OP Malhotra Class 12 Maths Solutions Chapter 22 Vectors (Continued) Ex 22(c)

Practicing OP Malhotra ISC Class 12 Solutions Chapter 22 Vectors (Continued) Ex 22(c) is the ultimate need for students who intend to score good marks in examinations.

S Chand Class 12 ICSE Maths Solutions Chapter 22 Vectors (Continued) Ex 22(c)

Question 1.
Determine (\(\vec{a}\), \(\vec{b}\), \(\vec{c}\)) if \(\vec{a}\) = 2 \(\hat{i}\) – 3 \(\hat{j}\), \(\vec{b}\) = \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\), \(\vec{c}\) = 3 \(\hat{i}\) – \(\hat{k}\).
Answer:
Given \(\vec{a}\) = 2 \(\hat{i}\) – 3 \(\hat{j}\);
\(\vec{b}\) = \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\);
\(\vec{c}\) = 3 \(\hat{i}\) – \(\hat{k}\)
∴ \(\vec{a}\) \(\vec{b}\) \(\vec{c}\)
= \(|\begin{array}{rrr}
2 -3 0
1 1 -1
3 0 -1
\end{array}|\)
Expanding along R₁ = 2(-1 – 0) + 3(-1 + 3) = -2 + 6 = 4

Question 2.
If \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) is an orthogonal unit vector triad in a right handed system, then prove that
(i) [\(\hat{i}\) \(\hat{j}\) \(\hat{k}\)] = [\(\hat{j}\) \(\hat{k}\) \(\hat{i}\)]
= [\(\hat{k}\) \(\hat{i}\) \(\hat{j}\)] = 1
(ii) [\(\hat{i}\) \(\hat{k}\) \(\hat{j}\)] = [\(\hat{j}\) \(\hat{i}\) \(\hat{k}\)]
= [\(\hat{k}\) \(\hat{j}\) \(\hat{i}\)] = -1
(iii) [\(\hat{i}\) \(\hat{k}\) \(\hat{j}\)] + [\(\hat{i}\) \(\hat{k}\) \(\hat{j}\)] = 0
(iv) [\(\hat{i}\) \(\hat{k}\) \(\hat{j}\)] = [\(\hat{k}\) \(\hat{j}\) \(\hat{i}\)]
= [\(\hat{j}\) \(\hat{k}\) \(\hat{i}\)] = -1
Answer:
[\(\hat{i}\) \(\hat{j}\) \(\hat{k}\)] = \(\hat{i}\) (\(\hat{j}\) × \(\hat{k}\)) = \(\hat{i}\) .\(\hat{i}\) = 1
[\(\hat{j}\) \(\hat{k}\) \(\hat{i}\)] = \(\hat{j}\).(\(\hat{k}\) × \(\hat{i}\)) = \(\hat{j}\) .\( \hat{j}\) = 1
[\(\hat{k}\) \(\hat{i}\) \(\hat{j}\)] = \(\hat{k}\).(\(\hat{i}\) × \(\hat{j}\)) = \(\hat{k}\) .\(\hat{k}\) = 1

(ii) [\(\hat{i}\) \(\hat{k}\) \(\hat{j}\)]
= \(\hat{i}\).(\(\hat{k}\) × \(\hat{j}\)) = \(\hat{i}\) (\(-\hat{i}\))
= -(\(\hat{i}\).\(\hat{i}\)) = -1
[\(\hat{j}\) \(\hat{i}\) \(\hat{k}\)]
= \(\hat{j}\).(\(\hat{i}\) × \(\hat{k}\))
= \(\hat{j}\).(\(-\hat{j}\)) = -(\(\hat{j}\).\(\hat{j}\)) = -1
[\(\hat{k}\) \(\hat{j}\) \(\hat{i}\)] = \(\hat{k}\).(\(\hat{j}\) × \(\hat{i}\)) = \(\hat{k}\) .(\(-\hat{k}\)) = -(\(\hat{k}\).\(\hat{k}\)) = -1

(iii) [\(\hat{i}\) \(\hat{j}\) \(\hat{k}\)] + [\(\hat{i}\) \(\hat{k}\) \(\hat{j}\)]
= \(\hat{i}\).(\(\hat{j}\) × \(\hat{j}\)) + \(\hat{i}\).(\(\hat{k}\) × \(\hat{j}\))
= \(\hat{i}\).\(\hat{i}\) + \(\hat{i}\).(\(-\hat{i}\))
= \(\hat{i}\).\(\hat{i}\) – \(\hat{i}\).\(\hat{i}\) = 1 – 1 = 0

(iv) [\(\hat{i}\) \(\hat{k}\) \(\hat{j}\)] + [\(\hat{k}\) \(\hat{j}\) \(\hat{i}\)] + [\(\hat{j}\) \(\hat{k}\) \(\hat{i}\)]
= \(\hat{i}\).(\(\hat{k}\) × \(\hat{j}\)) + \(\hat{k}\).(\(\hat{j}\) × \(\hat{i}\)) + \(\hat{j}\).(\(\hat{k}\) × \(\hat{i}\))
= \(\hat{i}\).(\(-\hat{i}\)) + \(\hat{k}\).(\(-\hat{k}\)) + \(\hat{j}\) .(\(\hat{j}\))
= -1 -1 + 1 = -1
[∵ \(\hat{i}\).\(\hat{i}\) = 1
= \(\hat{j}\).\(\hat{j}\) = \(\hat{k}\).\(\hat{k}\) and \(\hat{i}\) × \(\hat{j}\) = \(\hat{k}\) ;
\(\hat{j}\) × \(\hat{k}\) = \(\hat{i}\) and \(\hat{k}\) × \(\hat{i}\) = \(\hat{j}\)]

Question 3.
If α and β are any vectors. Prove that \(\vec{\beta}\).(\(\vec{\alpha}\) × \(\vec{\beta}\)) = \(\overrightarrow{0}\).
Answer:
Let \(\vec{\alpha}\) = α₁ \(\hat{i}\) + α₂ \(\hat{j}\) + α₃ \(\hat{k}\) and \(\vec{\beta}\) = β₁ \(\hat{i}\) + β₂ \(\hat{j}\) + β₃ \(\hat{k}\)
∴ \(\vec{\alpha}\) × \(\vec{\beta}\)
= \(|\begin{array}{ccc}
\hat{i} \hat{j} \hat{k}
\alpha_1 \alpha_2 \alpha_3
\beta_1 \beta_2 \beta_3
\end{array}|\)
= \(\hat{i}\)(α₂ β₃ – α₃ β₂) – \(\hat{j}\)(α₁ β₃ – α₁ β₃) + \(\hat{k}\)(α₁ β₂ – α₂ β₁)
∴ \(\vec{\beta}\).(\(\vec{\alpha}\) × \(\vec{\beta}\))
= (β₁ \(\hat{i}\) + β₂ \(\hat{j}\) + β₃ \(\hat{k}\)).[(α₂ β₃ – α₃ β₂) \(\hat{i}\) + (α₃ β₁ – α₁ β₃) \(\hat{j}\) + (α₁ β₂ – α₂ β₁) \(\hat{k}\)
= β₁(α₂ β₃ – α₃ β₂) + β₂(α₃ β₁ – α₁ β₃) + β₃(α₁ β₂ – α₂ β₁) = 0

Question 4.
Find the volume of the parallelogram whose edges are represented by the vectors
(i) \(\vec{a}\) = 2 \(\hat{i}\) – 3 \(\hat{j}\) + \(\hat{k}\), \(\vec{b}\) = \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\) and \(\vec{c}\) = 2 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)
(ii) \(\vec{a}\) = 11 \(\hat{i}\), \(\vec{b}\) = 2 \(\hat{j}\), \(\vec{c}\) = 13 \(\hat{k}\)
Answer:
(i) Given \(\vec{a}\) = 2 \(\hat{i}\) – 3 \(\hat{j}\) + \(\hat{k}\); \(\vec{b}\) = \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\) and \(\vec{c}\) = 2 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)
∴ Volume of parallelopiped = V = |[\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)]|
Here
\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)
= |
2 -3 1
1 -1 2
2 1 -1
|;
Expanding along R₁
= 2(1 – 2) + 3(-1 – 4) + 1(1 + 2) = -2 – 15 + 3 = -14
∴ reqd. volume of parallelopiped = |-14| = 14 cubic units.

(ii) Given \(\vec{a}\) = 11 \(\hat{i}\); \(\vec{b}\) = 2 \(\hat{j}\) and \(\vec{c}\) = 13 \(\hat{k}\)
∴ \(\vec{a}\) \(\vec{b}\) \(\vec{c}\)
= \(\vec{a}\) .(\(\vec{b}\) × \(\vec{c}\))
= 11 \(\hat{i}\).(2 \(\hat{j}\) × 13 \(\hat{k}\))
= 286 \(\hat{i}\).(\(\hat{j}\) × \(\hat{k}\))
= 286(\(\hat{i}\). \(\hat{i}\)) = 286
Thus required volume of parallelopiped = |\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)|
= 286 cubic units.

Question 5.
The volume of a parallelogram whose edges are represented by -12 \(\hat{i}\) + λ\(\hat{k}\), 3 \(\hat{j}\) – \(\hat{k}\), 2 \(\hat{i}\) + \(\hat{j}\) – 15 \(\hat{k}\) is 546 . Find the value of λ.
Answer:
Let \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are the position vectors of the edges of parallelopiped.
Then
\(\vec{a}\) = -12 \(\hat{i}\) + λ\(\hat{k}\) ;
\(\vec{b}\) = 3 \(\hat{j}\) – \(\hat{k}\) ;
\(\vec{c}\) = 2 \(\hat{i}\) + \(\hat{j}\) – 15 \(\hat{k}\)
Here, \(\vec{a}\) \(\vec{b}\) \(\vec{c}\)
= |-12 0 λ 0 3 -1 2 1 -15|;
Expanding along R = -12(-45 + 1) + 0 + λ(0 – 6) = 528 – 6λ
∴ Volume of parallelopiped = |\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)| = |528 – 6 λ|
also given volume of parallelopiped = 546
⇒ |528 – 6λ|= 546
⇒ 528 – 6 λ = 546
⇒ 6λ = -18
⇒ λ = -3 [Since volume be always be positive]

Question 6.
A(1, 1, 1), B(2, 1, 3), C(3, 2, 2) andr D(3, 3, 4).
Find the volume of the parallelo- gram of which the segments A B, A C and A D are coterminous edges.
Answer:
Given P.V. of A = \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\);
P.V. of B = 2 \(\hat{i}\) + \(\hat{j}\) + 3 \(\hat{k}\);
P.V. of C = 3 \(\hat{i}\) + 2 \(\hat{j}\) + 2 \(\hat{k}\)
∴ \(\overrightarrow{\mathrm{AB}}\) = P.V. of B – P.V of A
= (2 \(\hat{i}\) + \(\hat{j}\) + 3 \(\hat{k}\)) – (\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\))
= \(\hat{i}\) + 0 \(\hat{j}\) + 2 \(\hat{k}\)
\(\overrightarrow{\mathrm{AC}}\) = P.V} . of C – P.V of A
= (3 \(\hat{i}\) + 2 \(\hat{j}\) + 2 \(\hat{k}\)) – (\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\))
= 2 \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)
\(\overrightarrow{\mathrm{AD}}\) = P.V. of D – P.V of A
= (3 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\)) – (\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\))
= 2 \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\)
∴ [\(\overrightarrow{\mathrm{AB}}\) \(\overrightarrow{\mathrm{AC}}\) \(\overrightarrow{\mathrm{AD}}\)]
= |1 0 2 2 1 1 2 2 3|;
Expanding along R₁
= 1(3 – 2) + 0 + 2(4 – 2) = 1 + 4 = 5
∴ Volume of parallelopiped having coterminous edges are \(\overrightarrow{\mathrm{AB}}\), \(\overrightarrow{\mathrm{AC}}\) and \(\overrightarrow{\mathrm{AD}}\)
= |[\(\overrightarrow{\mathrm{AB}}\) \(\overrightarrow{\mathrm{AC}}\) \(\overrightarrow{\mathrm{AD}}\)]| = 5 cubic units

Question 7.
Prove that [\(\hat{i}\) – \(\hat{j}\), \(\hat{j}\) – \(\hat{k}\), \(\hat{k}\) – \(\hat{i}\)] = 0.
Answer:
[\(\hat{i}\) – \(\hat{j}\), \(\hat{j}\) – \(\hat{k}\), \(\hat{k}\) – \(\hat{i}\)]
= |
1 -1 0
0 1 -1
-1 0 1
|; expanding along R₁
=1(1 – 0) + 1(0 – 1) = 1 – 1 = 0

Question 8.
Show that the following vectors are coplanar :
\(\vec{a}\) = \(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\),
\(\vec{b}\) = \(\hat{i}\) – 3 \(\hat{j}\) + 7 \(\hat{k}\),
\(\vec{c}\) = 5 \(\hat{i}\) + 6 \(\hat{j}\) + 5 \(\hat{k}\)
Answer:
Given
\(\vec{a}\) = \(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\) ;
\(\vec{b}\) = 3 \(\hat{i}\) + 2 \(\hat{j}\) + 7 \(\hat{k}\) and
\(\vec{c}\) = 5 \(\hat{i}\) + 6 \(\hat{j}\) + 5 \(\hat{k}\)
∴ \(\vec{a}\) \(\vec{b}\) \(\vec{c}\)
= |1 2 -1
3 2 7
5 6 5|;
Expanding along R₁
= 1(10 – 42) – 2(15 – 35) – 1(18 – 10) = -52 + 40 – 8 = 0
Thus vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are coplanar.

