+(-1)^2+2^2}}\) >
i.e., < \(\frac{2}{3}\), \(\frac{-1}{3}\), \(\frac{2}{3}\) >

Question 18.
Find the direction ratios of a line perpendicular to the two lines determined by the pairs of points (2, 3, -4), (-3, 3, -2) and (-1, 4, 2), (3, 5, 1).
Answer:
D ratios of given lines are < -3, -2, 3 – 3, -2 + 4 > and < 3 + 1, 5 – 4, 1 – 2 >
i.e., < -5, 0, 2 > and < 4, 1, -1 >
Let the direction ratios of required line be < a, b, c >.
Since the required line be ⊥ to given lines. Then -5 a + 0 b + 2 c = 0
4 a + b – c = 0
on solving eqn. (1) and eqn. (2), by cross multilplication method, we have
\(\frac{a}{0-2}\) = \(\frac{b}{8-5}\)
= \(\frac{c}{-5-0}\) i.e., \(\frac{a}{-2}\) = \(\frac{b}{3}\) = \(\frac{c}{-5}\)
∴ direction ratios of required line be < -2, 3, -5 >

Question 19.
For what value of x will the line through (4, 1, 2) and (5, x, 0) be parallel to the line through (2, 1, 1) and (3, 3, -1).
Answer:
D ratios of the line AB through the points A(4, 1, 2) and B(5, x, 0) are < 5 – 4, x – 1, 0 – 2 > i.e., < 1, x – 1, -2 >
also direction ratio of line (1) through the points C(2, 1, 1) and D(3, 3, -1) are < 3 – 2, 3 – 1, -1 – 1 > i.e., < 1, 2, -2 >
Since both lines are parallel
∴ direction ratios of both lines are proportional.
∴ \(\frac{1}{1}\) = \(\frac{x-1}{2}\)
= \(\frac{-2}{-2}\)
⇒ x – 1 = 2 ⇒ x = 3

Question 20.
For what value of x will the lines in Problem 19 be perpendicular?
Answer:
Now both lines AB and CD are perpendicular.
∴ 1(1) + (x – 1) 2 + (-2)(-2) = 0
⇒ 1 + 2 x – 2 + 4 = 0 ⇒ 2 x + 3 = 0
[∵ a₁ a₂ + b₁ b₂ + c₁ c₂ = 0]

Question 21.
Show that the points (4, 7, 8), (2, 3, 4), (-1, -2, 1) and (1, 2, 5) are the vertices of a parallelogram.
Answer:
Let ABCD be the quadailateral.
OP Malhotra Class 12 Maths Solutions Chapter 23 Three Dimensional Geometry Ex 23(a) 1
Mid point of AC = (\(\frac{4-1}{2}\), \(\frac{7-2}{2}\), \(\frac{8+1}{2}\))
= (\(\frac{3}{2}\), \(\frac{5}{2}\), \(\frac{9}{2}\))
and mid point of BD = (\(\frac{2+1}{2}\), \(\frac{3+2}{2}\), \(\frac{4+5}{2}\)),
i.e., (\(\frac{3}{2}\), \(\frac{5}{2}\), \(\frac{9}{2}\))
Thus diagonals AC and BD bisect each other.
D ratios of line AB are < 2 – 4, 3 – 7, 4 – 8 > i.e., < -2, -4, -4 >
D ratios of side DC are < -1 – 1, -2 – 2, 1 – 5 > i.e., < -2, -4, -4 >
∴ AB || DC.
similarly direction ratios of side AD are < 1 – 4, 2 – 7, 5 – 8 >
i.e., < -3, -5, -3 > and direction ratios of side BC are < -1 – 2, -2 – 3, 1 – 4 >
i.e., < -3, -5, -3 >
Thus AD || BC
Hence, A, B, C and D are the vertices of parallelogram.

Question 22.
Show that the points (5, -1, 1), (7, -4, 7), (1, -6, 10), and (-1, -3, 4) are the vertices of a rhombus.
Answer:
Let A, B, C and D are the given vertices of quadrilateral.
OP Malhotra Class 12 Maths Solutions Chapter 23 Three Dimensional Geometry Ex 23(a) 2
D ratios of side AB} are :
< 7 – 5, -4 + 1, 7 – 1 > i.e., < 2, -3, 6 >
D ratios of side AD are :
< -1 – 5, -3 + 1, 4 – 1 > i.e., < -6, -2, 3 >
Here D ratios of line AC are
< 1 – 5, -6 + 1, 10 – 1 > i.e., < -4, -5, 9 >
and D ratio of diagonal BD are
< -1 – 7, -3 + 4, 4 – 7 > i.e., < -8, 1, -3 >
Here a₁ a₂ + b₂ b₂ + c₁c₂ = (-4)(-8) + (-5) 1 + 9(-3) = 0
∴ both diagonals intersect each other at right angle.
Also mid point of AC= (\(\frac{5+1}{2}\), \(\frac{-1-6}{2}\), \(\frac{1+10}{2}\))
i.e., (3, \(\frac{-7}{2}\), \(\frac{11}{2}\))
and mid point of BD = (\(\frac{7-1}{2}\), \(\frac{-4-3}{2}\), \(\frac{4+7}{2}\))
i.e., (3, \(\frac{-7}{2}\), \(\frac{11}{2}\))
∴ both diagonals bisect each other at right angles.
|AB| = \(\sqrt{(7-5)^2+(-4+1)^2+(7-1)^2}\) = \(\sqrt{4+9+36}\) = 7
|AD| = \(\sqrt{(-1-5)^2+(-3+1)^2+(4-1)^2}\) = 7
|DC| = \(\sqrt{(7-1)^2+(-4+6)^2(10-7)^2}\) = \(\sqrt{36+4+9}\) = 7
|DC| = \(\sqrt{(1+1)^2+(-6+3)^2+(10-4)^2}\) = 7
∴ |AB| = |BC| = |AD| = |DC|
Also,
|AC| = \(\sqrt{(1-5)^2+(-6+1)^2+(10-1)^2}\) = \(\sqrt{16+25+81}\) = \(\sqrt{122}\)
|BD| = \(\sqrt{(-1-7)^2+(-3+4)^2+(4-7)^2}\) = \(\sqrt{64+1+9}\) = \(\sqrt{74}\)
∴ |AC| ≠ |BD|
Clearly A, B, C and D are the vertices of rhombus.

Question 23.
Find the foot of the perpendicular drawn from the point A(1, 0, 3) to the join of the points B(4, 7, 1) and C(3, 5, 3).
Answer:
Let D be the foot of ⊥ drawn from A(1, 0, 3) to BC.
Let point D divides the line BC in the ratio K : 1

OP Malhotra Class 12 Maths Solutions Chapter 23 Three Dimensional Geometry Ex 23(a) 3
∴ Coordinates of point D are (\(\frac{3{~K}+4}{{~K}+1}\), \(\frac{5{~K}+7}{{~K}+1}\), \(\frac{3{~K}+1}{{~K}+1}\))
∴ direction ratios of line AD are
< \(\frac{3{~K}+4}{{~K}+1}\) – 1, \(\frac{5{~K}+7}{{~K}+1}\) – 0, \(\frac{3{~K}+1}{{~K}+1}\) – 3 >
i.e., < \(\frac{2{~K}+3}{{~K}+1}\), \(\frac{5{~K}+7}{{~K}+1}\), \(\frac{-2}{{~K}+1}\) >
Also direction ratios of line BC are < 3 – 4, 5 – 7, 3 – 1 >
i.e., < -1, -2, 2 > Since line AD is ⊥ to line BC.
∴ (\(\frac{2{~K}+3}{{~K}+1}\))(-1) + (\(\frac{5{~K}+7}{{~K}+1}\))(-2) + (\(\frac{-2}{{~K}+1}\)) 2 = 0
⇒ -2K – 3 – 10K – 14 – 4 = 0
⇒ -12K – 21 = 0
⇒ K = \(-\frac{7}{4}\)
Thus required coordinates of foot of ⊥ D are (\(\frac{\frac{-21}{4}+4}{-\frac{7}{4}+1}\), \(\frac{\frac{-35}{4}+7}{-\frac{7}{4}+1}\), \(\frac{\frac{-21}{4}+1}{-\frac{7}{4}+1}\))
i.e., (\(\frac{5}{2}\), \(\frac{7}{3}\), \(\frac{17}{3}\))
Thus required coordinates of foot of ⊥ D are (\(\frac{\frac{-21}{4}+4}{\frac{7}{4}+1}\), \(\frac{\frac{-35}{4}+7}{-\frac{7}{4}+1}\), \(\frac{\frac{-21}{4}+1}{-\frac{7}{4}+1}\))
i.e., (\(\frac{5}{2}\), \(\frac{7}{3}\), \(\frac{17}{3}\))

Question 24.
A(1, 0, 4) and B(0, -11, 3), C(2, -3, 1) are three points and D is the foot of the perpendicular from A on B C. Find the coordinates of D.
Answer:
Let the point D divide the line BC in the ratio K : 1.
Then coordinates of point D are
OP Malhotra Class 12 Maths Solutions Chapter 23 Three Dimensional Geometry Ex 23(a) 4
(\(\frac{2 K}{K+1}\), \(\frac{-3 K-11}{K+1}\), \(\frac{K+3}{K+1}\))
∴ D ratios of line AD are
< \(\frac{2{~K}}{{~K}+1}\) – 1,\(\frac{-3{~K}-11}{{~K}+1}\) + 1 – 0, \(\frac{{K}+3}{{~K}+1}\) – 4 >
i.e., < \(\frac{{K}-1}{{~K}+1}\), \(\frac{-3{~K}-11}{{~K}+1}\), \(\frac{-3{~K}-1}{{~K}+1}\) >
D ratios of line BC are < 2, -3 + 11, 1 – 3 >
OP Malhotra Class 12 Maths Solutions Chapter 23 Three Dimensional Geometry Ex 23(a) 5

Question 25.
Calculate the cosine of the angle A of the triangle with vertices A(1,-1, 2), B(6, 11, 2),
Answer:
OP Malhotra Class 12 Maths Solutions Chapter 23 Three Dimensional Geometry Ex 23(a) 6
D ratios of side AB are
< 6 – 1, 11 + 1, 2 – 2 > i.e., < 5, 12, 0 >
D ratios of line AC are
< 1 – 1, 2 + 1, 6 – 2 > i.e., < 0, 3, 4 >
Here a_1 = 5 ; b_1 = 12 ; c_1 = 0
Here a_1 = 5 ; b_1 = 12 ; c_1 = 0
a_2 = 0 ; b_2 = 3 ; c_2 = 4
∴ cos A = \(\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\)
= \(\frac{5(0)+12(3)+0(4)}{\sqrt{5^2+12^2+0^2} \sqrt{0^2+3^2+4^2}}\)
= \(\frac{36}{13 \times 5}\)
= \(\frac{36}{65}\)

