OP Malhotra Solutions – ICSE Solutions

OP Malhotra Class 10 Solutions – S Chand Class 10 Maths Solutions ICSE

October 15, 2024

Class 10 ICSE Maths Solutions S Chand – OP Malhotra Class 10 ICSE Solutions

Unit I Commercial Mathematics

OP Malhotra Class 10 Solutions Chapter 1 GST (Goods and Service Tax)

Chapter 1 GST (Goods and Service Tax) Ex 1

S Chand Class 10 Maths Solutions ICSE Chapter 2 Banking

Chapter 2 Banking Ex 2

Class 10 ICSE Maths Solutions S Chand Chapter 3 Shares and Dividends

Unit II Algebra

ICSE Class 10 Maths Solutions S Chand Chapter 4 Linear Inequations in One Variable

Chapter 4 Linear Inequations in One Variable Ex 4

ICSE Class 10 Maths Solutions OP Malhotra Chapter 5 Quadratic Equations

ICSE Class 10 S Chand Maths Solution Chapter 6 Ratio and Proportion

ICSE S Chand Maths Class 10 Solutions Chapter 7 Factor Theorem – Factorization

Chapter 7 Factor Theorem – Factorization Ex 7

S Chand Class 10 ICSE Maths Solutions Chapter 8 Matrices

OP Malhotra Class 10 Maths Solutions Chapter 9 Arithmetic and Geometric Progression

Class 10 ICSE Maths S Chand Solutions Chapter 10 Reflection

S Chand Maths Class 10 ICSE Solutions Pdf Chapter 11 Coordinate Geometry

Unit III Geometry

ICSE Class 10 Maths S Chand Solutions Chapter 12 Similar Triangles

S Chand ICSE Class 10 Maths Solutions Chapter 13 Loci

‘S Chand Maths Class 10 Solutions Pdf ICSE Chapter 14 Circle

Unit IV Mensuration

OP Malhotra Class 10 ICSE Solutions Chapter 15 Three Dimensional Solids

Unit V Trigonometry

S Chand ICSE Maths Class 10 Solutions Chapter 16 Trigonometrical Identities and Tables

S Chand Maths Class 10 Solutions ICSE Pdf Chapter 17 Heights and Distances

Chapter 17 Heights and Distances Ex 17

Unit VI Statistics

ICSE Maths Solutions Class 10 S Chand Chapter 18 Arithmetic Mean, Median, Mode and Quartiles

S Chand Class 10 Maths ICSE Solutions Chapter 19 Histogram and Ogive

Class 10 ICSE Maths Solutions OP Malhotra Chapter 20 Probability

Also Read:

ML Aggarwal Class 10 Solutions PDF

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 = \(\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\)
= \(\frac{(2 \hat{i}-\hat{j}+\hat{k}) (3 \hat{i}-4 \hat{j}-\hat{k})}{\sqrt{2^2+(-1)^2+1^2} \sqrt{3^2+(-4)^2+(-1)^2}}\)
= \(\frac{2(3)-1(-4)+1(-1)}{\sqrt{4+1+1} \sqrt{9+16+1}}\)
= \(\frac{9}{\sqrt{6} \sqrt{26}}\)

OP Malhotra Class 12 Maths Solutions Chapter 12 Maxima and Minima Ex 12(a)

September 9, 2024

Practicing OP Malhotra Maths Class 12 Solutions Chapter 12 Maxima and Minima Ex 12(a) is the ultimate need for students who intend to score good marks in examinations.

