Utilizing OP Malhotra ISC Class 12 Solutions Chapter 21 Vectors Ex 21(d) as a study aid can enhance exam preparation.
S Chand Class 12 ICSE Maths Solutions Chapter 21 Vectors Ex 21(d)
Question 1.
If \({\overrightarrow{OA}}\) = 2 \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\)
and \({\overrightarrow{OB}}\) = 5 \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\), then find the vectors
\({\overrightarrow{A B}}\) and \({\overrightarrow{B A}}\).
Answer:
Given \({\overrightarrow{OA}}\) = 2 \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\);
\({\overrightarrow{OB}}\) = 5 \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\)
∴ \({\overrightarrow{AB}}\) = \({\overrightarrow{OB}}\) – \({\overrightarrow{OA}}\)
= (5 \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\)) – (2 \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\))
= 3 \(\hat{i}\) + 3 \(\hat{j}\)
\({\overrightarrow{BA}}\) = \({\overrightarrow{OA}}\) – \({\overrightarrow{OB}}\)
= (2 \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\)) – (5 \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\))
= -3 \(\hat{i}\) – 3 \(\hat{j}\)
Question 2.
If \(\vec{a}\) = 4 \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\),
\(\vec{b}\) = 2 \(\hat{i}\) + \(\hat{j}\) – 7 \(\hat{k}\),
\(\vec{c}\) = -3 \(\hat{i}\) – 4 \(\hat{j}\) + 2 \(\hat{k}\)
and \(\vec{d}\) = \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\),
then find the value of 3 \(\vec{a}\) + 2 \(\vec{b}\) – 4 \(\vec{c}\) – \(\vec{d}\).
Answer:
Given \(\vec{a}\) = 4 \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\) ;
\(\vec{b}\) = 2 \(\hat{i}\) + \(\hat{j}\) – 7 \(\hat{k}\);
\(\vec{c}\) = -3 \(\hat{i}\) – 4 \(\hat{j}\) + 2 \(\hat{k}\)
and \(\vec{d}\) = \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)
3 \(\vec{a}\) + 2 \(\vec{b}\) – 4 \(\vec{c}\) – \(\vec{d}\)
= 3(4 \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)) + 2(2 \(\hat{i}\) + \(\hat{j}\) – 7 \(\hat{k}\))
-4(-3 \(\hat{i}\) – 4 \(\hat{j}\) + 2 \(\hat{k}\)) – (\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\))
= (12 + 4 + 12 – 1) \(\hat{i}\) + (3 + 2 + 16 – 1) \(\hat{j}\) +(3 – 14 – 8 – 1) \(\hat{k}\)
= 27 \(\hat{i}\) + 20 \(\hat{j}\) – 20 \(\hat{k}\)
Question 3.
Find sum of the vectors 3 \(\hat{i}\) + 7 \(\hat{j}\) – 4 \(\hat{k}\) and \(\hat{i}\) – 5 \(\hat{j}\) – 8 \(\hat{k}\) and hence find the unit vector along the sum of these vectors.
Answer:
Let \(\vec{a}\) = 3 \(\hat{i}\) + 7 \(\hat{j}\) – 4 \(\hat{k}\) and \(\vec{b}\)
= \(\hat{i}\) – 5 \(\hat{j}\) – 8\(\hat{k}\)
∴ \(\vec{a}\) + \(\vec{b}\)
= (3 \(\hat{i}\) + 7 \(\hat{j}\) – 4 \(\hat{k}\)) + (\(\hat{i}\) – 5 \(\hat{j}\) – 8 \(\hat{k}\))
= 4 \(\hat{i}\) + 2 \(\hat{j}\) – 12 \(\hat{k}\)
⇒ |\(\vec{a}\) + \(\vec{b}\)|
= \(\sqrt{4^2+2^2+(-12)^2}\)
= \(\sqrt{16+4+144}\)
= \(\sqrt{164}\)
= 2 \(\sqrt{41}\)
Thus required unit vector along the sum of two vectors
= \(\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}\)
= \(\frac{4 \hat{i}+2 \hat{j}-12 \hat{k}}{2 \sqrt{41}}\)
= \(\frac{2 \hat{i}+\hat{j}-6 \hat{k}}{\sqrt{41}}\)
= \(\frac{2}{\sqrt{41}}\) \(\hat{i}\) + \(\frac{1}{\sqrt{41}}\) \(\hat{j}\) – \(\frac{6}{\sqrt{41}}\) \(\hat{k}\)
Question 4.
Find by vector method the perimeter of the triangle whose vectors are (-1, -1, 9), (3, 1, 5) and (0, -5, 1).