Question 26.
If A, B, C, D are the points (6, -6, 0), (-1, -7, 6), (3, -4, 4), (2, -9, 2) respectively, prove that A B is perpendicular to C D.
Answer:
Here direction ratios of line A B are
< -1, -6, -7 + 6, 6 – 0 >
i.e., < -7, -1, 6 >
Direction ratios of line CD are
Here
< 2 -3, -9 + 4, 2 – 4 > i.e., < -1, -5, -2 >
and
a₁ = -7 ; b₁ = -1 ; c₁ = 6
Now
a₂ = -1 ; b₂ = -5 ; c₂ = -2
a₁ a₂ + b₁ b₂ + c₁ c₂
= (-7)(-1) + (-1)(-5) + 6(-2) = 7 + 5 – 12 = 0

Question 27.
Find the angle between any two diagonals of a cube.
Answer:
Let a be the length of edge of the cube and let one corner of the cube be at (0,0,0).
Ther the diagonals of cube are OP, AR, BS and CQ
OP Malhotra Class 12 Maths Solutions Chapter 23 Three Dimensional Geometry Ex 23(a) 7
Let us consider the two diagonals OP and AR
∴ d ‘ratios of OP and AR are proportional to < a – 0, a – 0, a – 0 >
i.e. < a, a, a > and < 0 – a, a – 0, a – 0 > i.e. < -a, a, a >
Let θ be the angle between OP and AR
∴ cosθ = \(\frac{a(-a)+a(a)+a(a)}{\sqrt{a^2+a^2+a^2} \sqrt{a^2+a^2+a^2}}\)
= \(\frac{a^2}{\sqrt{3} a \sqrt{3} a}\)
= \(\frac{1}{3}\)
∴ θ = cos^-1 (\(\frac{1}{3}\))
Similarly the angle between other pairs of diagonals be cos ^-1 (\(\frac{1}{3}\))

OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l)

September 6, 2024

Well-structured S Chand ISC Maths Class 12 Solutions Chapter 8 Differentiation Ex 8(l) facilitate a deeper understanding of mathematical principles.

S Chand Class 12 ICSE Maths Solutions Chapter 8 Differentiation Ex 8(l)

Find the second derivative of the following functions :

Question 1.
(i) x²
(ii) a^x
(iii) ax³ + bx² + cx + d
(iv) log x
(v) 1/\(\sqrt{x}\)
(vi) x/\(\sqrt{x-1}\)
(vii) sin^-1 x
Solution:
(i) Let y = x² ; Diff. both sides w.r.t. x,
\(\frac { dy }{ dx }\) = 2x ; Again diff. both sides w.r.t. x
∴ \(\frac{d^2 y}{d x^2}\) = 2

(ii) Let y = a^x ; Diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = a^x log a; Again diff. both sides w.r.t. x ; we have
∴ \(\frac{d^2 y}{d x^2}\) = a²(log a)²

(iii) Let y = ox³ + bx² + cx + d; Diff. both sides w.r.t. x
\(\frac { dy }{ dx }\) = 3ax² + 2bx + c; Diff. again w.r.t. x
\(\frac { d²y }{ dx² }\) = 6ax + 2b

(iv) Let y = log x ; Diff. both sides w.r.t. x
\(\frac { dy }{ dx }\) = \(\frac { 1 }{ x }\) ; Diff. again w.r.t. x; we have
\(\frac{d^2 y}{d x^2}=-\frac{1}{x^2}\)

OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 1

Question 2.
(i) e^x + sin x
(ii) e^-x sin x
Solution:
(i) Let y = e^x + sin x ;
Diff. both sides w.r.t. x; we have
\(\frac { dy }{ dx }\) = e^x + cos x ;
Diff. again both sides w.r.t. x; we have
\(\frac{d^2 y}{d x^2}=e^x-\sin x\)

(ii) Let y = e^-x sin x ;
Diff. both sides w.r.t. x,
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 2

Question 3.
(i) If y = 2 sin x + 3 cos x, prove that y + \(\frac{d^2 y}{d x^2}\) = 0.
(ii) If y = a + bx², prove that x.\(\frac{d^2 y}{d x^2}=\frac{d y}{d x}\)
(iii) If y = tan x + sec x, prove that \(\frac{d^2 y}{d x^2}=\frac{\cos x}{(1-\sin x)^2}\).
(iv) If y = 500, e^7x + 600 e^-7x, show that \(\frac{d^2 y}{d x^2}\) = 49 y.
(iv) If e^y (1 + x) = 1, show that \(\frac{d^2 y}{d x^2}=\left(\frac{d y}{d x}\right)^2\).
Solution:
(i) Given y = 2 sin x + 3 cos x …(1)
Diff. both sides w.r.t. x; we have
\(\frac { dy }{ dx }\) = 2 cos x – 3 sin x ;
Again diff. both sides w.r.t. x
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 3

(iii) Given y = tan x + sec x … (1)
Diff. eqn. (1) both sides w.r.t. x; we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 4

(iv) Given y = 500 e^7x + 600 e^-7x …(1)
Diff. eqn. (1) both sides w.r.t. x; we have
\(\frac { dy }{ dx }\) = 3500 e^7x – 4200 e^-7x
Again diff. both sides w.r.t. x
\(\frac { dy }{ dx }\) = 7 x 3500 e^7x + 4200 x 7 e^-7x
= 49[500 e^7x + 600 e^-7x] = 49 y [using eqn. (1)]

(v) Given e^y (1 + x) = 1 ⇒ e^y = \(\frac { 1 }{ 1 + x }\)
Taking logorithm both sides w.r.t. x, we have
y = log\(\left(\frac{1}{1+x}\right)\) = – log(1 + x)
Diff. both sides w.r.t. x ; we have
\(\frac { dy }{ dx }\) = – \(\left(\frac{1}{1+x}\right)\) … (1)
Again diff. both sides w.r.t. x
\(\frac{d^2 y}{d x^2}=\frac{1}{(1+x)^2}=\left(\frac{d y}{d x}\right)^2\) [using eqn. (1)]

Question 4.
If y = tan x, prove that \(\frac{d^2 y}{d x^2}=2 y \frac{d y}{d x}\).
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 5

Question 5.
If y = \(\frac{\log x}{x}\), prove that \(\frac{d^2 y}{d x^2}=\frac{2 \log x-3}{x^3}\).
Solution:
Given y = \(\frac{\log x}{x}\)
Diff. both sides w.r.t. x; we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 5a

Question 6.
(i) If y = tan^-1 x, prove that
(1 + x²) \(\frac{d^2 y}{d x^2}+2 x \frac{d y}{d x}\) = 0.
(ii) If y = sin^-1x, then show that
(1 + x²) \(\frac{d^2 y}{d x^2}-x \frac{d y}{d x}\) = 0.
Solution:
(i) Given y = tan^-1 x;
Diff. both sides w.r.t. x, we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 6

(ii) Given y = sin^-1 x;
Diff. both sides w.r.t. x, we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 7

Question 7.
If y = \(e^{\tan ^{-1} x}\), prove that
\(\left(1+x^2\right) \frac{d^2 y}{d x^2}+(2 x-1) \frac{d y}{d x}\) = 0.
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 8

Question 8.
If y = x^x, prove that
\(\frac{d^2 y}{d x^2}-\frac{1}{y}\left(\frac{d y}{d x}\right)^2-\frac{y}{x}\) = 0
Solution:
Given y = x^x, … (1)
Taking logarithm on eqn. (1); we have
log y = x log x …(2)
Diff. eqn. (2) w.r.t. x, we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 9

Question 9.
If y = \(\frac{\sin ^{-1} x}{\sqrt{1-x^2}}\), prove that
\(\left(1-x^2\right) \frac{d^2 y}{d x^2}-3 x \frac{d y}{d x}-y\) = 0.
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 9a

Question 10.
If y = (tan^-1 x)², prove that
\(\left(x^2+1\right)^2 \frac{d^2 y}{d x^2}+2 x\left(x^2+1\right) \frac{d y}{d x}\) = 2.
Solution:
Given y = (tan^-1x)²,
Diff. both sides w.r.t. x, we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 10

Question 11.
If y = sin (m sin^-1 x) show that (1 – x²)\(\frac{d^2 y}{d x^2}-x \frac{d y}{d x}+m^2 y\)y = 0
Solution:
Given y = sin (m sin^-1x) … (1)
Diff. eqn (1) w.r.t. x, we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 11

Question 12.
If y = (A + Bx)e^3x, prove that y” + 6y’ + 9y + 2 = 2.
Solution:
Given y = (A + Bx)e^-3x …(1)
Diff. eqn. (1) w.r.t. x, we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 12

Question 13.
If x^myⁿ = (x + y)^m+n, prove that \(\frac{d^2 y}{d x^2}\) = 0.
Solution:
Given x^myⁿ = (x + y)^m+n
Taking logaritum on both sides, we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 13

Question 14.
If y = ae^mx + be^-mx, prove that \(\frac{d^2 y}{d x^2}-m^2 y\) = 0.
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 14

Question 15.
If y = a cos (log x) + b sin (log x), prove that \(x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}+y\) = 0.
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 15

Question 16.
Find \(\frac{d^2 y}{d x^2}\) when
(i) x = t², y = t³.
(ii) x = at², y = 2at.
(iii) x = a cos θ, y = b sin θ
(iv) x = cos t, y = sin t
Solution:
(i) Let x = t² … (1)
& y = t³ … (2)
Diff. eqn. (1) & (2) w.r.t. t; we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 16

(ii) x = at² … (1)
& y = 2at … (2)
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 17

(iii) Let x = a cos θ …(1)
& y = b sin θ …(2)
Diff. eqn. (1) & (2) w.r.t. θ; we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 18

Question 17.
Find \(\frac{d^2 y}{d x^2}\) when θ = \(\frac { π }{ 2 }\):
(i) x = a(θ + sin θ), y = a(1 – cos θ)
(ii) x = a(1 – cos θ), y = a(θ + sin θ).
Solution:
(i) Let x = a(θ + sin θ) …(1)
& y = a(1 – cos θ) …(2)
Diff. eqn. (1) & (2) w.r.t. θ; we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 19

(ii) Given x = a(1 – cos θ) …(1)
& y = a(θ + sin θ) …(2)
Diff. eqn. (1) & (2) w.r.t. θ; we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 20

Question 18.
If x = a sec³θ, y = a tan³θ, find \(\frac{d^2 y}{d x^2}\) at θ = \(\frac { π }{ 4 }\).
Solution:
Let x = a sec³θ …(1)
& y = a tan³θ …(2)
Diff. eqn. (1) & (2) w.r.t. θ; we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 21

Question 19.
If x = cos θ + θ sin θ, y = sin θ – θ cos θ, 0 < θ < \(\frac { π }{ 2 }\), prove that \(\frac{d^2 y}{d x^2}=\frac{\sec ^3 \theta}{\theta}\).
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 22