S Chand Class 12 ICSE Maths Solutions Chapter 12 Maxima and Minima Ex 12(a)

Question 1.
Find the turning values of the following functions, distinguishing in each case whether the value is a maximum, minimum, or inflexional:
(i) 4x³ + 19x² – 14x + 3
(ii) 2x³ + 3x² – 12x + 7
(iii) 3x⁴ + 8x³+ 6x²
(iv) x³ – 2x² – 4x – 1
Solution:
(i) Let y = 4x³ + 19x² – 14x + 3
∴ \(\frac { dy }{ dx }\) = (12x² + 38x – 14)
∴ \(\frac{d^2 y}{d x^2}\) = 24x + 38
for maxima/minima, \(\frac { dy }{ dx }\) = 0
⇒ 12x² + 38x – 14 = 0
⇒ \(x=\frac{-19 \pm \sqrt{361+168}}{12}\)
= \(\frac{-19 \pm \sqrt{529}}{12}\)
⇒ x = \(\frac{-19 \pm 23}{12}\) = \(\frac { 4 }{ 12 }\), \(\frac { -42 }{ 12 }\)
i.e. x = \(\frac { 1 }{ 3 }\), \(\frac { -7 }{ 2 }\)
when x = \(\frac { 1 }{ 3 }\); \(\frac{d^2 y}{d x^2}\) = 24 × \(\frac { 1 }{ 3 }\) + 38 = 46 > 0
∴ x = \(\frac { 1 }{ 3 }\) be a point of minima.
∴ min value of y = 4\(\left(\frac{1}{3}\right)^3\) + 19\(\left(\frac{1}{3}\right)^2\) – \(\frac { 14 }{ 3 }\) + 3
= \(\frac { 4 }{ 27 }\) + \(\frac { 19 }{ 9 }\) – \(\frac { 14 }{ 3 }\) + 3
= \(\frac{4+57-126+81}{27}\) = \(\frac{16}{27}\)
when x = –\(\frac{7}{2}\); \(\frac{d^2 y}{d x^2}\) = 24\(\left(-\frac{7}{2}\right)\) + 38
= – 84 + 38 = -46 < 0
∴ x = –\(\frac { 7 }{ 2 }\) be a point of maxima
& maximum value of y
= 4\(\left(-\frac{7}{2}\right)^3\) + 19\(\left(-\frac{7}{2}\right)^2\) – 14\(\left(-\frac{7}{2}\right)\) + 3
= \(\frac{-343}{2}\) + \(\frac{931}{4}\) + 49 + 3
= \(\frac{-686+931+208}{4}\) = \(\frac{453}{4}\) = 113\(\frac{1}{4}\)

(ii) Let y = 2x³ + 3x² – 12x + 7
∴ \(\frac{d y}{d x}\) = 6x² + 6x – 12
& \(\frac{d^2 y}{d x^2}\) = 12x + 6
for manima/minima, \(\frac{d y}{d x}\) = 0
⇒ 6(x² + x – 2) = 0
⇒ x = 1, – 2
∴ \(\left(\frac{d^2 y}{d x^2}\right)_{x=1}\) = 12 + 6 = 18 > 0
Thus x = 1 be a point of minima
& min value = 2 + 3 – 12 + 7 = 0
& \(\left(\frac{d^2 y}{d x^2}\right)_{x=-2}\) = -24 + 6 = – 18 < 0
∴ x = -2 be a point of maxima
& Maximum value
= 2(-2)³ + 3(-2)² – 12(-2) + 7
= -16 + 12 + 24 + 7
= 27

(iii) Let y = 3x⁴ + 8x³ + 6x²
∴ \(\frac{d y}{d x}\) = 12x³ + 24x² + 12x
& \(\frac{d^2 y}{d x^2}\) = 36x² + 48x + 12
for maxima/minma, \(\frac{d y}{d x}\) = 0
⇒ 12x(x² + 2x + 1) = 0
⇒ x (x + 1)² = 0
⇒ x = 0, -1
Now \(\left(\frac{d^2 y}{d x^2}\right)_{x=0}\) =12 > 0
∴ x = 0 be a point of minima.
∴ min value = 0 + 0 + 0 = 0
& \(\left(\frac{d^2 y}{d x^2}\right)_{x=-1}\) = 36 – 48 + 12 = 0
Now, \(\frac{d^3 y}{d x^3}\) = 72x + 48
∴ \(\left(\frac{d^3 y}{d x^3}\right)_{x=-1}\) = -72 + 48 = -24 ≠ 0
∴ x = -1 be a point of inflexion