Question 9.
If \(\vec{a}\) = 2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\), \(\vec{b}\) = \(\hat{i}\) – 3 \(\hat{j}\) – 5 \(\hat{k}\), \(\vec{c}\) = 3 \(\hat{i}\) – 4 \(\hat{j}\) – 4 \(\hat{k}\) determine [\(\vec{a}\), \(\vec{b}\), \(\vec{c}\)] and intercept the result.
Answer:
Here [\(\vec{a}\), \(\vec{b}\), \(\vec{c}\)]
= |2 -1 1 1 -3 -5 3 -4 -4|;
Expanding along R₁
= 2(12 – 20) + 1(-4 + 15) + 1(-4 + 9) = -16 + 11 + 5 = 0
Thus \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are coplanar vectors.

Question 10.
Find the value of λ such that the following vectors are coplanar :
(i) 2 \(\hat{i}\) – 4 \(\hat{j}\) + 5 \(\hat{k}\), \(\hat{i}\) – λ\(\hat{j}\) + \(\hat{k}\), 3 \(\hat{i}\) + 2 \(\hat{j}\) – 5 \(\hat{k}\)
(ii) \(\vec{a}\) = \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\),
\(\vec{b}\) = 2 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\),
\(\vec{c}\) = λ\(\hat{i}\) – \(\hat{j}\) + λ\(\hat{k}\)
Answer:
(i) Let \(\vec{a}\) = 2 \(\hat{i}\) – 4 \(\hat{j}\) + 5 \(\hat{k}\);
\(\vec{b}\) = \(\hat{i}\) – λ\(\hat{j}\) + \(\hat{k}\) and \(\vec{c}\)
= 3 \(\hat{i}\) + 2 \(\hat{j}\) – 5 \(\hat{k}\)
Here [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)]
= |2 -4 5 1 – λ 1 3 2 -5|;
Expanding along R₁
= 2(5λ – 2) + 4(-5 – 3) + 5(2 + 3λ)
= 10λ – 4 – 32 + 10 + 15λ = 25λ – 26
Since \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are coplanar.
∴ [\(\vec{a}\), \(\vec{b}\), \(\vec{c}\)] = 0
⇒ 25 λ – 26 = 0
⇒ λ = \(\frac{26}{25}\)

(ii) Given \(\vec{a}\) = \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\);
\(\vec{b}\) = 2 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\);
\(\vec{c}\) = λ\(\hat{i}\) – \(\hat{j}\) + λ\(\hat{k}\)
Here
\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)
= |
1 -1 1
2 1 -1
λ -1 λ|;
Expanding along R₁
= 1(λ – 1) + 1(2λ + λ) + 1(-2 – λ)
= λ – 1 + 3λ – 2 – λ
= 3λ – 3
Since \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are coplanar.
∴ [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = 0
⇒ 3 λ – 3 = 0⇒ λ = 1

Question 11.
Show that the four points having position vectors \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\), 6 \(\hat{i}\) + 1 \(\hat{1}\) \(\hat{j}\) + 2 \(\hat{k}\) \(\hat{i}\) + 2 \(\hat{j}\) + 6 \(\hat{k}\), \(\hat{i}\) + \(\frac{1}{2}\) \(\hat{j}\) + 4 \(\hat{k}\) are coplanar.
Answer:
Let A, B, C and D are given points then these points are coplanar iff any one of the following triads are of coplanar vectors. \(\overrightarrow{\mathrm{AB}}\), \(\overrightarrow{\mathrm{AC}}\), \(\overrightarrow{\mathrm{AD}}\) ; \(\overrightarrow{\mathrm{AB}}\), \(\overrightarrow{\mathrm{BC}}\), \(\overrightarrow{\mathrm{CD}}\) ; \(\overrightarrow{\mathrm{BC}}\), \(\overrightarrow{\mathrm{BA}}\), \(\overrightarrow{\mathrm{BD}}\) etc.
\(\overrightarrow{\mathrm{AB}}\) = P.V. of B – P.V of A
= (6 \(\hat{i}\) + 11 \(\hat{j}\) + 2 \(\hat{k}\)) – (\(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\))
= 5 \(\hat{i}\) + 12 \(\hat{j}\) + 0 \(\hat{k}\)
\(\overrightarrow{\mathrm{AC}}\) = P.V. of C – P.V. of A
= (\(\hat{i}\) + 2 \(\hat{j}\) + 6 \(\hat{k}\)) – (\(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\))
= 0 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\)
\(\overrightarrow{\mathrm{AD}}\) = P.V. of D – P.V. of A
= (\(\hat{i}\) + \(\frac{\hat{j}}{2}\) + 4 \(\hat{k}\)) – (\(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\))
= 0 \(\hat{i}\) + \(\frac{3}{2}\) \(\hat{j}\) + 2 \(\hat{k}\)
Here
[\(\overrightarrow{\mathrm{AB}}\), \(\overrightarrow{\mathrm{AC}}\), \(\overrightarrow{\mathrm{AD}}\)]
= | 5 12 0
0 3 4
0 \(\frac{3}{2}\) 2|;
Expanding along R₁
= 5(6 – 6) – 12(0 – 0) + 0 = 0
Thus \(\overrightarrow{\mathrm{AB}}\), \(\overrightarrow{\mathrm{AC}}\) and \(\overrightarrow{\mathrm{AD}}\) are coplanar vectors. Therefore, the given points A, B, C and D are coplanar.

Question 12.
(i) Show that the points A(-1, 4, -3), B(3, 2, -5), C(-3, 8, -5) and (-3, 2, 1) are coplanar.
(ii) Show that the four points \(-\hat{a}\) + 4 \(\hat{b}\) – 3 \(\hat{k}\), 3 \(\hat{a}\) + 2 \(\hat{b}\) – 5 \(\hat{c}\), -3 \(\hat{a}\) + 8 \(\hat{b}\) – 5 \(\hat{c}\) and -3\(\hat{a}\) + 2 \(\hat{b}\) + \(\hat{c}\) are coplanar.
Answer:
(i) Here P.V. of A = \(-\hat{i}\) + 4 \(\hat{j}\) – 3 \(\hat{k}\);
P.V. of B = 3 \(\hat{i}\) + 2 \(\hat{j}\) – 5 \(\hat{k}\);
P.V. of C = -3 \(\hat{i}\) + 8 \(\hat{j}\) – 5 \(\hat{k}\);
P.V. of D = -3 \(\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\)
∴ \(\overrightarrow{\mathrm{AB}}\) = P.V. of B – P.V of A
=(3 \(\hat{i}\) + 2 \(\hat{j}\) – 5 \(\hat{k}\))-(-\(\hat{i}\) + 4 \(\hat{j}\) – 3 \(\hat{k}\))
= 4 \(\hat{i}\) – 2 \(\hat{j}\) – 2 \(\hat{k}\)
\(\overrightarrow{\mathrm{AC}}\) = P.V. of C – P.V. of A
= (-3 \(\hat{i}\) + 8 \(\hat{j}\) – 5 \(\hat{k}\)) – (\(-\hat{i}\) + 4 \(\hat{j}\) – 3 \(\hat{k}\))
= -2 \(\hat{i}\) + 4 \(\hat{j}\) – 2 \(\hat{k}\)
\(\overrightarrow{\mathrm{AD}}\) = P.V. of D – P.V of A
= (-3 \(\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\)) – (\(-\hat{i}\) + 4 \(\hat{j}\) – 3 \(\hat{k}\))
= -2 \(\hat{i}\) – 2 \(\hat{j}\) + 4 \(\hat{k}\)
Here
[\(\overrightarrow{\mathrm{AB}}\), \(\overrightarrow{\mathrm{AC}}\), \(\overrightarrow{\mathrm{AD}}\)]
= | 4 -2 -2
-2 4 -2
-2 -2 4|;
Expanding along R₁
= 4(16 – 4) + 2(-8 – 4) – 2(4 + 8)
= 48 – 24 – 24 = 0
∴ \(\overrightarrow{\mathrm{AB}}\), \(\overrightarrow{\mathrm{AC}}\) and \(\overrightarrow{\mathrm{AD}}\) are coplanar and hence given points A, B, C and D are coplanar.

(ii) Let A, B, C and D are given points and \(\vec{\alpha}\), \(\vec{\beta}\), \(\vec{\gamma}\) and \(\vec{\delta}\) be their respective position vectors.
∴ \(\overrightarrow{\mathrm{AB}}\) = \(\vec{\beta}\) – \(\vec{\alpha}\)
= (3 \(\vec{a}\) + 2 \(\vec{b}\) – 5 \(\vec{c}\)) – (\(-\vec{a}\) + 4 \(\vec{b}\) – 3 \(\vec{c}\))
= 4 \(\vec{a}\) – 2 \(\vec{b}\) – 2 \(\vec{c}\)
\(\overrightarrow{\mathrm{AC}}\) = \(\vec{\gamma}\) – \(\vec{\alpha}\)
= (-3 \(\vec{a}\) + 8 \(\vec{b}\) – 5 \(\vec{c}\)) – (\(-\vec{a}\) + 4 \(\vec{b}\) – 3 \(\vec{c}\))
= -2 \(\vec{a}\) + 4 \(\vec{b}\) + 2 \(\vec{c}\)
\( \overrightarrow{\mathrm{AD}}\) = \(\vec{\delta}\) – \(\vec{\alpha}\)
= (-3 \(\vec{a}\) + 2 \(\vec{b}\) + \(\vec{c}\)) – (\(-\vec{a}\) + 4 \(\vec{b}\) – 3 \(\vec{c}\))
= -2 \(\vec{a}\) – 2 \(\vec{b}\) + 4 \(\vec{c}\)
Here \(\overrightarrow{\mathrm{AB}}\), \(\overrightarrow{\mathrm{AC}}\), \(\overrightarrow{\mathrm{AD}}\)]
= |4 -2 -2
-2 4 -2
-2 -2 4| ;
Expanding along R₁
= 4(16 – 4) + 2(-8 – 4) -2(4 + 8)
= 48 – 24 – 24 = 0
Here
[\(\overrightarrow{\mathrm{AB}}\), \(\overrightarrow{\mathrm{AC}}\), \(\overrightarrow{\mathrm{AD}}\)]
= |4 -2 -2
-2 4 -2
-2 -2 4|;
Expanding along R₁
= 4(16 – 4) + 2(-8 – 4) – 2(4 + 8)
= 48 – 24 – 24 = 0
Thus \(\overrightarrow{\mathrm{AB}}\), \(\overrightarrow{\mathrm{AC}}\) and \(\overrightarrow{\mathrm{AD}}\) are coplanar.
∴ given points A, B, C and D are coplanar.