Question 20.
If x = cos θ, y = sin³θ, show that \(\frac{d^2 y}{d x^2} \cdot\left(\frac{d y}{d x}\right)^2=3 \sin ^2 \theta\left(5 \cos ^2 \theta-1\right)\) .
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 23

Question 21.
If f(x) = \(\left|\begin{array}{ccc}
\sec \theta & \tan ^2 \theta & 1 \\
\theta \sec \theta & \tan x & x \\
1 & \tan x-\tan \theta & 0
\end{array}\right|\), then f'(θ) is
(a) 0
(b) – 1
(c) independent of θ
(d) None of these.
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 24

Question 22.
If y = \(\left|\begin{array}{ccc}
f(x) & g(x) & h(x) \\
l & m & n \\
a & b & c
\end{array}\right|\), prove that \(\frac{d y}{d x}=\left|\begin{array}{ccc}
f(x) & g(x) & h(x) \\
l & m & n \\
a & b & c
\end{array}\right|\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 25

Examples

Question 1.
If y = \(\log \sqrt{\frac{1-\cos x}{1+\cos x}}, \text { find } \frac{d y}{d x}\).
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 26

Question 2.
If y = (cos x)^{cos x}, find \(\frac { dy }{ dx }\).
Solution:
Given y = (cos x)^{cos x} ;
Taking logarithm or both sides, we have
log y = log (cos x)^{cos x}
log y = cos x log cos x
Diff both sides w.r.t x ; we have
\(\frac{1}{y} \frac{d y}{d x}\) = cos x \(\frac{1}{\cos x}\) (- sin x) + log cos x (- sinx)
\(\frac { dy }{ dx }\) = y [- sinx – sinx log cos x]
= (cos x)^{cos x} (- sinx) [1 + log cos x]
= – sinx (cos x)^{cos x} [log e + log cos x]
= – sinx (cos x)^{cos x} log (e cos x)
[∵ log a + log b = log ab]

Question 3.
If y = e^x log (tan 2x), find \(\frac { dy }{ dx }\).
Solution:
Given y = e^x log (tan 2x) ;
Diff. both sides w.r.t x ; we have
\(\frac { dy }{ dx }\) = e^x \(\frac{1}{\tan 2 x}\) sec²2x.2 + log (tan 2x).e^x
= e^x [2cot 2x sec²2x + log (tan 2x)]

Question 4.
If y = \(\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\), prove that \(\frac{d y}{d x}=\frac{2}{1+x^2}\).
Solution:
Given y = \(\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\) ;
put x = tan θ ⇒ θ = tan^-1x
∴ y = tan^-1\(\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right)\) = tan^-1(tan 2θ)
⇒ y = 2θ = 2 tan^-1 x
DifF. both sides w.r.t x ; we have
\(\frac{d y}{d x}=\frac{2}{1+x^2}\)

Question 5.
If y = \(e^{m \cos ^{-1} x}\), prove that
\(\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}=m^2 y\).
Solution:
Given y = \(e^{m \cos ^{-1} x}\)
Diff both sides w.r.t x ; we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 27

Question 6.
If x^yy^x = 5, show that \(\frac{d y}{d x}=-\left(\frac{\log y+\frac{y}{x}}{\log x+\frac{x}{y}}\right)\).
Solution:
Given x^yy^x = 5;
Taking logarithm on both sides; we get
log x^y + log y^x = log 5
⇒ y log x + x log y = log 5 :
Diff both sides w.r.t. x ; we get
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 28

Question 7.
If x = a sin³t and y = a cos³t, find \(\frac { dy }{ dx }\).
Solution:
Given x = a sin³t; …(1)
y = a cos³t …(2)
Diff eqn (1) & eqn (2) w.r.t. t, we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 29

Question 8.
If sin (xy) + cos (xy) = 1 and tan (xy) ≠ 1, then show that \(\frac{d y}{d x}=-\frac{y}{x}\).
Solution:
Given sin (xy) + cos (xy) = 1 …(1)
Diff eqn (1) & eqn (2) w.r.t. x, we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 30

Question 9.
If x^py^q = (x + y)^p+q, prove that \(\frac{d y}{d x}=\frac{y}{x}\).
Solution:
Given x^py^q = (x + y)^p+q;
Taking logarithm on both sides; we have
p log x + q log y = (p + q) log(x + y)
Diff both sides w.r.t x; we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 31

Question 10.
If y = \(e^{\sin \left(x^2\right)}\), find \(\frac { dy }{ dx }\).
Solution:
Given y = \(e^{\sin \left(x^2\right)}\);
Diff both sides w.r.t. x, we have
\(\frac{d y}{d x}=e^{\sin \left(x^2\right)} \cos x^2 \cdot 2 x\)

Question 11.
If y = \(\frac{\sin ^{-1} x}{\sqrt{1-x^2}}\), prove that (1 – x²) \(\frac { dy }{ dx }\) – xy = 1.
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 32

Question 12.
If e^x+y = xy, show that \(\frac{d y}{d x}=\frac{y(1-x)}{x(y-1)}\).
Solution:
Given e^x+y = xy ;
Taking logarithm on both sides, we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 33

Question 13.
If sin y = x sin (a + y), show that \(\frac{d y}{d x}=\frac{\sin ^2(a+y)}{\sin a}\).
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 34

Question 14.
Find \(\frac { dy }{ dx }\) if y = tan^-1\(\frac{\sqrt{1+x^2}-1}{x}\).
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 35

Question 15.
If y = \(\sqrt{\frac{1-\cos x}{1+\cos x}}\), find \(\frac { dy }{ dx }\).
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 36

Question 16.
Using a suitable substitution, find the derivative of tan^-1\(\frac{4 \sqrt{x}}{1-4 x}\) with respect to x
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 37

Question 17.
Find the derivative of sin x2 with respect to x³.
Solution:
Let y = sin x² & z = x³ …(2)
So we want to diff. y w.r.t. z i.e. to find \(\frac { dy }{ dx }\)
diff. eqn. (1) & eqn. (2) both sales w.r.t. – x ; we have
\(\frac { dy }{ dx }\) = cos x². 2x ; \(\frac { dz }{ dx }\) = 3x²
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}=\frac{2 x \cos x^2}{3 x^2}=\frac{2 \cos x^2}{3 x}\)

Question 18.
Using a suitable substitution and the derivative of tan^-1\(\sqrt{\frac{a-x}{a+x}}\) with respect to x.
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 38

Question 19.
If y = x^x, prove that \(\frac{d^2 y}{d x^2}-\frac{1}{y}\left(\frac{d y}{d x}\right)^2-\frac{y}{x}\) = 0.
Solution:
Given y = x^x … (1)
Taking logarithm on eqn. (1); we have log y = x log x …(2)
Diff. eqn. (2) w.r.t. x, we have
\(\frac{1}{y} \frac{d y}{d x}=x \cdot \frac{1}{x}\) + log x = 1 + log x
∴ \(\frac{d y}{d x}=y(1+\log x)\) … (3)
Diff. eqn. (3) both sides w.r.t. x; we have
\(\frac{d^2 y}{d x^2}=\frac{d y}{d x}(1+\log x)+\frac{y}{x}\)
⇒ \(\frac{d^2 y}{d x^2}=\frac{1}{y}\left(\frac{d y}{d x}\right)^2+\frac{y}{x}\)
[using eqn. (3)]

Question 20.
If e^y (x + 1) = 1, then show that \(\frac{d^2 y}{d x^2}=\left(\frac{d y}{d x}\right)^2\).
Solution:
Given e^y (x + 1) = 1 ⇒ e^y = \(\frac { 1 }{ 1+x }\)
Taking logorithm both sides w.r.t. x, we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 39

Question 21.
If y = (cot^-1 x)², show that
\(\left(1+x^2\right)^2 \frac{d^2 y}{d x^2}+2 x\left(1+x^2\right) \frac{d y}{d x}\) = 2.
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 40

Question 22.
If y = \(\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}\), prove that (1 – x²) \(\frac{d y}{d x}=x+\frac{y}{x}\).
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 41

Question 23.
If log y = tan^-1x, prove that
(1 + x²) \(\frac{d^2 y}{d x^2}+(2 x-1) \frac{d y}{d x}\) = 0.
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 42

Question 24.
If y = cos (sin x), show that \(\frac{d^2 y}{d x^2}+\tan x \frac{d y}{d x}+y \cos ^2 x\) = 0
Solution:
Given = cos (sin x) …(1)
diff. both sides w.r.t. x ; we have
\(\frac { dy }{ dx }\) = – sin (sin x) cos x …(2)
again diff. both sides w.r.t. x ; we have
\(\frac{d^2 y}{d x^2}\) = – [sin (sin x) (- sin x) + cos² x cos (sin x)]
⇒ \(\frac{d^2 y}{d x^2}=-\frac{\sin x}{\cos x} \frac{d y}{d x}-y \cos ^2 x\) [using (1) and (2)]
⇒ \(\frac{d^2 y}{d x^2}+\tan x \frac{d y}{d x}+y \cos ^2 x\) = 0

Question 25.
If y = sec (tan^-1 x), then \(\frac { dy }{ dx }\) is equal to
(a) \(\frac{x}{\sqrt{1+x^2}}\)
(b) \(x \sqrt{1+x^2}\)
(c) \(\sqrt{1+x^2}\)
(d) \(\frac{1}{\sqrt{1+x^2}}\)
Solution:
Let y = sec (tan^-1 x)
⇒ y = sec (sec^-1 \(\sqrt{1+x^2}\)) ⇒ y = \(\sqrt{1+x^2}\)
Diff both sides w.r.t. x, we get
\(\frac{d y}{d x}=\frac{1}{2}\left(1+x^2\right)^{-\frac{1}{2}} \times 2 x=\frac{x}{\sqrt{1+x^2}}\)

Question 26.
Differentiate sin (sin 2x).
(a) 2 cos 2x cos 2x
(b) 2 cos 2x cos (sin 2x)
(c) 2 cos 2x sin 2x
(d) cos 2x cos (sin 2x)
Solution:
Let y = sin (sin 2x)
\(\frac { dy }{ dx }\) = cos (sin 2x) \(\frac { d }{ dx }\) sin 2x = cos (sin 2x) 2 cos 2x

Question 27.
If x = ct and y = \(\frac { c }{ t }\), find \(\frac { dy }{ dx }\), at t = 2.
(a) 4
(b) 0
(c) \(\frac { 1 }{ 4 }\)
(d) – \(\frac { 1 }{ 4 }\)
Solution:
Given x = ct … (1)
and y = \(\frac { c }{ t }\) … (2)
Diff. both eqns. (1) and (2) w.r.t. t, we get
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 43