(iv) Let y = x³ – 2x² – 4x – 1
∴ \(\frac{d y}{d x}\) = 3x² – 4x – 4
& \(\frac{d^2 y}{d x^2}\) = 6x – 4
For maxima/minima, \(\frac{d y}{d x}\) = 0
⇒ 3x² – 4x – 4 = 0
⇒ (x – 2)(3x + 2) = 0
⇒ x = 2, –\(\frac{2}{3}\)
Now, \(\left(\frac{d^2 y}{d x^2}\right)_{x=2}\) = 6 × 2 – 4 = 8 > 0
∴ x = 2 be a point of minima & min value of f(x) = f(2)
= 2³ – 2 × 2² – 4 × 2 – 1
= 8 – 8 – 8 – 1 = – 9
∴ at \(\left(\frac{d^2 y}{d x^2}\right)\)x = –\(\frac{2}{3}\) = 6\(\left(-\frac{2}{3}\right)\) – 4 = -8 < 0
∴ x = –\(\frac{2}{3}\) be a point of maxima
& maximum value = – \(\frac{8}{27}\) – \(\frac{8}{9}\) + \(\frac{8}{3}\) – 1
= \(\frac{-8-24+72-27}{27}\) = \(\frac{13}{27}\)

Question 2.
If V = 2x² (6 – x), where x is (positive, determine the greatest value of V.
Solution:
Given V = 2x²(6x – x)
∴ \(\frac{dV}{dx}\) = 2(12x – 3x²)
& \(\frac{d^2 \mathrm{~V}}{d x^2}\) = 2(12 – 6x)
for maxima/minima, \(\frac{d V}{d x}\) = 0
⇒ 2(12x – 3x²) = 0
⇒ 2x(12 – 3x) = 0
⇒ x = 0, 4
∴ \(\left(\frac{d^2 \mathrm{~V}}{d x^2}\right)_{x=0}\) = 2(12 – 0) = 24 > 0
∴ x = 0 be a point of minima & min value = 0
& \(\left(\frac{d^2 \mathrm{~V}}{d x^2}\right)_{x=4}\) = 2(12 – 24) = -24 < 0
∴ x = 4 be a point of maxima
& greatest value of V = 2 × 4²(6 – 4) = 64

Question 3.
Find the co-ordinates of the turning points on the curve y = x³ – 3x² – 9x + 7, distinguishing between maximum and minimum points.
Solution:
Given y = x³ – 3x² – 9x + 7
∴ \(\frac{dy}{dx}\) = 3x² – 6x – 9x
& \(\frac{d^2 \mathrm{~V}}{d x^2}\) = 6x – 6
for maxima/minima, \(\frac{dV}{dx}\) = 0
⇒ 3(x² – 2x – 3) = 0
⇒ (x + 1)(x – 3) = 0
⇒ x = -1, 3
∴ \(\left(\frac{d^2 y}{d x^2}\right)_{x=-1}\) = – 6 – 6 = – 12 < 0 ∴ x = -1 be a point of mixima & maximum value of y = – 1 – 3 + 9 + 7 = 12 & \(\left(\frac{d^2 y}{d x^2}\right)_{x=3}\) = 18 – 6 = 12 > 0
∴ x = 3 is a point of minima
& minimum value = 27 – 27 – 27 + 7 = -20
thus the turning point are (-1, 12) & (3, – 20).

Question 4.
Find the minimum value of \(\left(x+\frac{4}{x^2}\right)\).
Solution:
Let y = x + \(\frac{4}{x^2}\)
∴ \(\frac{dy}{dx}\) = 1 – \(\frac{8}{x^3}\) & \(\frac{d^2 y}{d x}\) = \(\frac{24}{x^4}\)
for maxima/minima, put \(\frac{dy}{dx}\) = 0
⇒ 1 – \(\frac{8}{x^3}\) = 0 ⇒ x³ = 8
⇒ (x – 2) (x² + 2x + 4) = 0
⇒ x = 2
while the other two values of x are non-real.
∴ \(\left(\frac{d^2 y}{d x^2}\right)_{x=2}\) = \(\frac{24}{2^4}\) = \(\frac{3}{2}\) > 0
∴ x = 2 be a point of minima & minimum value of y = 2 + \(\frac{4}{2^2}\) = 3