Answer:
Let he P.V’s of the vertices A, B and C of ∆ABC are
Question 5.
If the position vectors of the angular points A, B, C and D of a quadrilateral be \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\), \(\hat{j}\) + \(\hat{k}\) – \(\hat{i}\), \(\hat{k}\) + \(\hat{i}\) – \(\hat{j}\), and \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\) respectively, then determine the lengths of its sides.
Answer:
Question 6.
Prove by vector method that the triangle whose vertices are (2, 4, -1), (4, 5, 1) and (3, 6, -3) in an isosceles triangle.
Answer:
Question 7.
The position vectors of two points A and B are 5 \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\) and \(\hat{3}\) + 2 \(\hat{j}\) + \(\hat{k}\) respectively. Find the direction cosines of \({\overrightarrow{A B}}\) and show that A B is parallel to X-Y plane.
Answer:
Given P.V of A = 5 \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)
P.V of B = 3 \(\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\)
∴ \({\overrightarrow{AB}}\) = P.V of B – P.V of A
= (3 \(\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\)) – (5 \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\))
= -2 \(\hat{i}\) + \(\hat{j}\) + 0 \(\hat{k}\)
= -2 \(\hat{i}\) + \(\hat{j}\)
∴ direction consines of \({\overrightarrow{AB}}\) are
< \(-\frac{2}{\sqrt{(-2)^2+1^2+0^2}}\), \(\frac{1}{\sqrt{(-2)^2+1^2+0^2}}\), 0 >
i.e. < \(-\frac{2}{\sqrt{5}}\), \(\frac{1}{\sqrt{5}}\), 0 >.
Equation of XOY plane be z = 0
∴ d consine’s of XOY plane are < 0, 0, 1 >
Clearly \(\frac{-2}{\sqrt{5}}\) × 0 + \(\frac{1}{\sqrt{5}}\) × 0 + 0 × 1 = 0
Thus \({\overrightarrow{AB}}\) is parallel to XOY plane.
Question 8.
Given that \(\vec{a}\) = (x + 4 y) \(\hat{i}\) + (2 x + y + 1) \(\hat{j}\) and \(\vec{b}\) = (y – 2 x + 2) \(\hat{i}\) + (2 x – 3 y – 1) \(\hat{j}\), find the value of x and y, if 3 \(\vec{a}\) = 2 \(\vec{b}\).
Answer:
Given \(\vec{a}\) = (x + 4 y) \(\hat{i}\) + (2 x + y + 1) \(\hat{j}\) and \(\vec{b}\) = (y – 2 x + 2) \(\hat{i}\) + (2 x – 3 y – 1) \(\hat{j}\)
Given 3 \(\vec{a}\) = 2 \(\vec{b}\)
⇒ 3(x + 4 y) \(\hat{i}\) + 3(2 x – y + 1) \(\hat{j}\)
= 2(y – 2 x + 2) \(\hat{i}\) + 2(2 x – 3 y – 1) \(\hat{j}\)
Comparing the coefficients of \(\hat{i}\) and \(\hat{j}\) on both sides; we have
3 x + 12 y = 2 y – 4 x + 4
⇒ 7 x + 10 y = 4
and 6 x + 3 y + 3 = 4 x – 6 y – 2
⇒ 2 x + 9 y = -5
on solving eqn. (1) and eqn. (2); we have x = 2 and y = -1
Question 9.
Prove that the vectors 2 i + 3 j – 6 k, 6 \(\hat{i}\) – 2 \(\hat{j}\) + 3 \(\hat{k}\) and 3 \(\hat{i}\) + 6 \(\hat{j}\) – 2 \(\hat{k}\) from the sides of an equilateral triangle.
Answer:
Let
\({\overrightarrow{AB}}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) – 6 \(\hat{k}\)
\({\overrightarrow{BC}}\) = 6 \(\hat{i}\) – 2 \(\hat{j}\) + 3 \(\hat{k}\)
\({\overrightarrow{CA}}\) = 3 \(\hat{i}\) + 6 \(\hat{j}\) – 2 \(\hat{k}\)
∴ |\(\overrightarrow{\mathrm{AB}}\)| = \(\sqrt{2^2+3^2+(-6)^2}\) = 7 ;
|\(\overrightarrow{\mathrm{BC}}\)| = \(\sqrt{6^2+(-2)^2+3^2}\) = 7
and |\(\overrightarrow{\mathrm{CA}}\)| = \(\sqrt{3^2+6^2+(-2)^2}\) = 7
and |\(\overrightarrow{\mathrm{CA}}\)| = \(\sqrt{3^2+6^2+(-2)^2}\) = 7
Thus, |\(\overrightarrow{\mathrm{AB}}\)| = |\(\overrightarrow{\mathrm{BC}}\)|
= |\(\overrightarrow{\mathrm{CA}}\)| = 7
Hence ∆ABC be an equilateral triangle.