Question 28.
If y = tan^-1\(\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right)\), then \(\frac { dy }{ dx }\) is equal to
(a) 0
(b) \(\frac { 1 }{ 2 }\)
(c) \(\frac { π }{ 4 }\)
(d) 1
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 43a

Question 29.
If y = tan^-1 x + cot^-1 x + sec^-1 x + cosec^-1 x, then \(\frac { dy }{ dx }\) is equal to
(a) \(\frac{x^2-1}{x^2+1}\)
(b) π
(c) 0
(d) 1
(e) \(\frac{1}{x \sqrt{x^2-1}}\)
Solution:
y = (tan^-1 x + cot^-1 x) + (sec^-1 x + cosec^-1 x)
⇒ y = \(\frac { π }{ 2 }\) + \(\frac { π }{ 2 }\)
∴ \(\frac { dy }{ dx }\) = 0

Question 30.
If y = sin^-1 \(\sqrt{1-x}\), then \(\frac { dy }{ dx }\) is equal to
(a) \(\frac{1}{\sqrt{1-x}}\)
(b) \(\frac{-1}{2 \sqrt{1-x}}\)
(c) \(\frac{1}{\sqrt{x}}\)
(d) \(\frac{-1}{2 \sqrt{x} \sqrt{1-x}}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 44

Question 31.
If x ≠ 0 and y = log, |2 x|, then \(\frac { dy }{ dx }\) is equal to
(a) \(\frac { 1 }{ x }\)
(b) \(\frac { -1 }{ x }\)
(c) ± \(\frac { 1 }{ 2x }\)
(d) None of these
Solution:
Given y = log | 2x | ; x ≠ 0, when x < 0 then | 2x | = – 2x ∴ y = log (- 2x) ⇒ \(\frac{d y}{d x}=\frac{-1}{2 x}(-2)=\frac{1}{x}\) when x > 0 Then |2x| = 2x
∴ y = log2x
⇒ \(\frac{d y}{d x}=\frac{1}{2 x} \times 2=\frac{1}{x}\)
Thus \(\frac { dy }{ dx }\) = \(\frac { 1 }{ x }\) ; when x ≠ 0

Question 32.
If x² + y² = 4, then y\(\frac { dy }{ dx }\) + x =
(a) 4
(b) 0
(c) 1
(d) – 1
Solution:
Given x² + y² = 4 ; diff. both sides w.r.t. x ;
2x + 2y \(\frac { dy }{ dx }\) = 0 ⇒ x + y\(\frac { dy }{ dx }\) = 0

Question 33.
If y = sin^-1 \(\left(2 x \sqrt{1-x^2}\right), \quad-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}\), then \(\frac { dy }{ dx }\) is equal t0
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 45

Question 34.
If y = tan^-1 \(\left(\frac{a-x}{1+a x}\right)\), then \(\frac { dy }{ dx }\)
(a) \(\frac{1}{1+x^2}\)
(b) \(\frac{a}{1+a x^2}\)
(c) – \(\frac{1}{1+x^2}\)
(d) \(\frac{x}{1+x^2}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 46

Question 35.
If y = log\(\left(\frac{x^2}{e^2}\right)\), then \(\frac{d^2 y}{d x^2}\) equals
(a) \(\frac{1}{1+x^2}\)
(b) – \(\frac{a}{1+a x^2}\)
(c) \(\frac{1}{1+x^2}\)
(d) – \(\frac{x}{1+x^2}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 47

Question 36.
If y = Ae^5x + Be^-5x, then \(\frac{d^2 y}{d x^2}\) is equal to
(a) 25y
(b) 5y
(c) – 25y
(d) 15y
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 48

Question 37.
If y = | cos x | + | sin x |, then \(\frac { dy }{ dx }\) at x = \(\frac { 2π }{ 3 }\) is
(a) \(\frac{1-\sqrt{3}}{2}\)
(b) 0
(c) \(\frac{1}{2}(\sqrt{3}-1)\)
(d) None of these
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 49

Find \(\frac { dy }{ dx }\) if y =

Question 38.
cosec x°
Solution:
Given y = cosec x° = cosec \(\frac { πx }{ 180 }\) [π rad = 180°]
∴ \(\frac { dy }{ dx }\) = – cot \(\frac { πx }{ 180 }\) cosec \(\frac { πx }{ 180 }\).\(\frac { π }{ 180 }\) = – \(\frac { π }{ 180 }\) cot x° cosec x°

Question 39.
cos x³
Solution:
Given y = cos x³
∴ \(\frac { dy }{ dx }\) = – sinx³ (3x²)

Question 40.
sin (sin 3x)
Solution:
Given y = sin (sin 3x)
∴ \(\frac { dy }{ dx }\) = cos (sin3x) \(\frac { d }{ dx }\) sin 3x = 3 cos (sin 3x) cos 3x

Question 41.
log (tan x)
Solution:
Given y = log (tan x)
∴ \(\frac { dy }{ dx }\) = ∴ \(\frac{1}{\tan x} \frac{d}{d x} \tan x=\frac{\sec ^2 x}{\tan x}=\frac{1}{\sin x \cos x}\) = sec x. cosec x

Question 42.
xy = c²
Solution:
Given xy = c² ; diff. both sides w.r.t. x
x\(\frac { dy }{ dx }\) + y.1 = 0 ⇒ \(\frac { dy }{ dx }\) = – \(\frac { y }{ x }\)

Question 43.
log (cos e^x)
Solution:
Given y = log (cos e^x)
∴ \(\frac{d y}{d x}=\frac{1}{\cos e^x} \frac{d}{d x} \cos e^x=\frac{1}{\cos e^x}\left\{-\sin e^x\right\} \frac{d}{d x} e^x=-e^x \tan e^x\)

Question 44.
cosec (cot \(\sqrt{x}\))
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 50

Question 45.
\(\tan ^{-1} \frac{1+\cos x}{\sin x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 51

Question 46.
If f(x) = x + 1, then write the value of \(\frac { d }{ dx }\) (fof) (x).
Solution:
Given f(x) = x + 1
∴ (fof) (x) = f(f (x)) = f(x + 1) = x + 1 + 1 = x + 2
∴ \(\frac { d }{ dx }\) (fof) (x) = 1

Question 47.
If f (x) = | cos x |, then f'(\(\frac { π }{ 4 }\)) is equal to …………..
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 52

Question 48.
If f (x) = x | x |, then f'(x) = ……………
Solution:
Given, f (x) = x | x | ; difF. both sides w.r.t. x ; we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 53

Question 49.
If f(x) = | cos x – sin x |, then f”(\(\frac { π }{ 3 }\)) = ………….
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 54

Question 50.
If y = a^x, then find \(\frac{d^2 y}{d x^2}\).
Solution:
Given y = a^x
∴ \(\frac{d y}{d x}=a^x \log a\)
∴ \(\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(a^x \log a\right)=\log a \cdot a^x \log a=a^x(\log a)^2\)

Question 51.
For the curve \(\sqrt{x}+\sqrt{y}=1, \frac{d y}{d x} \text { at }\left(\frac{1}{4}, \frac{1}{4}\right)\) is …………
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 55

Question 52.
Write the derivative of sin x w.r.t. cos x.
Solution:
Let y = sin x
and z = cos x
We want to find \(\frac { dy }{ dz }\)
Diff. both eqns. (1) and (2) w.r.t. x ; we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 56

Question 53.
Derivative of x² w.r.t x³ is ……………
Solution:
Let y = x²
and z = x³
We want to find \(\frac { dy }{ dz }\)
Diff. eqns. (1) and (2) w.r.t. x ; we have
OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l) 57

OP Malhotra Class 12 Maths Solutions Chapter 23 Three Dimensional Geometry Ex 23(b)

September 6, 2024

Effective ISC Class 12 Maths OP Malhotra Solutions Chapter 23 Three Dimensional Geometry Ex 23(b) can help bridge the gap between theory and application.

S Chand Class 12 ICSE Maths Solutions Chapter 23 Three Dimensional Geometry Ex 23(b)

Question 1.
Find the equations of a line passing through the point (-1, 2, 3) and having direction ratios proportional to -4, 5, 6.
Answer:
We know that, eqn. of line through the point (x₁, y₁, z₁) and having direction ratios < a, b, c> be given by \(\frac{x-x_1}{a}\) = \(\frac{y-y_1}{b}\) = \(\frac{z-z_1}{c}\)
∴ required eqn. of line through the point (-1, 2, 3) and having direction ratio < -4, 5, 6 > be
\(\frac{x-1}{-4}\) = \(\frac{y-2}{5}\) = \(\frac{z-3}{6}\)

Question 2.
Find the equations of a line passing through the point (2, -3, 0) and having direction cosines \(-\frac{1}{7}\), \(\frac{4}{7}\), \(-\frac{6}{7}\)
Answer:
Thus required eqn. of line passing through the point (2, -3, 0) and having direction cosines < \(-\frac{1}{7}\), \(\frac{4}{7}\), \(\frac{-6}{7}\) > by given by \(\frac{x-2}{\frac{-1}{7}}\) = \(\frac{y+3}{\frac{4}{7}}\) = \(\frac{z-0}{\frac{-6}{7}}\) or \(\frac{x-2}{-1}\) = \(\frac{y+3}{4}\) = \(\frac{z-0}{-6}\)

Question 3.
Find the equations of a line passing through the points (2, 3, 4) and (4, 6, 5).
Answr:
We know that eqn. of line passing through the points (2, 3, 4) and (4, 6, 5) is given by
\(\frac{x-2}{4-2}\) = \(\frac{y-3}{6-3}\) = \(\frac{z-4}{5-4}\)
i.e., \(\frac{x-2}{2}\) = \(\frac{y-3}{3}\) = \(\frac{z-4}{1}\)
[∵ eqn. of line through the points (x₁, y₁, z₁) and (x₂, y₂, z₂) be given by
\(\frac{x-x_1}{x_2-x_1}\)
= \(\frac{y-y_1}{y_2-y_1}\) = \(\frac{z-z_1}{z_2-z_1}\)

Question 4.
Find the equations of a line passing through the points (3, -2, -5) and (3, -2, 6).
Answer:
The required equation of line passing through the points (3,-2,-5) and (3,-2,6) is given by
\(\frac{x-3}{3-3}\)
= \(\frac{y+2}{-2+2}\)
= \(\frac{z+5}{6+5}\)
i.e., \(\frac{x-3}{0}\)
= \(\frac{y+2}{0}\) = \(\frac{z+5}{11}\)

Question 5.
Find the coordinates of the point, where the line through (5, 1, 6) and (3, 4, 1) crosses the
(i) y z-plane
(ii) the x y-plane
(iii) the x-plane.
Sol.
eqn. of line passing through the points (5, 1, 6) and (3, 4, 1) is given by
\(\frac{x-5}{3-5}\) = \(\frac{y-5}{4-1}\)
= \(\frac{z+6}{1-6}\)
i.e., \(\frac{x-5}{-2}\)
= \(\frac{y-1}{3}\)
= \(\frac{z+6}{-5}\)

(i) Since line (1) crosses y z plane
∴ x = 0
∴ from (1); \(\frac{0-5}{-2}\) = \(\frac{y-1}{3}\) = \(\frac{z-6}{-5}\)
⇒ 2 y – 2 = 15
⇒ y = \(\frac{17}{2}\)
and 2(6 – z) = 25
⇒ 6 – z = \(\frac{25}{2}\)
⇒ z = \(-\frac{13}{2}\)
Thus the required coordinates of point be (0, \(\frac{17}{2}\), \(\frac{-13}{2}\)).