Question 5.
If y = \(\frac { 1 }{ 4 }\)x⁴ – \(\frac { 2 }{ 3 }\)x³ + \(\frac { 1 }{ 2 }\)x² + \(\frac { 11 }{ 2 }\), show that the ordinate at the point x = 1 is neither a maximum nor a minimum, though \(\frac{dy}{dx}\) = 0, when x = 1.
Solution:
Given y = \(\frac{x^4}{4}\) – \(\frac{2}{3}\)x³ + \(\frac{1}{2}\)x² + \(\frac{11}{2}\)
\(\frac{dy}{dx}\) = x³ – 2x² + x
& \(\frac{d^2 y}{d x^2}\) = 3x² – 4x + 1
for maxima/minima, we put \(\frac{dy}{dx}\) = 0
⇒ x (x² – 2x – 1) = 0
⇒ x (x – 1)² = 0
⇒ x = 0, 1
∴ \(\left(\frac{d^2 y}{d x^2}\right)_{x=0}\) = 1 > 0 ∴ x = 0 be a point of minima
∴ \(\left(\frac{d^2 y}{d x^2}\right)_{x=1}\) = 3 – 4 + 1 = 0
& \(\frac{d^3 y}{d x^3}\) = 6x – 4
∴ at x = 1, \(\frac{d^3 y}{d x^3}\) = 6 – 4 = 2 ≠ 0
Thus, x = 1, be a point of neither maxima nor minima.

Question 6.
Find the points at which the function f given by f(x) = (x – 2)⁴ (x + 1)³ has
(i) local maxima
(ii) local minima
(iii) point of inflexion
Solution:
Given, f(x) = (x – 2)⁴(x + 1)³ Diff. both sides w.r.t. x; we have
f ‘ (x) = (x – 2)⁴ 3 (x + 1)² + (x + 1)³ 4(x – 2)³
= (x + 1)² (x – 2)³ [3(x – 2)+ 4(x + 1)]
= (x + 1)² (x – 2)³ (7x – 2)
For critical points, f ‘ (x) = 0
⇒ (x + 1)² (x – 2)³ (7x – 2) = 0
⇒ x = -1, 2, \(\frac{2}{7}\)

Case – I : at x = -1
When x slightly < – 1 ⇒ x + 1 < 0
also x < -1 < 2 ⇒ x – 2 < 0 ∴ f ‘ (x) = (+ve) (- ve)(- ve) = + ve When x slightly > – 1
⇒ x + 1 > 0, x – 2 < 0
∴ f ‘ (x) = (+ ve) (- ve) (- ve) = + ve
So f ‘ (x) does not changes its sign as we move from slightly < – 1 to slightly > – 1 .
∴ x = – 1 be a point of neither maxima nor minima.
Hence x = -1 be a point of inflexion.

Case-II : at x = 2
When x slightly < 2
⇒ x – 2 < 0 but 7x – 2 > 0
∴ f ‘ (x) = (+ Ve) (- Ve) (+ Ve) = – Ve
When x slightly > 2
⇒ x – 2 > 0 and 7x – 2 > 0
∴ f ‘ (x) = (+ Ve) (+ Ve) (+ Ve) = + Ve
Thus, f ‘ (x) changes its sign from -ve to +ve as we move from slightly <2 to slightly >2
∴ x = 2 is a point of minima.

Case-III : at x = \(\frac{2}{7}\)
When x slightly < \(\frac{2}{7}\)
⇒ 7x – 2 < 0 and x – 2 < 0 ∴ f ‘ (x) = (+ ve) (- ve) (- ve) = + ve ⇒ 7 x – 2 > 0 and x – 2 < 0
∴ f ‘ (x) = (+ve) (-ve) (+ve) = -ve
Thus, f ‘ (x) changes its sign from + ve to – ve as we move from slightly < \(\frac{2}{7}\) to slightly > \(\frac{2}{7}\).
∴ x = \(\frac{2}{7}\) be a point of local maxima.