Question 10.
Prove that the points 2 i – j + k, \(\hat{i}\) – 4 \(\hat{j}\) – 5 \(\hat{k}\), 3 \(\hat{i}\) – 4 \(\hat{j}\) – 4 \(\hat{k}\) are the vertices of a right-angled triangle.
Answer:
Let the P.V’s of vertices A, B, C of ∆A B C
Question 11.
If the points \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\), 2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\) and 3 \(\hat{i}\) + 7 \(\hat{j}\) + p \(\hat{k}\) are collinear, then find the value of p.
Answer:
Let the points are A, B and C whose P.V’s are \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\),
2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\) and 3 \(\hat{i}\) + 7 \(\hat{j}\) + p \(\hat{k}\) Since A, B, C are callinear
∴ \({\overrightarrow{AB}}\) = λAC for non-zero scalar λ
P.V of B – P.V of A = λ[P.V of C – P.V of A]
⇒ (2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\)) – (\(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\))
= λ[(3 \(\hat{i}\) + 7 \(\hat{j}\) + p \(\hat{k}\)) – (\(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\))]
⇒ \(\hat{i}\) + 4 \(\hat{j}\) + 3 \(\hat{k}\)
= λ[2 \(\hat{i}\) + 8 \(\hat{j}\) + (p – 1) \(\hat{k}\)]
on comparing the coefficients of \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) on both sides we have
1 = 2λ
⇒ λ = \(\frac{1}{2}\)
and 3 = λ(p – 1)
⇒ 3 = \(\frac{1}{2}\)(p – 1)
⇒ 6 = p – 1
⇒ p = 7
Thus required value of p be 7 .
Question 12.
If the vertices of a triangle are the points
\(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\), 2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\),
3 \(\hat{i}\) + 3 \(\hat{j}\) – 4 \(\hat{k}\)
what are the vectors determined by its sides ? i, j, k are unit vectors parallel to the axes of coordinates.
Answer:
Let A, B and C are the points whose position vectors are
\(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\),
2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\)
and 3 \(\hat{i}\) + 3 \(\hat{j}\) – 4 \(\hat{k}\)
i.e. P.V of A = \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\);
P.V of B = 2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\) and
P.V of C = 3 \(\hat{i}\) + 3 \(\hat{j}\) – 4 \(\hat{k}\)
\({\overrightarrow{AB}}\) = P.V of B – P.V of A
= (2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\)) – (\(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\))
= \(\hat{i}\) + 4 \(\hat{j}\) + 2 \(\hat{k}\)
BC = P.V of C – P.V of B
= (3 \(\hat{i}\) + 3\(\hat{j}\) – 4 \(\hat{k}\)) – (2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\))
= \(\hat{i}\) + 0 \(\hat{j}\) – 8 \(\hat{k}\)
\({\overrightarrow{CA}}\) = P.V of A – P.V of C
= (\(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\)) – (3 \(\hat{i}\) + 3 \(\hat{j}\) – 4 \(\hat{k}\))
= -2 \(\hat{i}\) – 4 \(\hat{j}\) + 6 \(\hat{k}\)
Hence the required vectors along the sides are \(\hat{i}\) + 4 \(\hat{j}\) + 2 \(\hat{k}\) ;
\(\hat{i}\) – 8 \(\hat{k}\) and -2 \(\hat{i}\) – 4 \(\hat{j}\) + 6 \(\hat{k}\)
Question 13.
Find the length and direction cosines of \({\overrightarrow{P Q}}\) where \({\overrightarrow{OP}}\) = -2 \(\hat{i}\) + 3 \(\hat{j}\) – 4 \(\hat{k}\) and \({\overrightarrow{OQ}}\) = 2 \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\).