Question 13.
Find the value of λ so that the four points with position vectors \(-\hat{j}\) + \(\hat{k}\), 2 \(\hat{i}\) – \(\hat{j}\) – \(\hat{k}\), \(\hat{i}\) + λ \(\hat{j}\) + \(\hat{k}\) and 3 \(\hat{j}\) + 3 \(\hat{k}\) are coplanar.
Answer:
Let the given points be A, B, C and D whose position vectors are \(-\hat{j}\) + \(\hat{k}\);
2 \(\hat{i}\) – \(\hat{j}\) – \(\hat{k}\);
\(\hat{i}\) + λ \(\hat{j}\) + \(\hat{k}\) and 3 \(\hat{j}\) + 3 \(\hat{k}\)
\(\overrightarrow{\mathrm{AB}}\) = P.V. of B – P.V. of A
= (2 \(\hat{i}\) – \(\hat{j}\) – \(\hat{k}\)) – (\(-\hat{j}\) + \(\hat{k}\))
= 2 \(\hat{i}\) + 0 \(\hat{j}\) – 2 \(\hat{k}\)
\(\overrightarrow{\mathrm{AC}}\) = P.V. of C – P.V. of A
= (\(\hat{i}\) + λ \(\hat{j}\) + \(\hat{k}\)) – (\(-\hat{j}\) + \(\hat{k}\))
= \(\hat{i}\) + (λ + 1) \(\hat{j}\) + 0 \(\hat{k}\)
\(\overrightarrow{\mathrm{AD}}\) = P.V. of D – P.V. of A
= (3 \(\hat{j}\) + 3 \(\hat{k}\)) – (\(-\hat{j}\) + \(\hat{k}\))
= 4 \(\hat{j}\) + 2 \(\hat{k}\)
Since the points A, B, C and D are coplanar.
∴ \(\overrightarrow{\mathrm{AB}}\), \(\overrightarrow{\mathrm{AC}}\) and \(\overrightarrow{\mathrm{AD}}\) are coplanar.
∴ \(\overrightarrow{\mathrm{AB}}\), \(\overrightarrow{\mathrm{AC}}\), \(\overrightarrow{\mathrm{AD}}\)] = 0
⇒ |2 0 -2 1λ + 1 0 0 4 2| = 0;
Expanding along R₁
⇒ 2[2(λ + 1) – 0] + 0 – 2[4 – 0] =0
⇒ 4(λ + 1) – 8 = 0
⇒ 4λ – 4 = 0
⇒ λ = 1

Question 14.
Show that \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are coplanar if \(\vec{b}\) + \(\vec{c}\), \(\vec{c}\) + \(\vec{a}\), \(\vec{a}\) + \(\vec{b}\) and only if are coplanar.
Answer:
Given \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are coplanar.
∴ \(\vec{a}\) \(\vec{b}\) \(\vec{c}\) = 0
T.P. \(\vec{b}\) + \(\vec{c}\), \(\vec{c}\) + \(\vec{a}\), \(\vec{a}\) + \(\vec{b}\) are coplanar
i.e. we want to prove that [\(\vec{b}\) + \(\vec{c}\), \(\vec{c}\) + \(\vec{a}\), \(\vec{a}\) + \(\vec{b}\)] = 0
Now [\(\vec{b}\) + \(\vec{c}\), \(\vec{c}\) + \(\vec{a}\), \(\vec{a}\) + \(\vec{b}\)]
= (\(\vec{b}\) + \(\vec{c}\)) .[(\(\vec{c}\) + \(\vec{a}\)) × (\(\vec{a}\) + \(\vec{b}\))]}
= (\(\vec{b}\) + \(\vec{c}\)).[\(\vec{c}\) × \(\vec{a}\) + \(\vec{c}\) × \(\vec{b}\) + \(\vec{a}\) × \(\vec{a}\) + \(\vec{a}\) × \(\vec{b}\)]
= (\(\vec{b}\) + \(\vec{c}\)) [\(\vec{c}\) × \(\vec{a}\) – \(\vec{b}\) × \(\vec{c}\) + \(\vec{a}\) × \(\vec{b}\)]
= \(\vec{b}\).(\(\vec{c}\) × \(\vec{a}\)) – \(\vec{b}\).(\(\vec{b}\) × \(\vec{c}\)) + \(\vec{b}\) .(\(\vec{a}\) × \(\vec{b}\)) + \(\vec{c}\).(\(\vec{c}\) × \(\vec{a}\)) – \(\vec{c}\).(\(\vec{b}\) × \(\vec{c}\)) + \(\vec{c}\).(\(\vec{a}\) × \(\vec{b}\))
= [\(\vec{b}\) \(\vec{c}\) \(\vec{a}\)] – [\(\vec{b}\) \(\vec{b}\) \(\vec{c}\)] + [\(\vec{b}\) \(\vec{a}\) \(\vec{b}\)] + [\(\vec{c}\) \(\vec{c}\) \(\vec{a}\)] – [\(\vec{c}\) \(\vec{b}\) \(\vec{c}\)] + [\(\vec{c}\) \(\vec{a}\) \(\vec{b}\)]
= [\(\vec{b}\) \(\vec{c}\) \(\vec{a}\)] + [\(\vec{c}\) \(\vec{a}\) \(\vec{b}\)]
[since scalar triple product containing two same vectors is 0 ]
= 2[\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)]
[∵ [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = [\(\vec{b}\) \(\vec{c}\) \(\vec{a}\)] = [\(\vec{c}\) \(\vec{a}\) \(\vec{b}\)].
Converse : \(\vec{b}\) + \(\vec{c}\), \(\vec{c}\) + \(\vec{a}\), \(\vec{a}\) + \(\vec{b}\) are coplanar.
T.P. \(\vec{a}\) \(\vec{b}\) \(\vec{c}\) are coplanar.
⇒ [\(\vec{b}\) + \(\vec{c}\), \(\vec{c}\) + \(\vec{a}\), \(\vec{a}\) + \(\vec{b}\)] = 0
⇒ 2[\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = 0
[∵ [\(\vec{b}\) + \(\vec{c}\), \(\vec{c}\) + \(\vec{a}\), \(\vec{a}\) + \(\vec{b}\)]
= 2[\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)]
⇒ [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = 0
∴ \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are coplanar.

Question 15.
Simplify :
(i) \(\vec{a}\).(\(\vec{a}\) × \(\vec{b}\))
(ii) (\(\vec{b}\) × \(\vec{c}\)) .[(\(\vec{c}\) + \(\vec{a}\)) × (\(\vec{a}\) + \(\vec{b}\))]
Answer:
(i) \(\vec{a}\) .(\(\vec{a}\) × \(\vec{b}\)) = (\(\vec{a}\) × \(\vec{a}\)).\(\vec{b}\)
= \(\overrightarrow{0}\) .\(\vec{b}\) = 0
[since dot and cross are interchangeable to each other]

(ii) Given \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are coplanar.
∴ [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = 0 ………….(1)
T.P. \(\vec{b}\) + \(\vec{c}\), \(\vec{c}\) + \(\vec{a}\), \(\vec{a}\) + \(\vec{b}\) are coplanar
i.e. we want to prove that [\(\vec{b}\) + \(\vec{c}\), \(\vec{c}\) + \(\vec{a}\), \(\vec{a}\) + \(\vec{b}\)] = 0
Now [\(\vec{b}\) + \(\vec{c}\), \(\vec{c}\) + \(\vec{a}\), \(\vec{a}\) + \(\vec{b}\)]
= (\(\vec{b}\) + \(\vec{c}\)) [(\(\vec{c}\) + \(\vec{a}\)) ×(\(\vec{a}\) + \(\vec{b}\))]
= (\(\vec{b}\) + \(\vec{c}\)) .[\(\vec{c}\) × \(\vec{a}\) + \(\vec{c}\) × \(\vec{b}\) + \(\vec{a}\) × \(\vec{a}\) + \(\vec{a}\) × \(\vec{b}\)]
= (\(\vec{b}\) + \(\vec{c}\)).[\(\vec{c}\) × \(\vec{a}\) – \(\vec{b}\) × \(\vec{c}\) + \(\vec{a}\) × \(\vec{b}\)]
= \(\vec{b}\) .(\(\vec{c}\) × \(\vec{a}\)) – \(\vec{b}\) .(\(\vec{b}\) × \(\vec{c}\)) + \(\vec{b}\) .(\(\vec{a}\) × \(\vec{b}\)) + \(\vec{c}\) .(\(\vec{c}\) × \(\vec{a}\)) – \(\vec{c}\) .(\(\vec{b}\) × \(\vec{c}\)) + \(\vec{c}\) .(\(\vec{a}\) × \(\vec{b}\))
= [\(\vec{b}\) \(\vec{c}\) \(\vec{a}\)] – [\(\vec{b}\) \(\vec{b}\) \(\vec{c}\)] + [\(\vec{b}\) \(\vec{a}\) \(\vec{b}\)]
+ [\(\vec{c}\) \(\vec{c}\) \(\vec{a}\)] – [\(\vec{c}\) \(\vec{b}\) \(\vec{c}\)] + [\(\vec{c}\) \(\vec{a}\) \(\vec{b}\)]
= [ \(\vec{b}\) \(\vec{c}\) \(\vec{a}\)] + [\(\vec{c}\) \(\vec{a}\) \(\vec{b}\)]
[since scalar triple product containing two same vectors is 0] [using eqn. (1)]
= 2[\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = 0
[∵ [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = [\(\vec{b}\) \(\vec{c}\) \(\vec{a}\)] = [\(\vec{c}\) \(\vec{a}\) \(\vec{b}\)].

Question 16.
Find a unit vector coplanar with \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\) and \(\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\) perpendicular to \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\).
Answer:
Let the required vector \(\vec{a}\) = x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)
since it is given that \(\vec{a}\) is coplanar with \(\vec{b}\) and \(\vec{c}\);
where \(\vec{b}\) = \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\) and
\(\vec{c}\) = \(\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\)
∴ [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = 0
⇒ |x y z 1 1 2 1 2 1| = 0;
Expanding along R₁;
⇒ x(1 – 4) – y(1 – 2) + z(2 – 1) = 0
⇒ -3 x + y + z = 0
also \(\vec{a}\) is ⊥ to \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)
∴ \(\vec{a}\) .(\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)) = 0
⇒ (x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)) .(\(\hat{i}\) +\(\hat{j}\) + \(\hat{k}\)) = 0
⇒ x + y + z = 0
eqn. (1) – eqn. (2) gives; -4 x = 0
⇒ x = 0 ∴ from (1); z = -y
∴ required vector \(\vec{a}\) = 0 \(\hat{i}\) + y \(\hat{j}\) – y \(\hat{k}\)
= y(\(\hat{j}\) – \(\hat{k}\))
Thus required unit vector = \(\frac{y(\hat{j}-\hat{k})}{\sqrt{y^2+y^2}}\)
= ± \(\frac{(\hat{j}-\hat{k})}{\sqrt{2}}\)

Question 17.
If are three non-coplanar vectors, then prove that = 0.
Answer:

Examples:

Question 1.
The diagonals A C and B D of a parallelogram A B C D are represented by the vectors and respectively. Find the area of the parallelogram.
Answer:
Given \(\overrightarrow{\mathrm{AC}}\) = \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\);
\(\overrightarrow{\mathrm{BD}}\) = \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)
∴ \(\overrightarrow{\mathrm{AC}}\) × \(\overrightarrow{\mathrm{BD}}\)
= |\(\hat{i}\) \(\hat{j}\) \(\hat{k}\)
1 1 1
1 -1 1|
= \(\hat{i}\)(1 + 1) – \(\hat{j}\)(1 – 1) + \(\hat{k}\)(-1 – 1)
= 2 \(\hat{i}\) + 0 \(\hat{j}\) – 2 \(\hat{k}\)
⇒|\(\overrightarrow{\mathrm{AC}}\) × \(\overrightarrow{\mathrm{BD}}\)|
= \(\sqrt{2^2+0^2+(-2)^2}\)
= \(\sqrt{8}\) = 2 \(\sqrt{2}\)
∴ required area of parallelogram
= \(\frac{1}{2}\)|\(\overrightarrow{\mathrm{AC}}\) × \(\overrightarrow{\mathrm{BD}}\)|
= \(\frac{1}{2}\) × 2 \(\sqrt{2}\)
= \(\sqrt{2}\) sq. units.