Question 7.
Discuss the maxima and minima of the expression \(\frac{6 x^3-45 x^2+108 x+2}{2 x^3-15 x^2+36 x+1}\).
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 12 Maxima and Minima Ex 12(a) Img 1

Question 8.
Find the turning values of the function – x³ + 12x² – 5, distinguishing whether the value is a maximum, minimum or inflexional.
Solution:
Let y = -x³ + 12 x² – 5
∴ \(\frac{dy}{dx}\) = -3x² + 24x
& \(\frac{d^2 y}{d x^2}\) = -6x + 24
for maxima minima, we put \(\frac{dy}{dx}\) = 0
⇒ -3x²+ 24x = 0 ⇒ -3x(x – 8) = 0
⇒ x = 0, 8
∴ \(\left(\frac{d^2 y}{d x^2}\right)_{x=0}\) = 24 > 0
∴ x = 0 be a point of minima & minimum value of y = 0 + 0 – 5 = – 5
& \(\left(\frac{d^2 y}{d x^2}\right)_{x=8}\) = -6 × 8 + 24 = -24 < 0
∴ x = 8 be a point of maxima.
Thus maximum value of y
= -512 + 768 – 5 = 251

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., < \(\frac{2}{7}\), \(\frac{3}{7}\), \(\frac{-6}{7}\) >

OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a)

September 5, 2024

Practicing OP Malhotra Maths Class 12 Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) is the ultimate need for students who intend to score good marks in examinations.

S Chand Class 12 ICSE Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a)

Find by applying L’Hospital’s Rule the following limits :

Question 1.
\(\lim _{x \rightarrow 2} \frac{x^3-8}{x^2-4}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 1

Question 2.
\(\lim _{x \rightarrow 3} \frac{x^3-x^2-18}{x-3}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 2

Question 3.
\(\lim _{x \rightarrow 0} \frac{\tan ^{-1} x}{x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 3

Question 4.
\(\lim _{x \rightarrow 0} \frac{e^{\sin x}-1}{x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 4

Question 5.
\(\lim _{x \rightarrow 1} \frac{x-1}{2 x^2-7 x+5}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 5

Question 6.
\(\lim _{x \rightarrow 0} \frac{(1+x)^5-1}{(1+x)^3-1}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 6

Question 7.
\(\lim _{x \rightarrow \frac{\pi}{2}} \frac{2 x-\pi}{\cos x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 7

Question 8.
\(\lim _{x \rightarrow a} \frac{\sin x-\sin \alpha}{x-\alpha}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 7a

Question 9.
\(\lim _{x \rightarrow 0} \frac{a^x-b^x}{e^x-1}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 8

Question 10.
\(\lim _{\alpha \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos \alpha}{\alpha-\frac{\pi}{4}}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 9

Question 11.
\(\lim _{x \rightarrow a} \frac{(x+2)^{5 / 3}-(a+2)^{5 / 2}}{x-a}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 10

Question 12.
\(\lim _{x \rightarrow 0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 11

Question 13.
If f(1) = 1. f'(1) = 2, then find \(\lim _{x \rightarrow 1} \frac{\sqrt{f(x)}-1}{\sqrt{x}-1}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 12

Question 14.
\(\lim _{x \rightarrow 0} \frac{2^x-1}{\sqrt{1+x}-1}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 13

Question 15.
\(\lim _{x \rightarrow 0} \frac{(1+x)^{1 / 2}-(1-x)^{1 / 2}}{x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 14

Question 16.
\(\lim _{x \rightarrow 0} \frac{(1+x)^n-1}{x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 15

Question 17.
\(\lim _{x \rightarrow 0} \frac{1-x^{-1 / 3}}{1-x^{-2 / 3}}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 16

Question 18.
\(\lim _{x \rightarrow 0} \frac{1-\cos x}{x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 17