(ii) Since line (1) crosses x y plane ∴ z = 0
i.e., putting z = 0 in eqn. (1); we have
\(\frac{x-5}{-2}\)
= \(\frac{y-1}{3}\) = \(\frac{0-6}{-5}\)
= \(\frac{6}{5}\)
⇒ 5 x – 25 = -12
⇒ x = \(\frac{13}{5}\)
and
5(y – 1) = 18
⇒ y = \(\frac{23}{5}\)
Thus the required coordinates of point be (\(\frac{13}{5}\), \(\frac{23}{5}\), 0)

(iii) Since line (1) crosses z x plane,
i.e., y = 0, putting y = 0 in eqn. (1); we have
\(\frac{x-5}{-2}\) = \(\frac{0-1}{3}\)
= \(\frac{z-6}{-5}\)
⇒ 3(x – 5) = 2
z-6 = \(\frac{5}{3}\)
⇒ z = \(\frac{23}{3}\)
and
z – 6 = \(\frac{5}{3}\) ⇒ z = \(\frac{23}{3}\)
Thus, the required coordinates of point be (\(\frac{17}{3}\), 0, \(\frac{23}{3}\))

Question 6.
The cartesian equations of a line are 6 x-2=3 y+1=2 z-2. Find the direction ratios.
Answer:
The eqn. of given line by 6 x – 2 = 3 y + 1 = 2 z – 2
⇒ 6(x – \(\frac{1}{3}\)) = 3(y + \(\frac{1}{3}\)) = 2(z – 1)
⇒ \(\frac{x-\frac{1}{3}}{1 / 6}\) = \(\frac{y+\frac{1}{3}}{1 / 3}\)
= \(\frac{z-1}{1 / 2}\)
⇒ \(\frac{x-\frac{1}{3}}{1}\)
= \(\frac{y+\frac{1}{3}}{2}\) = \(\frac{z-1}{3}\)
∴ direction ratios of line (1) are < 1, 2, 3 >

Question 7.
Find the cartesian equations of a line which
(i) passes through the point (1, 2, 3) and parallel to the line
\(\frac{-x-2}{1}\) = \(\frac{y+3}{7}\) = \(\frac{2 z-6}{3}\)
(ii) passes through the point (1,3,-2) and is parallel to the line given by
\(\frac{x+1}{3}\) = \(\frac{y+4}{5}\) = \(\frac{z+3}{-6}\)
(iii) through the point (2, -1, 1) and parallel to the line joining the points (-1, 4, 1) and (1 , 2 , 2 ).
Answer:
(i) eqn. of given line be
\(\frac{-x-2}{1}\) = \(\frac{y+3}{7}\) = \(\frac{2 z-6}{3}\)
⇒ \(\frac{x+2}{-1}\) = \(\frac{y+3}{7}\) = \(\frac{z-3}{3 / 2}\)
⇒ \(\frac{x+2}{-2}\) = \(\frac{y+3}{14}\) = \(\frac{z-3}{3}\)
∴ direction ratios of given line are < -2, 14, 3 >
Thus direction ratio of the line which is parallel to given line be < -2, 14, 3 >
Thus the required cartesian eqn. of line passes through the point (1, 2, 3) and having direction ratios < -2, 14, 3 > be \(\frac{x-1}{-2}\)
= \(\frac{y-2}{14}\)
= \(\frac{z-3}{3}\)
(ii) eqn. of given line be \(\frac{x+1}{3}\) = \(\frac{y-4}{5}\)
= \(\frac{z+3}{-6}\)
∴ D ratios of given line ( 1 ) be < 3, 5, -6 >
Thus D ratios of line parallel to line (1) be < 3, 5, -6 >.
Therefore eqn. of line passing through the point (1,3,-2) and having direction ratios < 3, 5, -6 > be given by \(\frac{x-1}{3}\) = \(\frac{y-3}{5}\)
= \(\frac{z+2}{-6}\)

(iii) Direction numbers of the line joining the points (-1, 4, 1) and (1, 2, 2) be < 1 + 1, 2 – 4, 2 – 1 >
i.e., < 2, -2, 1 >
∴ direction ratios of the line || to given line are proportional to < 2, -2, 1 >
Thus the required eqn. of line through the point (2, -1, 1) and having direction ratios < 2, -2, 1 > will be given by \(\frac{x-2}{2}\)
= \(\frac{y+1}{-2}\) = \(\frac{z-1}{1}\)

Question 8.
Prove that the points A(-1, 3, 2), B(-4, 2, -2) and C(5, 5, 10) are collinear.
Answer:
The eqn. of line through the points A(-1, 3, 2) and B(-4, 2, -2) be given by
\(\frac{x+1}{-4+1}\) = \(\frac{y-3}{2-3}\) = \(\frac{z-2}{-2-2}\)
i.e., \(\frac{x+1}{-3}\) = \(\frac{y-3}{-1}\) = \(\frac{z-2}{-4}\)
The point C(5, 5, 10) lies on line (1) if (5, 5, 10) satisfies eqn. (1)
i.e., if \(\frac{5+1}{-3}\) = \(\frac{5-3}{-1}\) = \(\frac{10-2}{-4}\)
if -2 = -2 = -2, which is true
Thus, the given points are collinear.

Question 9.
Find the value of X, for which the points A(2, 1, 3), B(5, 0, 5) and C(-4, λ,-1) are collinear.
Answer:
The eqn. of line passing through the points A(2, 1, 3) and B(5, 0, 5) is given by
\(\frac{x-2}{5-2}\) = \(\frac{y-1}{0-1}\) = \(\frac{2-3}{5-3}\)
i.e., \(\frac{x-2}{3}\) = \(\frac{y-1}{-1}\) = \(\frac{2-3}{2}\)
Since the given points A, B and C are collinear.
∴ Point C(-4, λ, -1) lies on line (1).
Thus C(-4, λ, -1) satisfies eqn. (1).
∴ \(\frac{-4-2}{3}\) = \(\frac{\lambda-1}{-1}\) = \(\frac{-1-3}{2}\)
⇒ -2 = \(\frac{\lambda-1}{-1}\) = -2
⇒ λ -1 = 2
⇒ λ = 3

Question 10.
Find the equations of a line passing through the point P(1, 2, 3) and having direction cosines \(\frac{2}{3}\), \(-\frac{2}{3}\), \(\frac{1}{3}\) Also find the coordinates of a point on the line at a distance of 6 units from P.
Answer:
The required eqn. of line passing through the point P(1, 2, 3) and having direction cosines < \(\frac{2}{3}\), \(\frac{-2}{3}\), \(\frac{1}{3}\) >
i.e., having direction ratios proportional to < 2, -2, 1 > is given by
\(\frac{x-1}{2}\) = \(\frac{y-2}{-2}\) = \(\frac{z-3}{1}\) = t (say)
So any point on line (1) be Q(2 t+1, -2 t+2, t+3)
Also it is given that |P Q| = 6 units
∴ \(\sqrt{(2 t+1-1)^2+(-2 t+2-2)^2+(t+3-3)^2}\) = 6
⇒ \(\sqrt{4 t^2+4 t^2+t^2}\) = 6;
on squaring; we have
⇒ 9 t² = 36
⇒ t² = 4
⇒ t = ± 2
When t = 2; coordinates of point Q be (5, -2, 5)
When t = -2; coordinates of point Q be (-3, 6, 1)

Question 11.
Find the values of p and q so that the points (p, q, 1), (-1, 4, -2) and (0, 2, -1) are collinear.
Answer:
Let the given points are A(p, q, 1), B(-1, 4, -2) and C(0, 2, -1)
∴ eqn. of line passing through the points A(p, q, 1) and B(-1, 4, -2) is given by
\(\frac{x-p}{-1-p}\) = \(\frac{y-q}{4-q}\) = \(\frac{z-1}{-2-1}\)
Since points must A, B and C are collinear
∴ point C(0, 2, -1) lies on eqn. (1).
∴ \(\frac{0-p}{-1-p}\) = \(\frac{2-q}{4-q}\)
= \(\frac{-1-1}{-3}\)
i.e., \(\frac{p}{1+p}\) = \(\frac{2-q}{4-q}\) = \(\frac{2}{3}\)
Now
\(\frac{p}{1+p}\) = \(\frac{2}{3}\)
⇒ 3p = 2 + 2 p
⇒ p = 2
\(\frac{2-q}{4-q}\) = \(\frac{2}{3}\)
⇒ 6 – 3 q = 8 – 2 q
⇒ q = -2
and
\(\frac{2-q}{4-q}\) = \(\frac{2}{3}\)
⇒ 6 – 3q = 8 – 2 q ⇒ q = -2

Question 12.
Write the following equations of a line in standard form and hence find the coordinates of a point on it and its direction cosines :
\(\frac{3-2 x}{4}\) = \(\frac{2 y-1}{2}\) = \(\frac{3+z}{2}\)
Answer:
Equation of given line be, given by
\(\frac{3-2 x}{4}\) = \(\frac{2 y-1}{2}\) = \(\frac{3+z}{2}\)
⇒ \(\frac{-2\left(x-\frac{3}{2}\right)}{4}\)
= \(\frac{2\left(y-\frac{1}{2}\right)}{2}\) = \(\frac{z+3}{2}\)
⇒ \(\frac{x-\frac{3}{2}}{-2}\) = \(\frac{y-\frac{1}{2}}{1}\)
= \(\frac{z+3}{2}\)
⇒ \(\frac{x-\frac{3}{2}}{-2}\)
= \(\frac{y-\frac{1}{2}}{1}\) = \(\frac{z-(-3)}{2}\)
which is the required line in standard form clearly line (1) passing through the point (\(\frac{3}{2}\), \(\frac{1}{2}\), – 3) and having direction ratios < -2, 1, 2 >
∴ direction cosines of line (1) are
< \(\frac{-2}{\sqrt{(-2)^2+1^2+2^2}}\), \(\frac{1}{\sqrt{(-2)^2+1^2+2^2}}\), \(\frac{2}{\sqrt{(-2)^2+1^2+2^2}}\) >
i.e., < \(\frac{-2}{3}\), \(\frac{1}{3}\), \(\frac{2}{3}\) >