Question 2.
Determine the value of λ so that the vectors 2 \(\hat{i}\) – 3 \(\hat{j}\) + \(\hat{k}\) \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\) and \(\hat{j}\) + λ\(\hat{k}\) are coplanar.
Answer:
Let \(\vec{a}\) = 2 \(\hat{i}\) – 3 \(\hat{j}\) + \(\hat{k}\) ;
\(\vec{b}\) = \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\) and
\(\vec{c}\) = \(\hat{j}\) + λ\(\hat{k}\)
Since \(\vec{a}\) and \(\vec{b}\), \(\vec{c}\) are coplanar.
∴ [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = 0
⇒ |2 -3 1
1 2 -3
0 1 λ| = 0;
Expanding along R₁
⇒ 2(2λ + 3) + 3(λ – 0) + 1(1 – 0) = 0
⇒ 4λ + 6 + 3 λ + 1 = 0
⇒ 7 λ + 7 = 0
⇒ λ = -1

Question 3.
The position vectors of the vertices A, B and C of a triangle A B C are \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\). Show that the vector area of the triangle is \(\frac{1}{2}\)[\(\vec{b}\) × \(\vec{c}\) + \(\vec{c}\) × \(\vec{a}\) + \(\vec{a}\) × \(\vec{b}\)]
Answer:
Let δ be the area of triangle and \(\hat{n}\) be the unit vector ⊥ to plane of the triangle considered in anticlockwise direction from \(\overrightarrow{\mathrm{AB}}\) to \(\overrightarrow{\mathrm{AC}}\).
So \(\overrightarrow{\mathrm{AB}}\), \(\overrightarrow{\mathrm{AC}}\), \(\hat{n}\) forms a right handed vector triad.
∴ Vector area of ∆ABC

= \(\frac{1}{2}\) b c sin \(\mathrm{A}\). \(\hat{n}\)
= \(\frac{1}{2}\)|\(\overrightarrow{\mathrm{AB}}\)||\(\overrightarrow{\mathrm{AC}}\)| sin \(\mathrm{A}\) \(\hat{n}\)
= \(\frac{1}{2}\)(\(\overrightarrow{\mathrm{AB}}\) × \(\overrightarrow{\mathrm{AC}}\))
= \(\frac{1}{2}\)[(\(\vec{b}\) – \(\vec{a}\)) × (\(\vec{c}\) – \(\vec{a}\))]
= \(\frac{1}{2}\)[\(\vec{b}\) × \(\vec{c}\) – \(\vec{b}\) × \(\vec{a}\) – \(\vec{a}\) ×\(\vec{c}\) + \(\vec{a}\) × \(\vec{a}\)]
= \(\frac{1}{2}\)[\(\vec{b}\) × \(\vec{c}\) + \(\vec{a}\) × \(\vec{b}\) + \(\vec{c}\) × \(\vec{a}\)]

Question 4.
Given \(\vec{a}\) = 3 \(\hat{i}\) – \(\hat{j}\), \(\vec{b}\) = 2 \(\hat{i}\) + \(\hat{j}\) – 3 \(\hat{k}\) express \(\vec{b}\) as \(\overrightarrow{b_1}\) + \(\overrightarrow{b_2}\) where \(\overrightarrow{b_1}\) is parallel to a and b_2 is perpendicular to \(\vec{a}\).
Answer:
Given \(\vec{a}\) = 3 \(\hat{i}\) – \(\hat{j}\) and \(\vec{b}\)
= 2 \(\hat{i}\) + \(\hat{j}\) – 3 \(\hat{k}\)
Now we want to express \(\vec{b}\) = \(\overrightarrow{b_1}\) + \(\overrightarrow{b_2}\)
Now \(\vec{b}_1\) is parallel to \(\vec{a}\)
⇒ \(\overrightarrow{b_1}\)
= λ \(\vec{a}\) for some non-zero scalar
λ \(\overrightarrow{b_1}\) = λ(3 \(\hat{i}\) – \(\hat{j}\))
Now \(\vec{b}_2\) is given to be ⊥ to \(\vec{a}\)
⇒ \(\overrightarrow{b_2}\) . \(\vec{a}\) = 0
⇒ (\(\vec{b}\) – \(\vec{b}_1\)) .\(\vec{a}\) = 0 [using (1)]
⇒ \(\vec{b}\) .\(\vec{a}\) – \(\vec{b}_1\) .\(\vec{a}\) = 0
⇒ (2 \(\hat{i}\) + \(\hat{j}\) – 3 \(\hat{k}\)) .(3 \(\hat{i}\) – \(\hat{j}\)) – λ(3 \(\hat{i}\) – \(\hat{j}\)) .(3 \(\hat{i}\) – \(\hat{j}\)) = 0
⇒ 6 – 1 – λ(3 + 1) = 0
⇒ 5 = 4 λ
⇒ λ = \(\frac{5}{4}\)
∴ from (2) ; \(\overrightarrow{b_1}\) = \(\frac{5}{4}\)(3 \(\hat{i}\) – \(\hat{j}\))
∴ from (1) ; \(\overrightarrow{b_2}\) = \(\vec{b}\) – \(\overrightarrow{b_1}\)
⇒ \(\overrightarrow{b_2}\) = (2 \(\hat{i}\) + \(\hat{j}\) – 3 \(\hat{k}\)) – (\(\frac{15}{4}\) \(\hat{i}\) – \(\frac{5}{4}\) \(\hat{j}\))
⇒ \(\overrightarrow{b_2}\) = \(-\frac{7}{4}\) \(\hat{i}\) + \(\frac{9}{4}\) \(\hat{j}\) – 3 \(\hat{k}\)
Thus, \(\vec{b}\) = \(\frac{5}{4}\)(3 \(\hat{i}\) – \(\hat{j}\)) + (\(-\frac{7}{4}\) \(\hat{i}\) + \(\frac{9}{4}\) \(\hat{j}\) – 3 \(\hat{k}\))

Question 5.
Find the volume of the parallelopiped whose edges are represented by
\(\vec{a}\) = 2 \(\hat{i}\) – 3 \(\hat{j}\) + 4 \(\hat{k}\), \(\vec{b}\) = \(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\) and \(\vec{c}\) = 3 \(\hat{i}\) –\(\hat{j}\) + 2 \(\hat{k}\)
Answer:
Given \(\vec{a}\) = 2 \(\hat{i}\) – 3 \(\hat{j}\) + 4 \(\hat{k}\) ;
\(\vec{b}\) = \(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\) and
\(\vec{c}\) = 3 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\)
Here, [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)]
= |2 -3 4 1 2 -1 3 -1 2|
= 2(4 – 1) + 3(2 + 3) + 4(-1 – 6)
= 6 + 15 – 28
= -7

Question 6.
Find the unit vector perpendicular to the plane of vectors \(\vec{a}\) and \(\vec{b}\) where
\(\vec{a}\) = \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\) and \(\vec{b}\) = \(\hat{i}\) + 2 \(\hat{j}\)
Answer:
Here \(\vec{a}\) × \(\vec{b}\) = |\(\hat{i}\) \(\hat{j}\) \(\hat{k}\) 1 -1 1 1 2 0|
= \(\hat{i}\)(0 – 2) – \(\hat{j}\)(0 – 1) + \(\hat{k}\)(2 + 1)
= -2 \(\hat{i}\) + \(\hat{j}\) + 3 \(\hat{k}\)
∴ |\(\vec{a}\) × \(\vec{b}\)|
= \(\sqrt{(-2)^2+1^2+3^2}\)
= \(\sqrt{14}\)
Thus, required unit vector ⊥ to plane of \(\vec{a}\) and \(\vec{b}\)
= \(\frac{\vec{a} × \vec{b}}{|\vec{a} ×\vec{b}|}\)
= \(\frac{-2 \hat{i}+\hat{j}+3 \hat{k}}{\sqrt{14}}\)

Question 7.
Three ectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are given which are mutually perpendicular and |\(\vec{a}\)| = |\(\vec{b}\)| = |\(\vec{c}\)| = 1. Show that the vector \(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\) is equally inclined to \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\).
Answer:
Since \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are mutually perpendicular.
∴ \(\vec{a}\) .\(\vec{b}\) = \(\vec{b}\).\(\vec{c}\)
= \(\vec{c}\) . \(\vec{a}\) = 0 and |\(\vec{a}\)| = |\(\vec{b}\)| = |\(\vec{c}\)| = 1
Let α, β and γ be the angle made by \(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\) with \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) respectively.
Now
(a + \(\vec{b}\) + \(\vec{c}\))^2
= ( \(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\)) .(\(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\))
= |\(\vec{a}\)|^2 + |\(\vec{b}\)|^2 + |\(\vec{c}\)|^2 + 2(\(\vec{a}\) . \(\vec{b}\) + \(\vec{b}\) . \(\vec{c}\) + \(\vec{c}\) . \(\vec{a}\))
= 1 + 1 + 1 = 3
|\(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\)| = \(\sqrt{3}\)
⇒ |\(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\)| = \(\sqrt{3}\)
∴ cos α = \(\frac{(\vec{a}+\vec{b}+\vec{c}) . \vec{a}}{|\vec{a}+\vec{b}+\vec{c}||\vec{a}|}\)
= \(\frac{|\vec{a}|^2+\vec{b} . \vec{a}+\vec{c} . \vec{a}}{\sqrt{3} . 1}\)
= \(\frac{1+0+0}{\sqrt{3}}\)
= \(\frac{1}{\sqrt{3}}\)
cos β = \(\frac{(\vec{a}+\vec{b}+\vec{c}) . \vec{b}}{|\vec{a}+\vec{b}+\vec{c}||\vec{b}|}\)
= \(\frac{\vec{a} . \vec{b}+|\vec{b}|^2+\vec{b} . \vec{c}}{\sqrt{3} \times 1}\)
= \(\frac{1}{\sqrt{3}}\)
cos γ = \(\frac{(\vec{a}+\vec{b}+\vec{c}) . \vec{c}}{|\vec{a}+\vec{b}+\vec{c}||\vec{c}|}\)
= \(\frac{\vec{a} . \vec{c}+\vec{b} . \vec{c}+|\vec{c}|^2}{\sqrt{3} \times 1}\)
= \(\frac{0+0+1}{\sqrt{3}}\)
= \(\frac{1}{\sqrt{3}}\)
∴ cos α = cos β = cos γ = \(\frac{1}{\sqrt{3}}\)
⇒ α = β = γ
Hence \(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\) is equally inclined to \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\).

