Students appreciate clear and concise ML Aggarwal Maths for Class 12 Solutions Chapter 1 Vectors Ex 1.2 that guide them through exercises.

## ML Aggarwal Class 12 Maths Solutions Section B Chapter 1 Vectors Ex 1.2

Very short answer type questions (1 to 14) :

Question 1.
(i) If $$\vec{a}$$ is a unit vector and $$(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})$$ = 15, then find $$|\vec{x}|$$.
(ii) If $$\vec{p}$$ is a unit vector and $$(\vec{x}-\vec{p}) \cdot(\vec{x}+\vec{p})$$= 80, then find $$|\vec{x}|$$.
Solution:
(i) Given $$(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})$$ = 15

(ii) Given $$\vec{p}$$ is a unit vector and $$(\vec{x}-\vec{p}) \cdot(\vec{x}+\vec{p})$$= 80

Question 2.
(i) If $$(\vec{a}-\vec{b}) \cdot(\vec{a}+\vec{b})$$ = 12 and $$|\vec{a}|=2|\vec{b}|$$, then find $$|\vec{a}| \text { and }|\vec{b}|$$.
(ii) If $$|\vec{a}|$$ = 3, $$|\vec{b}|$$ = 4 and $$\vec{a} \cdot \vec{b}$$ = 1, then find $$(\vec{a}-\vec{b})^2$$.
(iii) If $$|\vec{a}|$$ = 2, $$|\vec{b}|$$ = 5 and $$\vec{a} \cdot \vec{b}$$ = 8, then find $$|\vec{a}-\vec{b}|$$.
(iv) If $$|\vec{a}|$$ = 3, $$|\vec{b}|$$ = 5 and $$\vec{a} \cdot \vec{b}$$ = – 8 then find $$|\vec{a}+\vec{b}|$$.
Solution:
(i) Given $$(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})$$ = 12

(ii) Given $$|\vec{a}|$$ = 3, $$|\vec{b}|$$ = 4
and $$\vec{a} \cdot \vec{b}$$ = 1

(iii) Given $$|\vec{a}|$$ = 2, $$|\vec{b}|$$ = 5
and $$\vec{a} \cdot \vec{b}$$ = 8

(iv) Given $$|\vec{a}|$$ = 3, $$|\vec{b}|$$ = 5
and $$\vec{a} \cdot \vec{b}$$ = – 8

Question 3.
Find the magnitude of each of the two vectors $$\vec{a} \text { and } \vec{b}$$, having same magnitude such that the angle between them is 60° and their scalar product is $$\frac{9}{2}$$.
Solution:
Given $$|\vec{a}|=|\vec{b}|$$ ;
θ = 60° ;
$$\vec{a} \cdot \vec{b}=\frac{9}{2}$$

Question 3 (old).
If the angle between two vectors $$\vec{a} \text { and } \vec{b}$$ of equal magnitude is 60° and their scalar product is $$\frac{1}{2}$$, then find their magnitudes. (NCERT)
Solution:
Given $$|\vec{a}|=|\vec{b}|$$ ……………..(1)
Let θ be the angle between a and b such that θ = 60°
and $$\vec{a} \cdot \vec{b}=\frac{1}{2}$$ [using (1)]
We know that
⇒ $$\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$$
⇒ $$\frac{1}{2}=|\vec{a}| \cdot|\vec{b}| \cdot \cos \frac{\pi}{3}$$
⇒ $$\frac{1}{2}=|\vec{a}|^2 \cdot \frac{1}{2}$$
⇒ $$|\vec{a}|=1=|\vec{b}|$$

Question 4.
(i) If $$|\vec{a}|$$ = 2, $$|\vec{b}|$$ = √3 and $$\vec{a} \cdot \vec{b}$$ = 3, then find the angle between $$\vec{a} \text { and } \vec{b}$$.
(ii) Find the angle between the vectors $$\vec{a} \text { and } \vec{b}$$ such that $$|\vec{a}|=|\vec{b}|$$ = 3 and $$\vec{a} \cdot \vec{b}$$ = 1.
(iii) Find the angle between two vectors having the same length √2 and scalar product – 1.
(iv) If $$|\vec{a}|$$ = √3, $$|\vec{b}|$$ = 2 and the angle between $$\vec{a} \text { and } \vec{b}$$ is 60°, find $$\vec{a} \cdot \vec{b}$$ = 1.
Solution:
(i) Given $$|\vec{a}|$$ = 3, $$|\vec{b}|$$ = 2
and $$\vec{a} \cdot \vec{b}$$ = 3
Let θ be the angle between $$\vec{a} \text { and } \vec{b}$$.
∴ $$\vec{a} \cdot \vec{b}$$ = $$|\vec{a}||\vec{b}|$$ cos θ
⇒ 3 = 3 × 2 × cos θ
⇒ cos θ = $$\frac{1}{2}$$ ; 0 ≤ θ ≤ π
∴ θ = $$\frac{\pi}{3}$$