Question 13.
Find the direction cosines of the line whose equations are \(\frac{x-2}{2}\) = \(\frac{2 y-5}{-3}\), z = -1.
Answer:
Given line can be written in symmetrical form as
\(\frac{x-2}{2}\) = \(\frac{2 y-5}{-3}\) = \(\frac{x+1}{0}\)
⇒ \(\frac{x-2}{2}\) = \(\frac{y-5 / 2}{-3 / 2}\) = \(\frac{z+1}{0}\)
⇒ \(\frac{x-2}{4}\) = \(\frac{y-5 / 2}{-3}\) = \(\frac{z+1}{0}\)
∴ Direction ratios of given line be < 4, -3, 0 >
∴ direction cosines of given line are; < \(\frac{4}{\sqrt{4^2+(-3)^2+0^2}}\), \(\frac{-3}{\sqrt{4^2+(-3)^2+0^2}}\), 0 >
i.e., < \(\frac{4}{5}\), \(\frac{-3}{5}\), 0 >

Question 14.
Find the equations of a line through A(1 ,-1, 5) and parallel to the line
\(\frac{x-2}{3}\) = \(\frac{y-5}{-2}\), z = -1
Answer:
Given eqn. of line can be written in symmetrical form as
\(\frac{x-2}{3}\) = \(\frac{y-5}{-2}\) = \(\frac{z+1}{0}\)
∴ direction ratios of line (1) are < 3, -2, 0 >
Thus direction ratios of the line || to line (1) are proportional to < 3, -2, 0 >
Thus, the required eqn. of line through A(1, -1, 5) and parallel to line (1) is given by
\(\frac{x-1}{3}\) = \(\frac{y+1}{-2}\) = \(\frac{z-5}{0}\)

Question 15.
The equation of a line is \(\frac{2 x-5}{4}\) = \(\frac{y+4}{3}\) = \(\frac{6-z}{6}\).
Find the direction cosines of a line parallel to the line.
Answer:
Given eqn. of line be \(\frac{2 x-5}{4}\) = \(\frac{y+4}{3}\) = \(\frac{6-z}{6}\)
eqn. (1) can be written in symmetrical form as :
\(\frac{2\left(x-\frac{5}{2}\right)}{4}\) = \(\frac{y+4}{3}\) = \(\frac{-(z-6)}{6}\)
i.e., \(\frac{x-\frac{5}{2}}{2}\) = \(\frac{y+4}{3}\) = \(\frac{z-6}{-6}\)
∴ direction ratios of line (1) are < 2, 3, -6 >
Thus the direction cosines of line (1) are ;
< \(\frac{2}{\sqrt{2^2+3^2+(-6)^2}}\), \(\frac{3}{\sqrt{2^2+3^2+(-6)^2}}\), \(\frac{-6}{\sqrt{2^2+3^2+(-6)^2}}\) >
i.e., < \(\frac{2}{7}\), \(\frac{3}{7}\), \(\frac{-6}{7}\) >

ICSE Solutions for Class 8 History and Civics – Traders to Rulers (II)

September 6, 2024

ICSE Solutions for Class 8 History and Civics – Traders to Rulers (II)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

ICSE Solutions Class 8 History & Civics Geography Biology Chemistry Physics Maths

Time To Learn
I. Fill in the blanks:

The British had to face the challenge of the rulers of Hyderabad and Mysore, Marathas after the conquest of Bengal.
The new Nizam of Hyderabad, Salabat Jung initially had cordial relations with the English.
Lord Wellesley made the Nizam of Hyderabad enter the Subsidiary Alliance system.
The English fought four wars with rulers of Mysore, known as Anglo-Mysore wars.
It took the British 43 years to defeat the Marathas in three Anglo-Maratha wars.
Lord Dalhousie annexed Awadh on basis of misgovernance.

II. Match the contents of Column A and Column B:
Column A Column B

Answer:

III. State whether the following statements are True or False:

The Subsidiary Alliance system was started by Lord Cornwallis.
False.
The capture of the port of Mahe by the British led to the Second Anglo-Mysore War in 1780
True.
Tipu won a great battle against his three enemies, the English, Nizam and the Marathas in 1792
False.
The war of succession among the Marathas in 1772 gave a chance to the English to interfere in Maratha politics.
True.
The Second Anglo-Maratha war was fought when Lord Wellesley was the Governor General of India.
True.
Jhansi and Nagpur were annexed under the pretext of misgovernance.
False.

IV. Answer the following questions:

Question 1.
What were the results of the First Anglo-Mysore War?
Answer:
The Marathas, the Nizam and the British marched towards Mysore. But Hyder won over the Marathas and the Nizam. The British were isolated. Haider won after some setbacks. He forced the British to sign a peace treaty in 1769, according to which both Haider and the British agreed to help each other in case of any attack by a third power.

Question 2.
How did the British force win the Third Anglo-Mysore War?
Answer:
In the Third Anglo-Mysore War (1790-1792) the company won the support of two other powers – the Nizam of Hyderabad and the Marathas. Tipu Sultan was completely defeated by these three joint powers and was forced to sign the Treaty of Seringapattam in 1792. According to the terms of this treaty Tipu had to surrender half of his territories which were to be shared between the British, the Nizam and the Marathas. Tipu also had to pay a war indemnity of ruppees 3.3 crore. Two of his sons were taken hostages.

Question 3.
What were the causes of the Fourth Anglo-Mysore War? What were its results?
Answer:
The Treaty of Seringapattam was short lived. Lord Wellesley, the Governor General wanted the ruler of Mysore, Tipu to accept the Subsidiary Alliance but Tipu refused to do so. Tipu sought French help. The British feared that the French might land in support of Tipu. So in 1799, the British went to war against Tipu. The Nizam joined the British but the Marathas remained neutral and the French support never came. Tipu was killed in the battle on 4th May 1799.

Question 4.
Name the four different centres of Maratha power.
Answer:
The four different centres of Maratha power were Bhonsle at Nagpur, Scindia at Gwalior, Peshwa at Pune and Holkar at Indore.

Question 5.
What were the causes and results of the First Anglo-Maratha War? When was it fought?
Answer:
In 1772, a war of succession followed for the seat of Peshwaship after the death of Peshwa Madhav Rao I between Raghoba and Madhav Rao II. The English sided with Raghoba. Marathas led by Nana Phadnavis supported Madhav Rao II. An indecisive war broke out and continued for seven years. It ended in 1782 with the signing of the Treaty of Salbai. Under this Madhav Rao II was recognised as Peshwa and the Marathas had to cede Salsette. Peace prevailed between the Marathas and the British for 20 years.

Question 6.
Who was the Governor-General of India during the Second Anglo-Maratha War? What was the significance of this battle?
Answer:
The Second Anglo-Maratha War was fought when Lord Wellesley was the Governor General of India. In 1803 Baji Rao II signed with the English East India Company a Subsidiary Alliance known as Treaty of Bassein. As a result the English installed Baji Rao II at Pune and helped to drive out the Holkars. The Maratha chiefs Scindia and Bhonsle refused to accept the system of Subsidiary Alliance and declared war against the British. But the British defeated the combined forces of Scindia and Bhonsle. They were forced to enter into the Subsidiary Alliance with the English. They ceded the territories of Ahmednagar, Broach, Cuttack and Balasore. This war gave a blow to the power and prestige of the Marathas.

Question 7.
Why was the Third Anglo-Maratha War fought? What were its results?
Answer:
The Third Anglo Maratha War was fought in (AD 1817-1818) because the Maratha chiefs were feeling humiliated after signing the Subsidiary Alliance with the British. Peshwa Baji Rao II began to make plans to unite the Marathas against the British. This war was fought during the Governor Generalship of Marquess Hastings. When Lord Hastings became aware of Baji Rao’s plans he forced him to sign the Treaty of Pune in 1817. According to it, Konkan was ceded and Baji Rao II renounced Maratha leadership. Scindia was forced to sign the Treaty of Gwalior and provide help to the British against the Pindaris.

The Marathas made their last attempt to regain their old prestige and independence. They declared war against English. Baji Rao II raised a huge army and attacked the British Residency at Poona in 1817. The war continued for about two years. Finally the Marathas were defeated and large parts of their territories were annexed by the British.

The Third Anglo – Maratha War led to the abolition of Peshwa’s hereditary office. The Maratha dream of building a Hindu empire was completely shattered and the British emerged as the unchallenged power in India.

Question 8.
State very briefly why the Marathas failed against the British.
Answer:
The Marathas were inferior to the English in material sources, military organisation, diplomacy and leadership. The Maratha state was despotic and feudal. It was a loose confederation of different powers – the Holkar, Bhonsle, Scindhia – headed by the Peshwa. The Marathas were careless about military intelligence whereas the British had superior espionage system. The Marathas did not possess any national sentiment. The internal jealously and selfish treachery among them triumphed over national interest. The Marathas lacked men of talent and leadership in later years. They were in disarray in all aspects.

Question 9.
Explain the doctrine of Lapse. Name the other methods used by Lord Dalhousie to expand the British power in India.
Answer:
According to Doctrine of Lapse if the ruler of a dependent state died without leaving a natural heir the state would automatically pass over to the British. The Doctrine of Lapse did not recognise adopted children as rightful heirs. Satara, Jaitpur, Baghat, Udaipur, Sambhalpur, Jhansi and Nagpur were annexed under the Doctrine of Lapse. Punjab was annexed through war. Awadh was annexed by Dalhousie on the basis of misgovemance.

V. Word Hunt

Given below is a grid. Look for names of places which Dalhousie annexed. You have to look vertically, horizontally and diagonally. In total there are 8 places. After you have found out these places, write down why these were annexed.
Answer:

VI. Picture Study –

This is a picture depicting the first Anglo-Sikh War.
ICSE Solutions for Class 8 History and Civics - Traders to Rulers (II) 5

Question 1.
What was the political condition of Punjab after the death of Maharaja Ranjit Singh?
Answer:
After the death of Maharaja Ranjit Singh there was a state of political instability in the Punjab. For six years there was conflict among various claimants to the throne.

Question 2.
When was the first Anglo-Sikh war fought?
Answer:
First Anglo Sikh War was fought in December 1845.

Question 3.
What were the causes and results of the First Anglo-Sikh War?
Answer:
The mother of the minor son of Ranjit Singh, Maharani Jindan Kaur instigated the Khalsa against the British. This led to the first Anglo-Sikh War.
The Sikhs were defeated and had to sign the Treaty of Lahore in 1846. The Sikhs lost territory and had to keep British Resident.