Question 8.
Write a unit vector to the plane of two vectors \(\vec{x}\) and \(\vec{y}\), where
\(\vec{x}\) = 5 \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\) and \(\vec{y}\) = \(-\hat{i}\) – 3 \(\hat{j}\) – \(\hat{k}\)
Answer:
Given \(\vec{x}\) = 5\( \hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\) and \(\vec{y}\)
= \(-\hat{i}\) – 3 \(\hat{j}\) – \(\hat{k}\)
∴ \(\vec{x}\) × \(\vec{y}\)
= |\(\hat{i}\) \(\hat{j}\) \(\hat{k}\)
5 2 -3
-1 -3 -1|
= \(\hat{i}\)(-2 – 9) – \(\hat{j}\)(-5 – 3) + \(\hat{k}\)(-15 + 2)
= -11 \(\hat{i}\) + 8 \(\hat{j}\) – 13 \(\hat{k}\)
⇒ |\(\vec{x}\) × \(\vec{y}\)|
= \(\sqrt{(-11)^2+8^2+(-13)^2}\)
= \(\sqrt{121+64+169}\) = \(\sqrt{354}\)
Thus required unit vector ⊥ to plane of \(\vec{x}\) and \(\vec{y}\)
= \(\frac{\vec{x} \times \vec{y}}{|\vec{x} \times \vec{y}|}\)
= \(\frac{-11 \hat{i}+8 \hat{j}-13 \hat{k}}{\sqrt{354}}\)

Question 9.
A particle acted upon by constant forces 4 \(\hat{i}\) – \(\hat{j}\) – 3 \(\hat{k}\) and 3 \(\hat{i}\) + \(\hat{j}\)– \(\hat{k}\) is displaced from the point \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\) to the point 5 \(\hat{i}\) – 4 \(\hat{j}\) – \(\hat{k}\). Find the total work done by the forces.
Answer:
Here \(\overrightarrow{\mathrm{F}}\) = (4 \(\hat{i}\) – \(\hat{j}\) – 3 \(\hat{k}\)) + (3 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\))
= 7 \(\hat{i}\) + 0 \(\hat{j}\) – 4 \(\hat{k}\)
∴ Displacement \(\vec{d}\) = (5 \(\hat{i}\) – 4 \(\hat{j}\) – \(\hat{k}\)) – (\(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\))
= 4 \(\hat{i}\) – 6 \(\hat{j}\) + 2 \(\hat{k}\)
∴ Required work done = \(\overrightarrow{\mathrm{F}}\) .\(\vec{d}\)
= (7 \(\hat{i}\) – 4 \(\hat{k}\)).(4 \(\hat{i}\) – 6 \(\hat{j}\) + 2 \(\hat{k}\))
= 7(4) + 0(-6) – 4(2)
= 28 – 8
= 20

Question 10.
Find the area of the parallelogram A B C D where diagonal A C and B D are represented by the vectors 3 \(\hat{i}\) + \(\hat{j}\) – 2 \(\hat{k}\) and \(\hat{i}\) – 3 \(\hat{j}\) + 4 \(\hat{k}\) respectively.
Answer:
Here \(\overrightarrow{\mathrm{AC}}\) = 3 \(\hat{i}\) + \(\hat{j}\) – 2 \(\hat{k}\) and
\(\overrightarrow{\mathrm{BD}}\) = \(\hat{i}\) – 3 \(\hat{j}\) + 4 \(\hat{k}\)
∴ \(\overrightarrow{\mathrm{AC}}\) × \(\overrightarrow{\mathrm{BD}}\)
= |\(\hat{i}\) \(\hat{j}\) \(\hat{k}\)
3 1 -2
1 -3 4|
= \(\hat{i}\)(4 – 6) – \(\hat{j}\)(12 + 2) + \(\hat{k}\)(-9 – 1)
= -2 \(\hat{i}\) – 14 \(\hat{j}\) – 10 \(\hat{k}\)
⇒ |\(\overrightarrow{\mathrm{AC}}\) × \(\overrightarrow{\mathrm{BD}}\)|
= \(\sqrt{(-2)^2+(-14)^2+(-10)^2}\)
= \(\sqrt{4+196+100}\)
= \(\sqrt{300}\)
= 10 \(\sqrt{3}\)
∴ required area of parallelopiped = \(\frac{1}{2}\)|\(\overrightarrow{\mathrm{AC}}\) × \(\overrightarrow{\mathrm{BD}}\)|
= 5 \(\sqrt{3}\) cubic units.

Question 11.
The vectors 2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\), \(\hat{i}\) – 3 \(\hat{j}\) – 5 \(\hat{k}\) and 3 \(\hat{i}\) – 4 \(\hat{j}\) – 4 \(\hat{k}\) are the position vectors of the vertices A, B and C respectively of the triangle A B C. Prove that A B C is a right-angled triangle.
Answer:
Let \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) respectively are the position vectors of points A, B and C of ∆ ABC.
∴ \(\vec{a}\) = 2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\) ;
\(\vec{b}\) = \(\hat{i}\) – 3 \(\hat{j}\) – 5 \(\hat{k}\) and
\(\vec{c}\) = 3 \(\hat{i}\) – 4 \(\hat{j}\) – 4 \(\hat{k}\)
\(\overrightarrow{\mathrm{AB}}\) = \(\vec{b}\) – \(\vec{a}\)
= \(-\hat{i}\) – 2 \(\hat{j}\) – 6 \(\hat{k}\) ;
\(\overrightarrow{\mathrm{BC}}\) = \(\vec{c}\) – \(\vec{b}\) = 2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\) ;
\(\overrightarrow{\mathrm{CA}}\) = \(\vec{a}\) – \(\vec{c}\)
= \(-\hat{i}\) + 3 \(\hat{j}\) + 5 \(\hat{k}\)
Here \(\overrightarrow{\mathrm{AB}}\) + \(\overrightarrow{\mathrm{BC}}\) + \(\overrightarrow{\mathrm{CA}}\)
= \(\overrightarrow{0}\)
Clearly A, B, C forms the vertices of triangle
|\(\overrightarrow{\mathrm{AB}}\)| = \(\sqrt{(-1)^2+(-2)^2+(-6)^2}\)
= \(\sqrt{41}\) ;
|\(\overrightarrow{\mathrm{CA}}\)| = \(\sqrt{(-1)^2+3^2+5^2}\)
= \(\sqrt{35}\)
|\(\overrightarrow{\mathrm{BC}}\)| = \(\sqrt{2^2+(-1)^2+1^2}\)
= \(\sqrt{6}\)
Clearly AB² = BC² + CA²
∴ Pythagoras theorem holds for ∆ ABC.
Thus ∆ ABC is right angled triangle.

Question 12.
If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\), \(\vec{d}\) are the position vectors of four points A, B, C, D respectively, and if \(\vec{b}\) – \(\vec{a}\) = \(\vec{c}\) – \(\vec{d}\), show that ABCD is a parallelogram.
Answer:
Given \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) and \(\vec{d}\) are the position vectors of four points A, B, C and D respectively.
Given \(\vec{b}\) – \(\vec{a}\) = \(\vec{c}\) – \(\vec{d}\)
⇒ \(\frac{\vec{b}+\vec{d}}{2}\) = \(\frac{\vec{c}+\vec{a}}{2}\)
Thus mid-point of \(\overrightarrow{\mathrm{BD}}\) = mid-point of \(\overrightarrow{\mathrm{AC}}\).
∴ diagonals \(\overrightarrow{\mathrm{AC}}\) and \(\overrightarrow{\mathrm{BD}}\) bisect each other.
Hence A B C D is a parallelogram.

Question 13.
If D, E, F are the mid-points of the sides B C, C A, A B respectively of a triangle A B C, show that the area of ∆DEF = \(\frac{1}{4}\) (area of ∆ ABC)
Answer:
Let \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are the position vectors of points A, B and C of ∆ ABC.
Since D, E and F respectively are the mid-points of sides BC, CA and AB of ∆ A B C.

Question 14.
The vectors \(\vec{a}\) = 3 \(\hat{i}\) + x \(\hat{j}\) – \(\hat{k}\) and \(\vec{b}\) = 2 \(\hat{i}\) + \(\hat{j}\) + y \(\hat{k}\) are mutually perpendi- cular. Given that |\(\vec{a}\)| = |\(\vec{b}\)|, find the values of x and y.
Answer:
Given \(\vec{a}\) = 3 \(\hat{i}\) + x \(\hat{j}\) – \(\hat{k}\) and
\(\vec{b}\) = 2 \(\hat{i}\) + \(\hat{j}\) + y \(\hat{k}\) since \(\vec{a}\) and \(\vec{b}\) are mutually perpendicular.
∴ \(\vec{a}\).\(\vec{b}\) = 0
⇒ (3 \(\hat{i}\) + x \(\hat{j}\) – \(\hat{k}\)).(2 \(\hat{i}\) + \(\hat{j}\) + y \(\hat{k}\)) = 0
⇒ 6 + x – y = 0
⇒ x – y = -6
Given |\(\vec{a}\)| = |\(\vec{b}\)|
⇒ \(\sqrt{9+x^2+1}\) = \(\sqrt{4+1+y^2}\)
⇒ \(\sqrt{10+x^2}\) = \(\sqrt{5+y^2}\);
on squaring both sides, we have
10 + x² = 5 + y²
⇒ x² – y² = -5
⇒ (x – y)(x + y) = -5
⇒ x + y = \(\frac{5}{6}\) [using (1)]
on adding (1) and (2); we have
2 x = -6 + \(\frac{5}{6}\) = \(-\frac{31}{6}\)
⇒ x = \(-\frac{31}{12}\)
y = \(\frac{5}{6}\) + \(\frac{31}{12}\)
= \(\frac{41}{12}\) from (2) ;
y = \(\frac{5}{6}\) + \(\frac{31}{12}\) = \(\frac{41}{12}\)

Question 15.
The position vectors of the vertices of a triangle are given as 2 \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\), \(\hat{i}\) – 3 \(\hat{j}\) – 5 \(\hat{k}\), and 3 \(\hat{i}\) – 4 \(\hat{j}\) – 4 \(\hat{k}\). Prove that it is a right-angled triangle.
Answer:

Question 16.
Find the volume of the parallelogram whose co-terminous edges are repre-sented by the vectors \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\), \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\) and \(\hat{i}\) + 3 \(\hat{j}\) – \(\hat{k}\).
Answer:
Let \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are the edges represented by the vectors \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\), \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\) and \(\hat{i}\) + 3 \(\hat{j}\) – \(\hat{k}\).
∴ [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = |1 1 1
1 -1 1
1 3 -1| ;
Expanding along R₁
= 1(1 – 3) – 1(-1 – 1) + 1(3 + 1) = -2 + 2 + 4 = 4

Question 17.
If \(\vec{a}\) and \(\vec{b}\) are two vectors such that |\(\vec{a}\) + \(\vec{b}\)| = |\(\vec{a}\)| prove that (2 \(\vec{a}\) + \(\vec{b}\)) is perpendicular to \(\vec{b}\).
Answer:
Given |\(\vec{a}\) + \(\vec{b}\)| = |\(\vec{a}\)|
⇒ |\(\vec{a}\) + \(\vec{b}\)|² = |\(\vec{a}\)|²
⇒ (\(\vec{a}\) + \(\vec{b}\)).(\(\vec{a}\) + \(\vec{b}\)) = |\(\vec{a}\)|²
[∵ \(\vec{a}\) . \(\vec{a}\) = \(\vec{a}\)² = |\(\vec{a}\)|²
⇒ \(\vec{a}\) . \(\vec{a}\) + \(\vec{a}\) \(\vec{b}\) + \(\vec{b}\) .\(\vec{a}\) + \(\vec{b}\).\(\vec{b}\) = |\(\vec{a}\)|²
⇒ |\(\vec{a}\)|² + 2 \(\vec{a}\) .\(\vec{b}\) + |\(\vec{b}\)|² = |\(\vec{a}\)|²
{[∵ \(\vec{a}\) . \(\vec{b}\) = \(\vec{b}\) . \(\vec{a}\)]}
⇒ 2 \(\vec{a}\) . \(\vec{b}\) + \(\vec{b}\) . \(\vec{b}\) = 0
⇒ (2 \(\vec{a}\) + \(\vec{b}\)) . \(\vec{b}\) = 0
Thus 2 \(\vec{a}\) + \(\vec{b}\) is perpendicular to \(\vec{b}\).