(ii) Given $$|\vec{a}|$$ = 3, $$|\vec{b}|$$ = 3 ;
and $$\vec{a} \cdot \vec{b}$$ = 1
Let θ be the angle between $$\vec{a} \text { and } \vec{b}$$.
∴ cos θ = $$\frac{(\vec{a} \cdot \vec{b})}{|\vec{a}||\vec{b}|}$$
= $$\frac{1}{3 \times 3}=\frac{1}{9}$$
⇒ θ = cos-1 ($$\frac{1}{9}$$)

(iii) Let θ be the angle between two vectors $$\vec{a} \text { and } \vec{b}$$ such that
$$|\vec{a}|=|\vec{b}|$$ = √2
and $$\vec{a} \cdot \vec{b}$$ = – 1
Since cos θ = $$\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$$
= $$\frac{-1}{\sqrt{2} \times \sqrt{2}}=-\frac{1}{2}$$
⇒ cos θ = – cos $$\frac{\pi}{3}$$
= cos (180° – 60°)
= cos 120°
∴ θ = 120°

(iv) Given $$|\vec{a}|$$ = √3 , $$|\vec{b}|$$ = 2 ;
and θ = 60°
where θ be the angle between $$\vec{a} \text { and } \vec{b}$$
∴ $$\vec{a} \cdot \vec{b}$$ = $$|\vec{a}||\vec{b}|$$ cos θ
= √3 × 2 × cos 60°
= 2√3 × $$\frac{1}{2}$$
= √3

Question 4 (old).
(i) If $$|\vec{a}|$$ = 3, $$|\vec{b}|$$ = 2 and $$\vec{a} \cdot \vec{b}$$ = 3, then find the angle between $$\vec{a} \text { and } \vec{b}$$.
(iv) Find the angle between the vectors $$\vec{a} \text { and } \vec{b}$$ with magnitudes 1 and 2 respectively, and $$\vec{a} \cdot \vec{b}$$ = 1
Solution:
(ii) Given $$|\vec{a}|$$ = 2, $$|\vec{b}|$$ = √3
and $$\vec{a} \cdot \vec{b}$$ = 3
Let θ be the angle between $$\vec{a} \text { and } \vec{b}$$.
Then $$\vec{a} \cdot \vec{b}$$ = $$|\vec{a}||\vec{b}|$$ cos θ
⇒ 3 = 2 × √3 × cos θ
⇒ cos θ = $$\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2}$$ ; 0 ≤ θ ≤ π
∴ θ = 30°.

(iv) Given $$|\vec{a}|$$ = 1, $$|\vec{b}|$$ = 2 ;
and $$\vec{a} \cdot \vec{b}$$ = 1
Let θ be the angle between $$\vec{a} \text { and } \vec{b}$$.
∴ cos θ = $$\frac{(\vec{a} \cdot \vec{b})}{|\vec{a}||\vec{b}|}$$
= $$\frac{1}{1 \cdot 2}=\frac{1}{2}$$
⇒ θ = $$\frac{\pi}{3}$$

Question 5.
Find the angle between the vectors $$2 \hat{i}-\hat{j}+\hat{k}$$ and $$3 \hat{i}+4 \hat{j}-\hat{k}$$. (NCERT Exemplar)
Solution:
Let $$\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$$
and $$\vec{b}=3 \hat{i}+4 \hat{j}-\hat{k}$$

Question 5 (old).
(i) Find the angle between the vectors $$\vec{a}=\hat{i}+\hat{j}-\hat{k}$$ and $$\vec{b}=\hat{i}-\hat{j}+\hat{k}$$.
Solution:
Given $$\vec{a}=\hat{i}+\hat{j}-\hat{k}$$
and $$\vec{b}=\hat{i}-\hat{j}+\hat{k}$$