Additional Questions

Rise of British Power in Bengal

A. Fill the in the blanks:

Madras, Bombay and Calcutta became the headquarters of the British settlements in the southern, western and eastern regions, respectively.
In 1717, the Mughal emperor granted the United East India Company the right to carry on duty-free trade in Bengal, Bihar and Orissa.
The French East India Company was established in 1664 C.E.
The British and the French fought the Carnatic Wars in India to establish their monopoly in trade.
Bengal in the 18th century was the richest and the most fertile province in India.
In 1757 Robert Clive recovered Calcutta which had been captured by Siraj-ud-Daulah in 1756.
Mir Jafar was deposed because he was unable to meet the demands of the British.
In 1765, Awadh was returned to Shuja-ud-Daulah but Kora and Allahabad were taken away and given to Shah Alam II.
Shah Alam II granted the Company the Diwani of Bengal, Bihar and Orissa in 1765.
Warren Hastings deposed and pensioned off the Nawab of Bengal and brought Bengal under the direct, and complete control of the Company.

B. Match the following:

Answer:

C. Choose the correct answer:

Question 1.
The English East India Company was established in the year 1600/1700/1800 CE.
Answer:
The English East India Company was established in the year 1600.

Question 2.
The English East India Company set up its first factory in Surat/Agra/Broach.
Answer:
The English East India Company set up its first factory in Surat.

Question 3.
The largest and the most prosperous European settlement in Bengal was the British settlement at Calcutta/Burdwan/ Murshidabad.
Answer:
The largest and the most prosperous European settlement in Bengal was the British settlement at Calcutta.

Question 4.
Alivardi Khan was succeeded by Mir Qasim/Siraj-ud- Daulah/Shuja-ud-Daula.
Answer:
Alivardi Khan was succeeded by Siraj-ud-Daulah.

Question 5.
Robert Clive hatched a plot with Mir Jafar/Mir Qasim/ Alivardi Khan to replace Siraj-ud-Daulah.
Answer:
Robert Clive hatched a plot with Mir Jafar to replace Siraj- ud-Daulah.

Question 6.
The Battle of Plassey was fought in 1757/1764/1772.
Answer:
The Battle of Plassey was fought in 1757.

Question 7.
The Dual Government in Bengal was introduced by Robert Clive/Warren Hastings/Lord Cornwallis.
Answer:
The Dual Government in Bengal was introduced by Robert Clive.

D. State whether the following are true or false:

The Carnatic Wars were fought between the British and the French.
True.
The employees of the Company were entitled to both private trade as well as duty-free trade.
False.
Correct: The employees of the Company were permitted to carry on private trade but they were not entitled to the Company’s privilege of duty-free trade.
The English East India Company was given the right to issue passes or dastaks for the free movement of their goods.
True.
The British army was defeated in the Battle of Buxar.
False
Correct : The British army won in the Battle of Buxar.
The Treaty of Allahabad was signed between the British Company and Mir Qasim.
False.
Correct: The Treaty of Allahabad was signed between the British and Shuja-ud-Daulah.
Warren Hastings laid the foundation of an organized system of government in Bengal.
True.

Answer the following questions in one or two words/ sentences:

Question 1.
Who granted the English East India Company the exclusive right to trade with the East ?
Answer:
The Queen of England, Elizabeth I, granted the Company the exclusive right to trade with the East.

Question 2.
Name the British trading settlements in
(a) Madras (b)Calcutta
Answer:
(a)
Madras was given to the British by a local ruler. They established a trading settlement which they fortified and named Fort St. George.
(b)
Calcutta In 1690 CE, a British trading settlement was established and fortified in Calcutta. It was named Fort William.

Question 3.
What important right did the Mughal emperor Farrukhsiyar grant the English East India Company ?
Answer:
In 1717, the Mughal Emperor, Farrukhsiyar, granted the Company the right to carry on duty-free trade in Bengal, Bihar and Orissa (now Odisha). The Company made enormous profits.

Question 4.
Why were European traders attracted to the Bengal province inthe 18th century?
Answer:
Bengal was the richest and the most fertile province in India at that time. It was known as the paradise of the earth and this province attracted traders from many European countries.

Question 5.
What privileges did the farman of 1717 confer on the English East India Company?
Answer:
By the ‘Farman” of 1717, the English East India company was granted the right to carry on duty-free trade in Bengal. They were allowed to export and import goods from and to Bengal without paying any taxes to the government. They were given the right to issue passes or dastaks for the free movement of their goods.

Question 6.
Why did the farman of 1717 become a bone of contention between the nawabs of Bengal and the British Company?
Answer:
The Nawabs of Bengal wanted British company to pay taxes on trade like the Indians. But the Britishers refused to do so. This created conflict between both the parties and resulted in wars between both of them.

Question 7.
Why did the British fortify their trade settlement in Calcutta?
Answer:
Siraj-ud-Daulah ordered the British to pay taxes to him like all other Indian merchants. The British refused to do so. This angered the young nawab. In anticipation of a war with the French, who had a trading settlement in Chandemagore, the British began to fortify Calcutta.

Question 8.
Why did Siraj-ud-Daulah attack Calcutta in 1756?
Answer:
When the Britishers tried to fortify Calcutta, Siraj-ud-Daulah ordered both the British and the French to dismantle their fortifycations and not to fight private wars on his territory. The French agreed but the British refused. This enraged the Nawab and he attacked Calcutta with a large army and captured Fort William.

Question 9.
What important trading right was granted to the English East India Company after its victory in the Battle of Plassey?
Answer:

The English East India Company was granted the undisputed right to free trade in Bengal, Bihar and Orissa.
The Company was given the zamindari of the 24 Parganas.

Question 10.
State the political significance of the Battle of Buxar.
Answer:

It gave them political influence and control over Awadh and the Mughal emperor.
It laid the foundation of British rule in India.

Question 11.
Name the Indian signatories of the Treaty of Allahabad.
Answer:
Shuja-ud-Daulah and Shah Alam II

Question 12.
In which year was the Dual Government abolished and by whom?
Answer:
In 1772, Warren Hastings abolished the Dual Government.

F.Answer the following questions briefly:

Question 1.
The Battle of Plassey was a major turning point in the history of India. In this context answer the following questions:
(a) Give an account of the events leading from the conspiracy to replace Siraj-ud-Daulah to his eventual defeat in the Battle of Plassey.
(b)State the results of the Battle of Plassey.
(c) Why is this battle considered a major turning point in the history of India?
Answer:
(a)
A major part of the nawab’s army under the command of Mir Jafar did not take any part in the battle. Realizing that he had been betrayed, the nawab fled from the battlefield. He was captured and put to death.
(b)
Results of Battle of Plasey:

The English East India company was granted the undisputed right to free trade in Bengal, Bihar and Orissa.
The Company w’as given the zamindari of the 24 parganas.
Mir Jafar paid the Company and its officials over 300 lakh rupees.

(c)
The Battle of Plassey was a major turning point in the histoiy of India.

It paved the way for the establishment of British rule in Bengal and, eventually, the rest of India.
It transformed a trading company into a political power.

It provided the British the vast resources of Bengal, which helped them to win the Third camatic war and other expeditions in India.

Question 2.
Mir Qasim was a competent ruler, determined to free himself from foreign control. In this context answer the following:
(a) What steps did Mir Qasim take to strengthen his position? Why did he abolish all duties on internal trade?
(b) Trace the events from the outbreak of war (1763) between Mir Qasim and the British up to the Battle of Buxar in 1764.
(c) Explain the importance of the Battle of Buxar.
Answer:
(a)
To strengthen his position, Mir Qasim improved the financial position of Bengal and raised a modern, disciplined and well-equipped army trained by the Europeans.The employers of the company misused their trade privileges. They sold their duty-free trade points to Indian merchants who also used them to carry on duty-free trade. This deprived the Nawab fo large revenues. To put an end to the corrupt practices of the British, Mir Qasim abolished all duties on internal trade.

(b)
The employees of the Company misused their trade privileges. They sold their duty-free trade permits to Indian merchants who also used them to carry on duty-free trade. This deprived the nawab of large revenues and was unfair to those local merchants who had to pay heavy duties.
To put an end to the corrupt practices of the British, Mir Qasim abolished all duties on internal trade. This made the British furious. They refused to accept an equal status with the Indian merchants.
In 1763, war broke out between Mir Qasim and the British. The nawab was defeated. Mir Jafar was reinstated on the throne.
Mir Qasim was determined to recover his throne. He escaped to Awadh, where he formed an alliance with Shuja-ud- Daulah, the nawab of Awadh, and the Mughal emperor, Shah Alam II.
The combined forces of the three allies clashed with the Company’s troops at Buxar in 1764, and were decisively defeated by the British.

(c)

The victory of the British in the Battle of Buxar firmly established them as masters of Bengal, Bihar and Orissa.
It gave them political influence and control over Awadh and the Mughal emperor.
It laid the foundation of British rule in India.
At this time, Robert Clive returned to India as the governor of Bengal.

Question 3.
With reference to the Treaty of Allahabad and its impact,answer the following:
(a) Mention the terms of agreement between Robert Clive and Shuja-ud-Daulah in this treaty.
(b) Explain how the treaty between Robert Clive and the Mughal emperor Shah Alam II legalized the English East India’s Company’s control over Bengal.
(c) Give an account of the events that followed the death of Mir Jafar, leading to the establishment of the Company as the real ruler of Bengal.
Answer:
(a)

Awadh was returned to Shuja-ud-Daulah. However, the two districts of Kora and Allahabad were taken away from the Nawab.
The nawab of Awadh had to pay a war indemnity of 50 lakh rupees to the Company.
The British agreed to defend the nawab of Awadh against his enemies. The nawab would have to pay for the cost of the British troops. Awadh became a buffer state between the British possessions in Bengal and the Marathas.

(b)
The British gave Shah Alam II the districts of Kora and Allahabad and an annual pension of 26 lakh rupees. In return, the emperor (the nominal head of the Mughal empire) granted the Company the Diwani of Bengal, Bihar and Orissa, e., the right to collect revenue from these provinces and judge civil cases. The Company’s control over Bengal was made legal.
(c)
After Mir Jafar’s death in 1765, his son was made-the nawab of Bengal. He had to sign a treaty with Clive,according to which he had to disband most of his army. He also had to transfer the ‘Nizamat’ powers (general administration and criminal justice) to a deputy nawab appointed by the British. The deputy nawab could not be dismissed by the nawab. The nawab was given an allowance of 53 lakh rupees which was subsequently reduced. Thus, the English East India Company became the real ruler of Bengal from 1765.