Question 18.
Find the unit vector perpendicular to the two vectors \(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\) and 2 \(\hat{i}\) + 3 \(\hat{j}\) + \(\hat{k}\).
Answer:
Let
\(\vec{a}\) = \(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\);
\(\vec{b}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) + \(\hat{k}\)
∴ \(\vec{a}\) × \(\vec{b}\)
= |\(\hat{i}\) \(\hat{j}\) \(\hat{k}\)
1 2 -1
2 3 1|
= \(\hat{i}\)(2 + 3) – \(\hat{j}\)(1 + 2) + \(\hat{k}\)(3 – 4)
= \(\hat{i}\) – 3 \(\hat{j}\) – \(\hat{k}\)
and |\(\vec{a}\) × \(\vec{b}\)|
= \(\sqrt{5^2+(-3)^2+(-1)^2}\)
= \(\sqrt{35}\)
Thus required unit vector ⊥ to \(\vec{a}\) and \(\vec{b}\)
= \(\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}\)
= \(\frac{5 \hat{i}-3 \hat{j}-\hat{k}}{\sqrt{35}}\)

Question 19.
If \(\vec{a}\) and \(\vec{b}\) are unit vectors such that 2 \(\vec{a}\) – 4 \(\vec{b}\) and 10 \(\vec{a}\) + 8 \(\vec{b}\) are perpendicular to each other, find the angle between the vectors \(\vec{a}\) and \(\vec{b}\).
Answer:

Question 20.
prove that \(\vec{a}\).(\(\vec{b}\) + \(\vec{c}\)) (\(\vec{a}\) + 2\(\vec{b}\) + 3\(\vec{c}\))
= [\(\vec{a}\)\(\vec{b}\)\(\vec{c}\)]
Answer:

Question 21.
The vectors -2 \(\hat{i}\) + 4 \(\hat{j}\) + 4 \(\hat{k}\) and -4 \(\hat{i}\) – 2 \(\hat{k}\) represent the diagonals BD and AC of a parallelogram ABCD. Find the area of the parallelogram.
Answer:

Question 22.
The vectors i + 3 j, 5 k and λ i – j are coplanar. Find the value of λ.
Answer:
Let \(\vec{a}\) = \(\hat{i}\) + 3 \(\hat{j}\); \(\vec{b}\) = 5 \(\hat{k}\) and \(\vec{c}\) = λ \(\hat{i}\) – \(\hat{j}\)
Here
\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)
= |1 3 0
0 0 5
λ -1 0| ;
Expanding along R₁
= 1(0 + 5) – 3(0 – 5 λ) + 0 = 5 + 15 λ
Since \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are coplanar.
∴ [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = 0
⇒ 5 + 15 λ = 0
⇒ λ = \(\frac{-1}{3}\)

Question 23.
If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are three vectors, show that (\(\vec{a}\) + \(\vec{b}\)) .(\(\vec{b}\) + \(\vec{c}\)) × (\(\vec{c}\) + \(\vec{a}\)) = 2[\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)]
Answer:

Question 24.
Find a unit vector perpendicular to the vector 4 i + 3 j + k and 2 i – j + 2 k. Determine the sine of the angle between these two vectors.
Answer:

Question 25.
Find \(\vec{a}\).\(\vec{b}\) if |\(\vec{a}\)| = 2, |\(\vec{b}\)| = 5 and |\(\vec{a}\) ×\(\vec{b}\)| = 8.
Answer:
Given |\(\vec{a}\)|= 2;
|\(\vec{b}\)| = 5 and |\(\vec{a}\) × \(\vec{b}\)| = 8
Since |\(\vec{a}\) × \(\vec{b}\)| = |\(\vec{a}\)||\(\vec{b}\)|
sinθ ⇒ sinθ
= \(\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}\)
= \(\frac{8}{2 \times 5}\) = \(\frac{4}{5}\)
∴ cosθ = \(\sqrt{1-\sin ^2 \theta}\)
= \(\sqrt{1-\frac{16}{25}}\) = \(\frac{3}{5}\)
Thus \(\vec{a}\).\(\vec{b}\)
= |\(\vec{a}\)||\(\vec{b}\)|
cos θ = 2 × 5 × \(\frac{3}{5}\) = 6

Question 26.
Find the value of λ for which the four points with position vectors 2 \(\hat{i}\) + 5 \(\hat{j}\) + \(\hat{k}\), \(-\hat{j}\) – 4 \(\hat{k}\), 3 \(\hat{i}\) + λ \(\hat{j}\) + 8 \(\hat{k}\) and -4 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\) are coplanar.
Answer:
Let A, B, C and D are four points whose position vectors are 2 \(\hat{i}[latex] + 5 [latex]\hat{j}\) + \(\hat{k}\), – \(\hat{j}\) – 4 \(\hat{k}\), 3 \(\hat{i}\) + λ \(\hat{j}\) + 8 \(\hat{k}\) and -4 \(\hat{i}\) + 3 \(\hat{j}\) + 3 \(\hat{k}\)
\(\overrightarrow{AB}}\) = P . V of B – P.V of A = \(-\hat{j}\) – 4 \(\hat{k}\) – (2 \(\hat{i}\) + 5 \(\hat{j}\) + \(\hat{k}\))
= -2 \(\hat{i}\) – 6 \(\hat{j}\) – 5 \(\hat{k}\)
\(\overrightarrow{AC}}\) = P . V of C – P.V . of A = 3 \(\hat{i}\) + λ \(\hat{j}\) + 8 \(\hat{k}\) – (2 \(\hat{i}\) + 5 \(\hat{j}\) + \(\hat{k}[latex])
= [latex]\hat{i}\) + (λ – 5) \(\hat{j}\) + 7 \(\hat{k}\)
\(\overrightarrow{AD}}\) = P. V of D – P.V of A
= (-4 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\)) – (2 \(\hat{i}\) + 5 \(\hat{j}\) + \(\hat{k}\))
= -6 \(\hat{i}\) – 2 \(\hat{j}\) + 3 \(\hat{k}\)
Now point A, B, C and D are coplanar.
∴ \(\overrightarrow{AB}}\), \(\overrightarrow{AC}}\) and \(\overrightarrow{AD}}\) are coplanar
⇒ [\(\overrightarrow{AB}}\), \(\overrightarrow{AC}}\), \(\overrightarrow{AD}}\)] = 0
∴ |-2 -6 -5
1 λ-5 7
-6 -2 3| = 0 ;
Expanding along R₁
⇒ -2[3(λ – 5) + 14] + 6[3 + 42] – 5[-2 + 6(λ – 5)] = 0
⇒ -2(3 λ – 1) + 6(45) – 5(6 λ – 32) = 0
⇒ -6 λ + 2 + 270 – 30 λ + 160 = 0
⇒ -36 λ + 432 = 0
⇒ λ = \(\frac{432}{36}\) = 12

Question 27.
For any three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) prove : \(\vec{a}\) – \(\vec{b}\) \(\vec{b}\) – \(\vec{c}\) \(\vec{c}\) – \(\vec{a}\)] = 0.
Answer:

Question 28.
If \(\vec{a}\) and \(\vec{b}\) are unit vectors and θ is the angle between them, then show that
|\(\vec{a}\) – \(\vec{b}\)| = 2 sin \(\frac{\theta}{2}\)
Answer:
Given |\(\vec{a}\)| = |\(\vec{b}\)| = 1 and
θ be the angle between \(\vec{a}\) and \(\vec{b}\)
Now
|\(\vec{a}\) – \(\vec{b}\)|^2 = (\(\vec{a}\) – \(\vec{b}\)).(\(\vec{a}\) – \(\vec{b}\))
= |\(\vec{a}|\)^2 – \(\vec{a}\) \(\vec{b}\) – \(\vec{b}\) .\(\vec{a}\) + |\(\vec{b}\)|²
= 1 – 2(\(\vec{a}\).\(\vec{b}\)) + 1
= 2 – 2(\(\vec{a}\). \(\vec{b}\))
= 2 – 2|\(\vec{a}\)||\(\vec{b}\)|
cosθ = 2 – 2
cosθ = 2(1 – cosθ)
= 2 × 2 sin² \(\frac{\theta}{2}\)
⇒ |\(\vec{a}\) – \(\vec{b}\)|
= 2 sin \(\frac{\theta}{2}\)

Question 29.
Find the value of λ for which the four points A, B, C, D will position vectors –\(\hat{j}\) – \(\hat{k}\), 4 \(\hat{i}\) + 5 \(\hat{j}\) + λ\(\hat{k}\), 3 \(\hat{i}\) + 9 \(\hat{j}\) + 4 \(\hat{k}\) and – 4 \(\hat{i}\) + 4 \(\hat{j}\) + 4 \(\hat{k}\) are coplanar.
Answer:

Question 30.
Prove that \(\vec{a}\).(\(\vec{b}\) + \(\vec{c}\)) × (\(\vec{a}\) + 2\(\vec{b}\) + 3\(\vec{c}\)) = [\(\vec{a}\).\(\vec{b}\).\(\vec{c}\)]
Answer:

Question 31.
Find the volume of a parallelopiped whose edges are represented by the vectors.
\(\vec{a}\) = 2 \(\hat{i}\) – 3 \(\hat{j}\) – 4 \(\hat{k}\), \(\vec{b}\) = \(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\), and
\(\vec{c}\) = 3 \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\)
Answer:
Given \(\vec{a}\) = 2 \(\hat{i}\) – 3 \(\hat{j}\) – 4 \(\hat{k}\) ;
\(\vec{b}\) = \(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\) and \(\vec{c}\)
= 3 \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\)
Here [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)]
= |2 -3 -4 1 2 -1 3 1 2| ;
Expanding along R₁ = 2(4 + 1) + 3(2 + 3) – 4(1 – 6)
= 10 + 15 + 20 = 45
∴ required volume of parallelopiped = 1
= |[\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)]|
= 45 cubic units

Question 32.
For any three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) show that \(\vec{a}\) – \(\vec{b}\), \(\vec{b}\) – \(\vec{c}\), \(\vec{c}\) – \(\vec{a}\) are coplanar.
Answer:

Question 33.
Find a unit vector perpendicular to each of the vector \(\vec{a}\) + \(\vec{b}\) and \(\vec{a}\) – \(\vec{b}\) where \(\vec{a}\) = 3 \(\hat{i}\) + 2 \(\hat{j}\) + 2 \(\hat{k}\), \(\vec{b}\) = \(\hat{i}\) + 2 \(\hat{j}\) – 2 \(\hat{k}\)
Answer:

Question 34.
If α, β, γ are the angles made by vector with coordinate axes, then sin² α + sin² β + sin² γ .
(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 + cos² γ = 1
⇒ (1 – sin² α) + (1 – sin² β) + (1 – sin² γ) = 1
⇒ sin ^2 α + sin ^2 β + sin ^2 γ = 3 – 1 = 2
∴ Ans. (d)

Question 35.
The value of λ, for which the vectors 3 \(\hat{i}\) – 6 \(\hat{j}\) + \(\hat{k}\) and 2 \(\hat{i}\) – 4 \(\hat{j}\) + λ \(\hat{k}\) are parallel is
(a) \(\frac{2}{3}\)
(b) \(\frac{3}{2}\)
(c) \(\frac{5}{2}\)
(d) \(\frac{2}{5}\)
Answer:
Let \(\vec{a}\) = 3 \(\hat{i}\) – 6 \(\hat{j}\) + \(\hat{k}\) and
\(\vec{b}\) = 2 \(\hat{i}\) – 4 \(\hat{j}\) + λ \(\hat{k}\)
Since \(\vec{a}\) and \(\vec{b}\) are parallel
∴ \(\vec{a}\) = m \(\vec{b}\)
⇒ (3 \(\hat{i}\) – 6 \(\hat{j}\) + \(\hat{k}\)) = m(2 \(\hat{i}\) – 4 \(\hat{j}\) + λ \(\hat{k}\))
⇒ 3 = 2 m ;-6 = -4 m and 1 = λ m
⇒ m = \(\frac{3}{2}\)
∴ λ = \(\frac{1}{m}\) = \(\frac{2}{3}\)
∴ Ans. (a)