Answer:
Given \({\overrightarrow{OP}}\) = -2 \(\hat{i}\) + 3 \(\hat{j}\) – 4 \(\hat{k}\) ;
\({\overrightarrow{OQ}}\) = 2 \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\)
∴ \({\overrightarrow{PQ}}\) = \({\overrightarrow{OQ}}\) – \({\overrightarrow{OP}}\)
= (2 \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\)) – (-2 \(\hat{i}\) + 3 \(\hat{j}\) – 4 \(\hat{k}\))
= 4 \(\hat{i}\) – 4 \(\hat{j}\) + 7 \(\hat{k}\)
Thus Direction ratios of \({\overrightarrow{P Q}}\) are < 4, -4, 7 >
∴ direction cosines of \({\overrightarrow{P Q}}\) are
\(\frac{4}{\sqrt{4^2+(-4)^2+7^2}}\),
\(\frac{-4}{\sqrt{4^2+(-4)^2+7^2}}\), \(\frac{7}{\sqrt{4^2+(-4)^2+7^2}}\) >
i.e. < \(\frac{4}{9}\), \(\frac{-4}{9}\), \(\frac{7}{9}\) >
∴ |\({\overrightarrow{PQ}}\)|
= \(\sqrt{4^2+(-4)^2+7^2}\)
= \(\sqrt{16+16+49}\) = 9
Question 14.
The vertices of a triangle are P(-2, 1, -3), Q(0, 4, 3), R(2, 5, 4). Find the lengths and direction cosines of the vectors \({\overrightarrow{P Q}}\) and \({\overrightarrow{P R}}\).
Answer:
Question 15.
A is the point (1, 3, -2), B is (3, -4, 1), C is (-1, 0, 2). Find the lengths of the sides of the triangle A B C and the cosines of the angles, A, B, C.
Answer:
Question 16.
Prove that the triangle with vertices A(1, 0, 1), B(2, -1, 4) and C(3, -4, 1) is right-angled.
Answer:
Question 17.
The position vectors of A, B and C are 2 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\), 3 \(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\) and \(\hat{i}\) + 4 \(\hat{j}\) – 3 \(\hat{k}\) respective show that A, B and C are collinear.
Answer:
Given,
P.V of A = 2 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)
P.V of B = 3 \(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\)
P.V of C = \(\hat{i}\) + 4 \(\hat{j}\) – 3 \(\hat{k}\)
∴ \({\overrightarrow{AB}}\) = P. V of B – P.V of A
= (3\(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\)) – (2 \(\hat{i}\) + \(\hat{j}\) – \(\hat{i}\))
= \(\hat{i}\) – 3 \(\hat{j}\) + 2 \(\hat{k}\)
\({\overrightarrow{BC}}\) = P. V of C – P.V of P
= (\(\hat{i}\) + 4 \(\hat{j}\) – 3 \(\hat{k}\)) – (3 \(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\))
= -2 \(\hat{i}\) + 6 \(\hat{j}\) – 4 \(\hat{k}\)
⇒ \({\overrightarrow{BC}}\) = -2(\(\hat{i}\) – 3 \(\hat{j}\) + 2 \(\hat{k}\))
= -2 \({\overrightarrow{AB}}\)
Thus \({\overrightarrow{BC}}\) and \({\overrightarrow{AB}}\) are parallel vectors and point B common to both vectors \({\overrightarrow{AB}}\) and \({\overrightarrow{BC}}\).
Therefore A, B and C are collinear vectors.
Question 18.
Show that the points (2, -1, 3), (3, -5, 1) and (-1, 11, 9) are collinear.
Answer:
Let A, B and C are the points whose position rectors are 2 \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\), 3 \(\hat{i}\) – 5 \(\hat{j}\) + \(\hat{k}\) and \(-\hat{i}\) + 11 \(\hat{j}\) + 9 \(\hat{k}\)
∴ \({\overrightarrow{AB}}\) = P.V of B – P.V of A
= (3 \(\hat{i}\) – 5 \(\hat{j}\) + \(\hat{k}\)) – (2 \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\))
= \(\hat{i}\) – 4 \(\hat{j}\) – 2 \(\hat{k}\)
\({\overrightarrow{BC}}\) = P.V of C – P.V of B
= (\(-\hat{i}\) + 11 \(\hat{j}\) + 9 \(\hat{k}\)) – (3 \(\hat{i}\) – 5 \(\hat{j}\) + \(\hat{k}\))
= -4 \(\hat{i}\) + 16 \(\hat{j}\) + 8 \(\hat{k}\)
= -4(\(\hat{i}\) – 4 \(\hat{j}\) – 2 \(\hat{k}\))
= -4 \({\overrightarrow{AB}}\)
Thus \({\overrightarrow{BC}}\) = -4 \({\overrightarrow{AB}}\)
∴ \({\overrightarrow{BC}}\) and \({\overrightarrow{AB}}\) are parallel vectors and point B is common to both the vectors \({\overrightarrow{BC}}\) and \({\overrightarrow{AB}}\). Thus A, B and C are collinear points.