Question 4.
With reference to the establishment of Dual Government in Bengal (1765-72) answer the following:

(a) Why was the government introduced in Bengal by Robert Clive referred to as ‘Dual Government’?
(b) What were the advantages and disadvantages of this system for the Company and the nawab respectively?
(c) The evils of the Dual Government led to the collapse of the administration and the economy. Explain.
Answer:
(a)
Robert Clive introduced Dual Government in Bengal in 1765. Bengal now had two masters-the nawab and the Company. The Nawab was responsible for general administration, maintenance of law and order and justice (i.e., criminal cases). The Company had military power and the right to collect and use the revenue of Bengal. This arrangement was known as Dual Government.
(b)
The Company enjoyed power without any responsibilities. The nawab, on the other hand, was burdened with the responsibility of administration without the resources necessary for running it efficiently e., responsibility without power.The revenue was collected by Indian officials appointed by the Company. The greed, corruption and oppression of these officials reduced the peasants to conditions of utter misery. The Company took no interest in the welfare of the people.
(c)
The conditions of the people worsened when Bengal was hit by a terrible famine in which one third of the population perished. Nobody cared, neither the Company nor the nawab, who in any case had neither the authority nor the resources to lessen the miseries of the people. The Company, through its power to nominate the deputy nawab, only interfered in the general administration without assuming any responsibility.The evils of the Dual Government began to manifest themselves. The administration and economy collapsed. In 1772, the Court of Directors of the Company appointed Warren Hastings as the Governor of Bengal. In 1773, by the Regulating Act, he was made the Governor General of British territories in India. The Governor General was now the most important functionary of the East India Company.

G Picture Study.

This picture portrays a momentous event in 1765, involving a British Governor and a Mughal emperor wherein the Mughal emperor is conveying the grant of the Diwani to the governor.
ICSE Solutions for Class 8 History and Civics - Traders to Rulers (II) 8

Identify the Mughal emperor and the British governor
Ans. Mughal Emperor – Shuja-ud-Daulah, British Governor – Robert Clive
What is the significance of this grant of the Diwani?
Ans. The significance of granting the company the Diwani i.e. the right to collect revenue from these provinces and judge civil cases.
Give a brief account of the battle that preceded this event. When did it take place?
Ans. Battle of Buxar in 1764.
What is the importance of this battle?
Ans. Mir Qasim was defeated and Mir Jafar was reinstated on the throne.

Additional Questions

Expansion of British Power in India
EXERCISES
A. Fill the in the blanks:

The five centres of Maratha power were:
Ans.
The five centres of Maratha power were:
- The Sindhias of Gwalior
- The Holkars of Indore
- The Bhonsle family of Nagpur
- The Gaekwad family of Baroda
- The Peshwa of Poona (Pune)
The Governors-General during the First and Second Anglo- Maratha Wars were Warren Hastings and Lord Wellesley, respectively.
The Second Anglo-Maratha War was a severe blow to the Sindhia and Bhonsle of the Marathas.
Lord Dalhousie adopted three methods to make the British the paramount power in India.
Jhansi was annexed by Dalhousie on the basis of the Doctrine of Lapse.
Dalhousie annexed Awadh on the grounds of Maladministration or Misgovernment.

B. Match the following:

Answer:

C. Choose the correct answer:

The First Anglo-Maratha War was fought during the Governor-Generalship of Lord Cornwallis/Lord Wellesley/ Warren Hastings.
Ans. The First Anglo-Maratha War was fought during the Governor- Generalship of Warren Hastings.
The Maratha chiefs were united under the leadership of Nana Phadnavis/Baji Rao II/Madhav Rao II during the First Anglo-Maratha War.
Ans. The Maratha chiefs were united under the leadership of Nana Phadnavis during the First Anglo-Maratha War.
After the Third Anglo-Maratha War the British placed a descendant of Shivaji on the throne of Nagpur/Satara/Jhansi.
Ans. After the Third Anglo-Maratha War the British placed a descendant of Shivaji on the throne of Satara.
Punjab was annexed by Lord Minto/Lord Dalhousie/Lord Wellesley in 1849.
Ans. Punjab was annexed by Lord Dalhousie in 1849.
The widespread resentment against annexations expressed itself in the Revolt of 1849/1861/1857.
Ans. The widespread resentment against annexations expressed itself in the Revolt of 1857

D. State whether the following are true or false:

After Hyder Ali’s death, his son Tipu Sultan continued the Anglo-Maratha Wars.
True.
The Subsidiary Alliance system was used by the British to bring Indian rulers under British control without any war.
True.
The Subsidiary Alliance proved very advantageous for the Indians.
False.
Correct : The Subsidiary Alliance proved very advantageous for the British.
The adopted son of Peshwa Baji Rao II was denied the pension that his father used to get from the British.
True.
By 1856, the English East India Company had brought the whole of India under its control.
True.

E. Answer the following questions in one or two words/ sentences:

Question 1.
What was the main objective of the Subsidiary Alliance system?
Answer:
The subsidiary Alliance system was a method perfected by Lord Wellesley to subjugate Indian powers without the cost and bother/ trouble of war.Any Indian ruler whose security was threatened was encouraged to seek help from and enter into an alliance with the British, who promised to protect the ruler from external attacks and internal revolts. The Indian ruler had to accept certain terms and conditions. ‘

Question 2.
What happened when the administration of a subsidiary state collapsed?
Answer:
When the administration collapsed, the British used it as an excuse for annexing the kingdom on grounds of misgovemment.

Question 3.
Why did Peshwa Baji Rao II sign the Subsidiary Alliance?
Answer:
In 1802, Peshwa baji Rao II, supported by Sindhia, was defeated by Holkar. Baji Rao II fled to Bassein where he signed the Subsidiary Treaty as a price for British protection and support. He was escorted back to Poona by British soldiers.

Question 4.
Why was Wellesley recalled from India during the Second Anglo-Maratha War?
Answer:
Wellesley was recalled from India because the government in England was unhappy with the enormous expenditure involved in Wellesley’s policy of wars and expansion.

Question 5.
How did the Subsidiary Alliance impact the
(a) economy
(b) administration in Awadh?
Answer:
(a)
Awadh had signed a Subsidiary Alliance with Wellesley. He had to protect the nawab from external invasions and internal rebellions. It made the nawabs complacent and unconcerned about the affairs of the state. The payement of annual subsidies to the Company exhausted the state treasury
(b)
When the administration was on the verge of collapse, Dalhousie struck. He brought charges of misgovemment or maladministration against the nawab. On those grounds he deposed the nawab and annexed Awadh in 1856.

F. Answer the following questions briefly:

Question 1.
With reference to the Subsidiary Alliance System perfected by Lord Wellesley, answer the following:
(a) Why did Indian rulers sign the Subsidiary Alliance? State two important military terms and conditions imposed on the Indian rulers by this treaty.
(b) The Subsidiary Alliance proved very beneficial for the British. Explain.
(c) Discuss the disastrous effects of the Subsidiary Alliance on the Indian states.
Answer:
(a)
Indian ruler whose security was threatened was encouraged to seek help from and enter into an alliance with the British,who promised to protect the ruler from external attacks and internal revolts. The Indian rulers had to accept certain terms and condition in return for
British protection under the Subsidiary Alliance like:

British troops would be permanently placed in the territory of the Subsidiary state.
The ruler would have to pay for the maintenance of the troops.
He could not employ any europeans in his service or dismiss those who were already there.
He could not form an alliance with any other power or declare war against any power without the permission of the British.
He would acknowledge the British Company as the paramount power.

(b)
The Subsidiary Alliance proved very advantageous for the British like.

The British maintained large armies at the expense of the Indian rulers.
The British acquired valuable territories as subsidiary payment. This led to the expansion of the
British empire in India and an increase in its resources.
The influence of European rivals, especially the French, was excluded from the courts of the Indian rulers.
The British controlled the foreign policy of the Subsidiary states.

(c)
The Subsidiary Alliance System had great effect on the subsidiary state like:

British troops would be permanently placed in the territory of the Subsidiary state.
The ruler would have to pay for the maintenance of the troops.
He could not employ any Europeans in his service or dismiss those who were already there.
He could not form an alliance with any other power or declare war against any power without the permission of the British. ‘
He would acknowledge the British Company as the paramount power.ses or dastaks for the free movement of their goods.

Question 2.
Dalhousie was a great expansionist and adopted a number of methods to build an all-India empire. In this context, answer the following questions:
(a) Mention the various methods adopted by Dalhousie and the territories annexed on the basis of these methods.
(b) Under what circumstances did a subordinate state automatically ‘lapse’ and pass into the hands of the British? How did the rulers react to this policy?
(c) Why did Nana Saheb become one of the leaders of the Revolt of 1857?
Answer:
(a)
Dalhousie adopted a number of methods to give the final touches to the work of empire-building in India. The methods he adopted were:

War-Punjab.
Doctrine of Lapse-Satara, Jhansi and Nagpur.
Annexation on grounds of maladministration-Awadh.

(b)
Lord Dalhousie brought several subordinate states under the direct rule of the Company by annexing them on the basis of the Doctrine of Lapse. According to the Doctrine of Lapse, all subordinate states (subsidiary states and states created by the British) where the rulers died without a natural male heir would automatically ‘lapse’, e., pass into the hands of the British. Rulers without heirs could not adopt sons, according to the age-old Hindu and Islamic traditions, without the permission of the Company.
(c)
Nana Saheb, the adopted son of Peshwa Baji Rao II, inherited his father’s personal property but was not given the pension that had been paid to his father. Nana Saheb became one of the important leaders of the Revolt of 1857.

Question 3.
With reference to the annexation of Awadh discuss:
(a) The effects of the Subsidiary Alliance on the administration in Awadh
(b)The political scenario in India by 1856
(c) Any three factors responsible for the success of the British over their Indian rivals
Answer:
(a)
Awadh had signed Subsidiary Alliance with Wellesley. He had to protect the nawab from external invasions and internal rebellions. It made the nawabs complacent and unconcerned about the affairs of the state. The payement of annual subsidies to the Company exhausted the state treasury. When the administration was on the verge of collapse, Dalhousie struck. He brought charges of misgovemment or maladministration against the nawab. On these grounds he deposed the nawab and annexed Awadh in 1856.
(b)
By 1856 the English East India Company had brought the whole of India under its control of the British. The British had eliminated all their rivals and established themselves as the paramount power in India.
(c)
The factors responsible for the success of the British are:

Lack of unity among Indian rulers.
Lack of organized and efficient administration in Indian states.
Superior military resources of the company.
Superior economic resources of the company.
Naval supremacy of the British.

G Picture study:

Name the Governor General.
Ans. Lord Dalhousie
What was the method adopted by him to subjugate the Indian territories ?
Ans. Governor General Dalhousie was a great imperialist who annexed several states on the basis of the Doctrine of Lapse.
Mention the terms and conditions under this method.
Ans. According to the Doctrine of Lapse, all subordinate states
(subsidiary states and states created by the British) where the rulers died without a natural male heir would automatically ‘lapse’, i.e. pass into the hands of the British. Rulers without heirs could not adopt sons, according to the age-old Hindu and Islamic traditions, without the permission of the Company.
Did this method have any advantage for the Indian rulers? Why ?
Ans. The annexation of these states caused widespread resentment among the Indian rulers and became a potent factor responsible for the outbreak of the Revolt of 1857.