Question 36.
The vectors from origin to the points A and B are \(\vec{a}\) = 2 \(\hat{i}\) – 3 \(\hat{j}\) + 2 \(\hat{k}\) and \(\vec{b}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) + \(\hat{k}\) respectively, then the area of triangle O A B is
(a) 340
(b) \(\sqrt{2}\) \(\frac{1}{2}\) \(\sqrt{23}\)
(c) \(\sqrt{229}\)
(d) \(\frac{1}{2}\) \(\sqrt{229}\)
Answer:
Given \(\overrightarrow{\mathrm{OA}}\) = \(\vec{a}\)
= 2 \(\hat{i}\) – 3 \(\hat{j}\) + 2 \(\hat{k}\) and
\(\overrightarrow{\mathrm{OB}}\) = \(\vec{b}\)
= 2 \(\hat{i}\) + 3 \(\hat{j}\) + \(\hat{k}\)
Then area of ∆OAB = \(\frac{1}{2}\)|\(\overrightarrow{\mathrm{OA}}\) × \(\overrightarrow{\mathrm{OB}}\)|
= \(\frac{1}{2}\)|\(\vec{a}\) × \(\vec{b}\)|
Now \(\vec{a}\) × \(\vec{b}\)
= |\(\hat{i}\) \(\hat{j}\) \(\hat{k}\)
2 -3 2
2 3 1|
= \(\hat{i}\)(-3 – 6) – \(\hat{j}\)(2 – 4) + \(\hat{k}\)(6 + 6)
= -9 \(\hat{i}\) + 2 \(\hat{j}\) + 12 \(\hat{k}\)
∴|\(\vec{a}\) × \(\vec{b}\)|
= \(\sqrt{(-9)^2+2^2+12^2}\)
= \(\sqrt{81+4+144}\)
= \(\sqrt{229}\)
Thus, required area of ∆OAB = \(\frac{1}{2}\) \(\sqrt{229}\) square units
∴ Ans. (d)

Question 37.
If |\(\vec{a}\)| = 10, |\(\vec{b}\)| = 2 and \(\vec{a}\).\(\vec{b}\) = 12, then the value of |\(\vec{a}\) × \(\vec{b}\)| is
(a) 5
(b) 10
(c) 14
(d) 16
Answer:
Given |\(\vec{a}\)| = 10 ;
|\(\vec{b}\)| = 2 ;
\(\vec{a}\) . \(\vec{b}\) = 12
Since \(\vec{a}\) .\(\vec{b}\) = 12
⇒ 12 = |\(\vec{a}\)||\(\vec{b}\)|
cosθ ⇒ cosθ = \(\frac{12}{10 \times 2}\)
= \(\frac{3}{5}\)
∴ sinθ = \(\sqrt{1-\cos ^2 \theta}\)
= \(\sqrt{1-(\frac{3}{5})^2}\) = \(\frac{4}{5}\)
Thus |\(\vec{a}\) × \(\vec{b}\)|
= |\(\vec{a}\)||\(\vec{b}\)|
sinθ = 10 × 2 × \(\frac{4}{5}\) = 16
∴ Ans. (d)

Question 38.
A vector of magnitude 7 units, parallel to the resultant of the vectors \(\vec{a}\) = 2 \(\hat{i}\) – 3 \(\hat{j}\) – 2 \(\hat{k}\) and \(\vec{b}\) = \(-\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\) is
(a) \(\frac{7}{\sqrt{3}}\)(\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\))
(b) (\(\hat{i}\) – \(\hat{j}\) – \(\hat{k}\))
(c) \(\frac{7}{\sqrt{3}}\)(\(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\))
(d) \(\frac{7}{\sqrt{3}}\)(\(\hat{i}\) – \(\hat{j}\) – \(\hat{k}\))
Answer:
Resultant of vector \(\vec{a}\) and \(\vec{b}\) = \(\vec{a}\) + \(\vec{b}\) =
\(\vec{d}\) = (2 \(\hat{i}\) – 3 \(\hat{j}\) – 2 \(\hat{k}\)) + (\(-\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\))
= \(\hat{i}\) – \(\hat{j}\) – \(\hat{k}\)
∴ Required unit vector = 7 \(\hat{d}\) = 7 \(\frac{\vec{d}}{|\vec{d}|}\)
= \(\frac{7(\hat{i}-\hat{j}-\hat{k})}{\sqrt{1^2+(-1)^2+(-1)^2}}\)
= \(\frac{7}{\sqrt{3}}\)(\(\hat{i}\) – \(\hat{j}\) – \(\hat{k}\))
∴ Ans. (d)

Question 39.
The area (in sq. units) of the parallelogram whose diagonals are along the vectors 8 \(\hat{i}\) – 6 \(\hat{j}\) and 3 \(\hat{i}\) + 4 \(\hat{j}\) – 12 \(\hat{k}\) is
(a) 65
(b) 52
(c) 26
(d) 20
Answer:

Question 40.
If \(\vec{a}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) – 5 \(\hat{k}\),
\(\vec{b}\) = m \(\hat{i}\) + n \(\hat{j}\) + 12 \(\hat{k}\) and \(\vec{a}\) × \(\vec{b}\) = 0 then (m, n) =
(a) (\(-\frac{24}{5}\), \(-\frac{36}{5}\))
(b) (\(-\frac{24}{5}\), \(\frac{36}{5}\))
(c) (\(\frac{24}{5}\), \(-\frac{36}{5}\))
(d) (\(\frac{24}{5}\), \(\frac{36}{5}\))
Answer:
\(\vec{a}\) × \(\vec{b}\) = |\(\hat{i}\) \(\hat{j}\) \(\hat{k}\)
2 3 -5
m n 12|
= \(\hat{i}\)(36 + 5 n) – \(\hat{j}\)(24 + 5 m) + \(\hat{k}\)(2 n – 3 m)
Since \(\vec{a}\) × \(\vec{b}\)
= \(\overrightarrow{0}\) ⇒ (36 + 5 n) \(\hat{i}\) – \(\hat{j}\)(24 + 5 m) + \(\hat{k}\)(2 n – 3 m)
= 0 \(\hat{i}\) + 0 \(\hat{j}\) + 0 \(\hat{k}\)
∴ 36 + 5 n = 0
⇒ n = \(-\frac{36}{5}\)
and 24 + 5 m = 0
⇒ m = \(-\frac{24}{5}\)
∴ Ans. (a)

Question 42.
If |\(\vec{a}\)| = 3, |\(\vec{b}\)| = 4 and the angle between \(\vec{a}\) and \(\vec{b}\) is 120°, then |4 \(\vec{a}\) + 3 \(\vec{b}\)| is equal to
(a) 25
(b) 7
(c) 13
(d) 12
Answer:
|4 \(\vec{a}\) + 3 \(\vec{b}\)|²
= (4 \(\vec{a}\) + 3 \(\vec{b}\)) .(4 \(\vec{a}\) + 3 \(\vec{b}\))
= 16 \(\vec{a}\)² + 9 \(\vec{b}\)² + 24(\(\vec{a}\).\(\vec{b}\))
[∵ \(\vec{a}\) . \(\vec{b}\)
= \(\vec{b}\).\(\vec{a}\)]
= 16|\(\vec{a}\)|² + 9|\(\vec{b}\)|² + 24|\(\vec{a}\)||\(\vec{b}\)|
cosθ = 16 × 3² + 9 × 4² + 24 × 3 × 4 × cos 120°
= 288 + 24 × 3 × 4(\(-\frac{1}{2}\)) = 144
⇒ |4 \(\vec{a}\) + 3 \(\vec{b}\)|= 12
Ans. (d)

Question 43.
If |\(\vec{a}\)| = 2, |\(\vec{b}\)| = 5 and |\(\vec{a}\) × \(\vec{b}\)|= 8, then \(\vec{a}\) .\(\vec{b}\) is equal to
(a) 3
(b) 4
(c) 5
(d) 6
Answer:
|\(\vec{a}\)| = 2 ;
|\(\vec{b}\)| = 5 ;
|\(\vec{a}\) × \(\vec{b}\)| = 8
⇒ |\(\vec{a}\)||\(\vec{b}\)|
sinθ = 8
⇒ sinθ = \(\frac{8}{2 \times 5}\) = \(\frac{4}{5}\)
∴ \(\vec{a}\) . \(\vec{b}\) = |\(\vec{a}\)||\(\vec{b}\)|
cosθ = 2 × 5 × \(\sqrt{1-\sin ^2 \theta}\)
=10 × \(\sqrt{1-(\frac{4}{5})^2}\)
=10 × \(\frac{3}{5}\) = 6
∴ Ans. (d)

Question 44.
If the vectors 2 \(\hat{i}\) + 2 \(\hat{j}\) + 6 \(\hat{k}\),
2 \(\hat{i}\) + λ \(\hat{j}\) + 6 \(\hat{k}\), 2 \(\hat{i}\) – 3 \(\hat{j}\) + \(\hat{k}\)
are coplanar, then the value of λ is
(a) -10
(b) 1
(c) 0
(d) 2
Answer:
Let \(\vec{a}\) = 2 \(\hat{i}\) + 2 \(\hat{j}\) + 6 \(\hat{k}\);
\(\vec{b}\) = 2 \(\hat{i}\) + λ \(\hat{j}\) + 6 \(\hat{k}\) ;
\(\vec{c}\) = 2 \(\hat{i}\) – 3 \(\hat{j}\) + \(\hat{k}\)
Given \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are coplanar.
∴ [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = 0
|2 2 6
2 λ 6
2 -3 1| = 0
⇒ 2(λ + 18) – 2(2 – 12) + 6(-6 – 2λ) = 0
⇒ 2λ + 36 + 20 – 36 – 12λ = 0
⇒ 20 – 10λ = 0
⇒ λ = 2
∴ Ans. (d)

Question 45.
If \(\vec{a}\).\(\vec{b}\) = \(\frac{1}{2}\)|\(\vec{a}\)||\(\vec{b}\)|, then the angle between \(\vec{a}\) . \(\vec{b}\) is
(a) 0°
(b) 30°
(c) 60°
(d) 90°
Answer:
Let θ be the angle between \(\vec{a}\) and \(\vec{b}\)
Then \(\vec{a}\). \(\vec{b}\) = \(\frac{1}{2}\)|\(\vec{a}\)||\(\vec{b}\)|
|\(\vec{a}\)||\(\vec{b}\)|
cos θ = \(\frac{1}{2}\)|\(\vec{a}\)||\(\vec{b}\)|
⇒ cosθ = \(\frac{1}{2}\)
⇒ θ = 60°
∴ Ans. (c)

Question 46.
Let \(\vec{a}\) = \(\hat{i}\) – 2 \(\hat{j}\) + 3 \(\hat{k}\). If \(\vec{b}\) is a vector sucb that \(\vec{a}\) .\(\vec{b}\) = 7 and
|\(\vec{a}\) – \(\vec{b}\)| = \(\sqrt{7}\), then |\(\vec{b}\)| equals
(a) 7
(b) 14
(c) \(\sqrt{7}\)
(d) 21
Answer:

Question 47.
If (2 \(\hat{i}\) + 6 \(\hat{j}\) + 27 \(\hat{k}\)) × (\(\hat{i}\) – p \(\hat{j}\) + q \(\hat{k}\)) = 0, then the value of p and q are
(a) p = 6, q = 27
(b) p = -3, q = \(\frac{27}{2}\)
(c) p = 6, q = \(\frac{27}{2}\)
(d) p = 3, q = 27
Answer:
Let \(\vec{a}\) = 2 \(\hat{i}\) + 6 \(\hat{j}\) + 27 \(\hat{k}\) ;
\(\vec{b}\) = \(\hat{i}\) – p \(\hat{j}\) + q \(\hat{k}\)
∴ \(\vec{a}\) × \(\vec{b}\) = 0
⇒ |\(\hat{i}\) \(\hat{j}\) \(\hat{k}\)
2 6 27
1 -p q| = 0
⇒ \(\hat{i}\)(6 q + 27 p) – \(\hat{j}\)(2 q – 27) + \(\hat{k}\)(-2 p – 6) = 0
= 0 \(\hat{i}\) + 0 \(\hat{j}\) + 0 \(\hat{k}\)
∴ 6 q + 27 p = 0 ;
2 q – 27 = 0
⇒ q = \(\frac{27}{2}\) and -2 p – 6 = 0
⇒ p = -3
∴ Ans. (b)

Question 48.
If the angle between \(\hat{i}\) + \(\hat{k}\) and \(\hat{i}\) + \(\hat{j}\) + a \(\hat{k}\) is \(\frac{\pi}{3}\), then the value of a is
(a) 0 or 2
(b) -4 and 0
(c) 0 or -2
(d) 2 or -2
Answer:

Question 49.
If \(\vec{u}\) = \(\hat{i}\) + 2 \(\hat{j}\), \(\vec{v}\) = -2 \(\hat{i}\) + \(\hat{j}\) and \(\vec{w}\) = 4 \(\hat{i}\) + 3 \(\hat{j}\). Find scalars x and y respectively such that \(\vec{w}\) = x \(\vec{u}\) + y \(\vec{v}\).
(a) 4, -2
(b) 2, -1
(c) 3, 5
(d) -5, 2
Answer:
Given \(\vec{w}\) = x \(\vec{u}\) + y \(\vec{v}\)
⇒ 4 \(\hat{i}\) + 3 \(\hat{j}\) = x(\(\hat{i}\) + 2 \(\hat{j}\)) + y(-2 \(\hat{i}\) + \(\hat{j}\))
⇒ 4 \(\hat{i}\) + 3 \(\hat{j}\) = (x – 2 y) \(\hat{i}\) + (2 x + y) \(\hat{j}\)
On comparing the coefficients of \(\hat{i}\) and \(\hat{j}\) on both sides ; we have
x-2 y=4
and 2 x + y = 3
and 2 x + y = 3
eqn. (1) +2 × eqn. (2) gives;
5 x = 10 ⇒ x = 2
∴ from (1); y = -1
∴ Ans. (b)

Question 50.
The number of vectors of unit lengths perpendicular to the vectors \(\vec{a}\) = \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\) and \(\vec{b}\) = \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\) is
(a) 2
(b) 1
(c) 3
(d) infinite
Ans.