Question 6.
Find
(i) $$(\vec{b}-\vec{a}) \cdot(3 \vec{a}+\vec{b})$$ where $$\vec{a}=\hat{i}-2 \hat{j}+5 \hat{k}$$, $$\overrightarrow{\vec{b}}=2 \hat{i}+\hat{j}-3 \hat{k}$$. $$(\vec{a}+3 \vec{b}) \cdot(2 \vec{a}-\vec{b})$$
(ii) where $$\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$$, $$\vec{b}=3 \hat{i}+2 \hat{j}-\hat{k}$$
Solution:
(i) Given $$\vec{a}=\hat{i}-2 \hat{j}+5 \hat{k}$$,
and $$\overrightarrow{\vec{b}}=2 \hat{i}+\hat{j}-3 \hat{k}$$

(ii) Given $$\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$$
and $$\vec{b}=3 \hat{i}+2 \hat{j}-\hat{k}$$

Question 7.
(i) Find the projection of $$\vec{a} \text { on } \vec{b}$$ if $$\vec{a} \cdot \vec{b}$$ = 8 and $$\vec{b}=2 \hat{i}+6 \hat{j}+3 \hat{k}$$.
(ii) Find the projection of $$\hat{i}+3 \hat{j}+7 \hat{k}$$ on the vector $$2 \hat{i}-3 \hat{j}+6 \hat{k}$$.
(iii) Find the projection of $$\hat{i}-\hat{j}$$ on the vector $$\hat{i}+\hat{j}$$.
(iv) Find λ when the projection of $$\hat{i}+\lambda \hat{j}+\hat{k}$$ on $$\hat{i}+\hat{j}$$ is √2 units.
Solution:
(i) Given $$\vec{a} \cdot \vec{b}$$ = 8
and $$\vec{b}=2 \hat{i}+6 \hat{j}+3 \hat{k}$$
∴ $$|\vec{b}|=\sqrt{2^2+6^2+3^2}$$
= $$\sqrt{49}$$ = 7
Thus required projection of $$\vec{a} \text { on } \vec{b}$$
= $$\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}=\frac{8}{7}$$

(ii) Let $$\vec{a}=\hat{i}+3 \hat{j}+7 \hat{k}$$
and $$\vec{b}=2 \hat{i}-3 \hat{j}+6 \hat{k}$$
∴ $$\vec{a} \cdot \vec{b}=(\hat{i}+3 \hat{j}+7 \hat{k}) \cdot(2 \hat{i}-3 \hat{j}+6 \hat{k})$$
= 1 (2) + 3 (- 3) + 7 (6)
= 2 – 9 + 42 = 35
and $$|\vec{b}|=\sqrt{2^2+(-3)^2+6^2}$$
= $$\sqrt{49}$$ = 7
∴ required projection of $$\vec{a} \text { on } \vec{b}$$ = $$\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$$
= $$\frac{35}{7}$$ = 5

(iii) Given $$\vec{a}=\hat{i}-\hat{j}$$
and $$\vec{b}=\hat{i}+\hat{j}$$
∴ $$\vec{a} \cdot \vec{b}$$ = 1 (1) – (1) (1) = 0
∴ scalar projection of $$\vec{a} \text { on } \vec{b}$$ = $$\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$$ = 0

(iv) Let $$\vec{a}=\hat{i}+\lambda \hat{j}+\hat{k}$$
and $$\vec{b}=\hat{i}+\hat{j}$$
∴ $$\vec{a} \cdot \vec{b}=(\hat{i}+\lambda \hat{j}+\hat{k}) \cdot(\hat{i}+\hat{j}+0 \hat{k})$$
= 1 (1) + λ (1) + 1 (0)
= 1 + λ
$$|\vec{b}|=\sqrt{1^2+1^2}=\sqrt{2}$$
We know that projection $$\vec{a} \text { on } \vec{b}$$ = $$\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$$
∴ √2 = $$\frac{1+\lambda}{\sqrt{2}}$$
⇒ 1 + λ = 2
⇒ λ = 1.