Question 19.
\(\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 18

Question 20.
\(\lim _{x \rightarrow 0} \frac{\tan 2 x-x}{3 x-\sin x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 19

Question 21.
\(\lim _{x \rightarrow 0} \frac{e^{x^2}-\cos x}{x^2}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 20

Question 22.
\(\lim _{x \rightarrow 0} \frac{x \cos x-\log (1+x)}{x^2}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 21

Question 23.
\(\lim _{x \rightarrow 0} \frac{\cos a x-\cos b x}{x^2}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 22

Question 24.
\(\lim _{x \rightarrow 0} \frac{x \cos x-\sin x}{x^2 \sin x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 23

Question 25.
\(\lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{\log (1+x)}{x^2}\right)\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 24

Question 26.
\(\lim _{x \rightarrow 0} \frac{y-\tan ^{-1} y}{y-\sin y}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 25

Question 27.
\(\lim _{x \rightarrow 0^{+}} \log _{\sin 2 x} \sin x\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 26

Question 28.
\(\lim _{x \rightarrow 0}\left(\frac{1}{x^2}-\frac{\cot x}{x}\right)\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(a) 27

OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(b)

September 5, 2024

Accessing OP Malhotra Maths Class 12 Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(b) can be a valuable tool for students seeking extra practice.

Find by applying L’Hospital’s Rule the following limits :

Question 1.
\(\lim _{x \rightarrow 0} \frac{x e^x-\log (1+x)}{x^2}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(b) 1

Question 2.
\(\lim _{x \rightarrow 0} \frac{x-\sin x \cos x}{x^3}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(b) 2

Question 3.
\(\lim _{x \rightarrow 0} \frac{e^x \sin x-x-x^2}{x^3}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(b) 3

Question 4.
\(\lim _{x \rightarrow \pi / 4} \frac{\sin \left(x+\frac{\pi}{4}\right)-1}{\log \sin 2 x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(b) 4

Question 5.
\(\lim _{x \rightarrow 0} \frac{e^x+e^{-x}+2 \cos x-4}{x^4}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(b) 5

Question 6.
\(\lim _{x \rightarrow 0} \frac{e^x-e^{-x}-2 \log (1+x)}{x \sin x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(b) 6

Question 7.
\(\lim _{x \rightarrow \frac{1}{2}} \frac{\cos ^2 \pi x}{e^{2 x}-2 e x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(b) 7

Question 8.
\(\lim _{x \rightarrow 0} \frac{\log \left(1-x^2\right)}{\log \cos x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(b) 8

Question 9.
\(\lim _{x \rightarrow 0} \frac{\tan x-x}{x^2 \tan x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(b) 9

Question 10.
\(\lim _{x \rightarrow 0} \frac{\sin x-\log \left(e^x \cos x\right)}{x \sin x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(b) 10

Question 11.
\(\lim _{x \rightarrow 0} \frac{\log \sin x}{\cot x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(b) 11

Question 12.
\(\lim _{x \rightarrow 0} \frac{\log \tan x}{\log x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(b) 12

Question 13.
\(\lim _{x \rightarrow 0} \frac{\log \tan 2 x}{\log \tan x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(b) 13

Question 14.
\(\lim _{x \rightarrow 0} \log (1-x) \cot \frac{\pi x}{2}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(b) 14

Question 15.
\(\lim _{x \rightarrow 0}\left(\frac{1}{x^2}-\cot ^2 x\right)\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(b) 15

Question 16.
\(\lim _{x \rightarrow \frac{\pi}{2}}(\sin x)^{\tan x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(b) 16

Question 17.
\(\lim _{x \rightarrow 0}(\cot x)^{\sin 2 x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(b) 17

Question 18.
\(\lim _{x \rightarrow 0}(1+\sin x)^{\cot x}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(b) 18

Question 19.
\(\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{1 / x^2}\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(b) 19

Question 20.
\(\lim _{x \rightarrow 0}\left(\frac{e^x-e^{-x}-2 x}{x-\sin x}\right)\)
Solution:
OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Ex 9(b) 20

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