Question 51.
If 2 \(\hat{i}\) + 3 \(\hat{j}\), \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\) and λ \(\hat{i}\) + 4 \(\hat{j}\) + 2 \(\hat{k}\) taken in order are coterminous edges of a parallelopiped of olume 2 cu. units, then value of λ is
(a) -4
(b) 2
(c) 3
(d) 4
Answer:
Given \(\vec{a}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) ;
\(\vec{b}\) = \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\);
\(\vec{c}\) = λ \(\hat{i}\) + 4 \(\hat{j}\) + 2 \(\hat{k}\)
Then volume of parallelopiped
= \(\vec{a}\) \(\vec{b}\) \(\vec{c}\)
= |2 3 0
1 1 1
λ 4 2|
= 2(2 – 4) – 3(2 – λ)
= -4 – 6 + 3 λ = 3 λ – 10
Given volume of parallelopiped =2 cu units
∴ 3λ – 10 = 2
⇒ 3λ = 12
⇒ λ =4
∴ Ans. (d)

Question 52.
X and Y are two points with position vectors 3 \(\vec{a}\) + \(\vec{b}\) and \(\vec{a}\) – 3 \(\vec{b}\) respectively. Write the position vector of a point Z which divides the line segment X Y in the ratio 2 : 1 externally.
Answer:
Thus the point Z divides the line segment XY in the ratio 2 : -1 internally.

Thus the P.V of point Z be
\(\frac{2(\vec{a}-3 \vec{b}) – 1(3 \vec{a} + \vec{b})}{2-1}\)
i.e. \(-\vec{a}\) – 7 \(\vec{b}\)

Question 53.
Write a unit vector in the direction of the sum of the vectors \(\vec{a}\) = 2 \(\hat{i}\) + 2 \(\hat{j}\) – 5 \(\hat{k}\) and \(\vec{b}\) = 2 \(\hat{i}\) + \(\hat{j}\) – 7 \(\hat{k}\).
Answer:

Question 54.
Find the area of the parallelogram where adjacent sides are the vector
\(\vec{a}\) = 2 \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\)
and \(\vec{b}\) = -3 \(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\).
Answer:
∴ \(\vec{a}\) × \(\vec{b}\) = |\(\hat{i}\) \(\hat{j}\) \(\hat{k}\)
2 2 3
-3 -2 1|
= \(\hat{i}\)(2 + 6) – \(\hat{j}\)(2 + 9) + \(\hat{k}\)(-4 + 6)
= 8 \(\hat{i}\) – 11 \(\hat{j}\) + 2 \(\hat{k}\)
∴ area of gm = |\(\vec{a}\) × \(\vec{b}\)|
= \(\sqrt{64+121+4}\)
= \(\sqrt{189}\)
= 3 \(\sqrt{21}\) sq. units

Question 55.
Find \(\vec{a}\) (\(\vec{b}\) × \(\vec{c}\)), if \(\vec{a}\) = 2 \(\hat{i}\) + \(\hat{j}\) + 3 \(\hat{k}\), \(\vec{b}\) = \(-\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\) and \(\vec{c}\) = 3 \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\)
Answer:
∴ \(\vec{a}\).(\(\vec{b}\) × \(\vec{c}\)) =
\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)
= |2 1 3
-1 2 1
3 1 2|
= 2(4 – 1) – 1(-2 – 3) + 3(-1 – 6)
= 6 + 5 – 21 = -10

Question 56.
(i) Find the projection of vector (2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)) on the vector (\(\hat{i}\) – 2 \(\hat{j}\) + 2 \(\hat{k}\)).
(ii) Find the projection of vector \(\hat{i}\) – \(\hat{j}\) on the vector \(\hat{i}\) + \(\hat{j}\).
Answer:

Question 57.
Find the angle between two vectors \(\vec{a}\) and \(\vec{b}\) with magnitudes 8 and 3 respectively and when |\(\vec{a}\) × \(\vec{b}\)|=12.
Answer:
Given |\(\vec{a}\)|= 8 ;|\(\vec{b}\)| = 3
Let θ be the required angle between \(\vec{a}\) and \(\vec{b}\).
Then |\(\vec{a}\) × \(\vec{b}\)| = 12
⇒ |\(\vec{a}\)||\(\vec{b}\)| sinθ = 12
⇒ 8 × 3 sinθ = 12
⇒ sin = \(\frac{12}{24}\)
= \(\frac{1}{2}\)
⇒ θ = \(\frac{\pi}{6}\)

Question 58.
Find the angle between the vectors
\(\vec{a}\) = \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\) and \(\vec{b}\) = \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)
Answer:
Let θ be the angle between \(\vec{a}\) and \(\vec{b}\)
Then cosθ = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\)
= \(\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\sqrt{1^2+1^2+1^2} \sqrt{1^2+(-1)^2+1^2}}\)
= \(\frac{1 \times 1+1 \times(-1)+1 \times 1}{\sqrt{3} \times \sqrt{3}}\)
= \(\frac{1}{3}\)
⇒ θ = cos^-1 \(\frac{1}{3}\)

Question 59.
For what value of λ are the vectors
\(\vec{a}\) = 2 \(\hat{i}\) + λ\(\hat{j}\) + \(\hat{k}\) and \(\vec{b}\) = \(\hat{i}\) – 2 \(\hat{j}\) + 3 \(\hat{k}\)
perpendicular to each other ?
Answer:
Since \(\vec{a}\) ⊥ \(\vec{b}\)
∴ \(\vec{a}\) .\(\vec{b}\) = 0
⇒ (2 \(\hat{i}\) + λ \(\hat{j}\) + \(\hat{k}\)).(\(\hat{i}\) – 2 \(\hat{j}\) + 3 \(\hat{k}\)) = 0
⇒ 2-2 λ + 3 = 0
⇒ 2 λ = 5
⇒ λ = \(\frac{5}{2}\)

Question 60.
If \(\vec{a}\) = x \(\hat{i}\) + 2 \(\hat{j}\) – z \(\hat{k}\) and \(\vec{b}\) = 3 \(\hat{i}\) – y \(\hat{j}\) + \(\hat{k}\)are two equal vectors, then write the value of x + y + z.
Answer:
Given \(\vec{a}\) = x \(\hat{i}\) + 2 \(\hat{j}\) – z \(\hat{k}\)
\(\vec{b}\) = 3 \(\hat{i}\) – y \(\hat{j}\) + \(\hat{k}\)
Given \(\vec{a}\) = \(\vec{b}\)
⇒ x \(\hat{i}\) + 2 \(\hat{j}\) – z \(\hat{k}\)
= 3 \(\hat{i}\) – y \(\hat{j}\) + \(\hat{k}\)
∴ x = 3 ; y = -2 ; z = -1
∴ x + y + z = 3 – 2 – 1 = 0

Question 61.
Find the vector of magnitude 5 units in the direction apposite to (2 \(\hat{i}\) + 3 \(\hat{j}\) – 6 \(\hat{k}\)).
Answer:
Let \(\vec{a}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) – 6 \(\hat{k}\)
and
|\(\vec{a}\)| = \(\sqrt{2^2+3^2+(-6)^2}\)
= \(\sqrt{4+9+36}\) = 7
∴ required unit vector of magnitude 5 in the direction opposite to \(\vec{a}\)
= \(-\frac{5 \vec{a}}{|\vec{a}|}\)
= \(-\frac{5}{7}\)(2 \(\hat{i}\) + 3 \(\hat{j}\) – 6 \(\hat{k}\))

Question 62.
If |\(\vec{a}\) × \(\vec{b}\)|² + |\(\vec{a}\).\(\vec{b}\)|² = 144 and |\(\vec{a}\)| = 4, then find |\(\vec{b}\)|.
Answer:
Since |\(\vec{a}\) × \(\vec{b}\)| = |\(\vec{a}\)||\(\vec{b}\)|
sinθ and |\(\vec{a}\).\(\vec{b}\)|
= |\(\vec{a}\)||\(\vec{b}\)|
cosθ given |\(\vec{a}\) × \(\vec{b}\)|² + |\(\vec{a}\) \cdot \(\vec{b}\)|² = 144
⇒ |\(\vec{a}\)|² |\(\vec{b}\)|²
(sin²θ + cos²θ) = 144
⇒ 4² |\(\vec{b}\)|² = 144
⇒ |\(\vec{b}\)|² = 9
⇒ |\(\vec{b}\)| = 3

Question 63.
Find a vector of magnitude 6 , which is perpendicular to both the vectors 2 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\) and 4 \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\).
Answer:

Question 64.
Find a unit vector is the direction of \(\overrightarrow{PQ}\), where P and Q have coordinates (5, 0, 8) and (3, 3, 2) respectively.
Answer:
P.V of P = 5 \(\hat{i}\) + 0 \(\hat{j}\) + 8 \(\hat{k}\);
P.V. of Q = 3 \(\hat{i}\) + 3 \(\hat{j}\) + 2 \(\hat{k}\)
∴ \(\overrightarrow{\mathrm{PQ}}\)
= P.V of Q – P.V of P
= (3 \(\hat{i}\) + 3 \(\hat{j}\) + 2 \(\hat{k}\)) – (5 \(\hat{i}\) + 0 \(\hat{j}\) + 8 \(\hat{k}\))
= -2 \(\hat{i}\) + 3 \(\hat{j}\) – 6 \(\hat{k}\)
Thus required unit vector = \(\frac{\overrightarrow{\mathrm{PQ}}}{|\overrightarrow{\mathrm{PQ}}|}\)
= \(\frac{-2 \hat{i}+3 \hat{j}-6 \hat{k}}{\sqrt{4+9+36}}\)
= \(-\frac{2}{7}\) \(\hat{i}\) + \(\frac{3}{7}\) \(\hat{j}\) – \(\frac{6}{7}\) \(\hat{k}\)

Question 65.
Find the magnitude of each of the two vectors \(\vec{a}\) and \(\vec{b}\), having the same magnitude such that the angle between them is 60° and their scalar product is \(\frac{9}{2}\).
Answer:
Given |\(\vec{a}\)| = |\(\vec{b}\)| we know that
cos θ = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\)
where θ be the angle between \(\vec{a}\) and \(\vec{b}\).
Here, given θ = 60^{\circ} and \(\vec{a}\). \(\vec{b}\)
= \(\frac{9}{2}\) ∴ from (1); cos 60°
= \(\frac{\frac{9}{2}}{|\vec{a}|^2}\)
⇒ \(\frac{1}{2}\)|\(\vec{a}\)|^2
= \(\frac{9}{2}\)
⇒ |\(\vec{a}\)|² = 9
⇒ |\(\vec{a}\)| = 3;
Thus |\(\vec{a}\)| = |\(\vec{b}\)| = 3