Question 8.
(i) If $$\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k}$$ and $$\vec{b}=\hat{i}+3 \hat{j}-5 \hat{k}$$, then show that the vectors $$\vec{a}+\vec{b}$$ and $$\vec{a}-\vec{b}$$ are perpendicular to each other. (NCERT)
(ii) Find λ if $$\vec{a}=3 \hat{i}-\hat{j}+4 \hat{k}$$ and $$\vec{b}=-\lambda \hat{i}+3 \hat{j}+3 \hat{k}$$ are perpendicular to each other.
(iii) For what value of λ are the vectors $$\vec{a}=2 \hat{i}+\lambda \hat{j}+\hat{k}$$ and $$\vec{b}=\hat{i}-2 \hat{j}+3 \hat{k}$$ perpendicular to each other ?
Solution:
(i) Given $$\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k}$$
and $$\vec{b}=\hat{i}+3 \hat{j}-5 \hat{k}$$

(ii) Given $$\vec{a}=3 \hat{i}-\hat{j}+4 \hat{k}$$
and $$\vec{b}=-\lambda \hat{i}+3 \hat{j}+3 \hat{k}$$
Since $$\vec{a}$$ is ⊥ to $$\vec{b}$$
∴ $$\vec{a} \cdot \vec{b}$$ = 0
⇒ $$(3 \hat{i}-\hat{j}+4 \hat{k}) \cdot(-\lambda \hat{i}+3 \hat{j}+3 \hat{k})$$ = 0
⇒ 3 (- λ) – 1 (3) + 4 (3) = 0
⇒ – 3λ + 9 = 0
⇒ λ = 3.

(iii) Given $$\vec{a}=2 \hat{i}+\lambda \hat{j}+\hat{k}$$
and $$\vec{b}=\hat{i}-2 \hat{j}+3 \hat{k}$$
Now $$\vec{a} \text { and } \vec{b}$$ are orthogonal or perpendicualr.
∴ $$\vec{a} \cdot \vec{b}$$ = 0
⇒ 2 (1) + λ (- 2) + 1 (3) = 0
⇒ 2λ = 5
⇒ λ = $$\frac{5}{2}$$

Question 9.
(i) If $$\vec{a} \text { and } \vec{b}$$ are unit vectors, then what is the angle between $$\vec{a} \text { and } \vec{b}$$ so that $$\sqrt{3} \vec{a}-\vec{b}$$ may be a unit vector. (NCERT Exemplar)
(ii) If $$|\vec{a}|$$ = 2, $$|\vec{b}|$$ = 3 and $$\vec{a} \cdot \vec{b}$$ = 4, then what is the value of $$|\vec{a}+2 \vec{b}|$$ ?
(iii) If $$|\vec{a}|$$ = 2, $$|\vec{b}|$$ = 3 and $$\vec{a} \cdot \vec{b}$$ = 4, find $$|2 \vec{a}-3 \vec{b}|$$.
(iv) If $$|\vec{a}|$$ = 2, $$|\vec{b}|$$ = 3 and $$|2 \vec{a}-\vec{b}|$$ = 4, then find $$|2 \vec{a}+\vec{b}|$$.
Solution:
(i) Given $$\vec{a} \text { and } \vec{b}$$ are unit vectors
∴ $$|\vec{a}|$$ = $$|\vec{b}|$$ = 1
since $$\sqrt{3} \vec{a}-\vec{b}$$ is a unit vector

(ii) Given $$|\vec{a}|$$ = 2, $$|\vec{b}|$$ = 3
and $$\vec{a} \cdot \vec{b}$$ = 4

= 22 + 4 (4) + 4 (9)
= 4 + 16 + 36 = 56
∴ $$|\vec{a}+2 \vec{b}|=\sqrt{56}=2 \sqrt{14}$$

(iii) Given $$|\vec{a}|$$ = 2, $$|\vec{b}|$$ = 3
and $$\vec{a} \cdot \vec{b}$$ = 4

(iv) Given $$|\vec{a}|$$ = 2,
$$|\vec{b}|$$ = 3
and $$|2 \vec{a}-\vec{b}|$$ = 4

Question 10.
If $$\vec{a}, \vec{b}$$ are unit vectors and c = $$|\vec{a}+\vec{b}|$$, d = $$|\vec{a}-\vec{b}|$$, then what is the value of c2 + d2 ?
Solution:
Given c = $$|\vec{a}+\vec{b}|$$

Question 11.
If $$\vec{a}, \vec{b}, \vec{c}$$ are three vectors such that $$\vec{c}=\vec{a}+\vec{b}$$ and $$\vec{a} \cdot \vec{b}$$ = 0, then show that c2 = a2 + b2.
Solution:
Given $$\vec{c}=\vec{a}+\vec{b}$$

Question 12.
(i) If $$\vec{a}, \vec{b} \text { and } \vec{c}$$ are unit vectors such that $$\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$$, then find the value of $$\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$$. (NCERT)
(ii) If $$\vec{a}, \vec{b}, \vec{c}$$ are three vectors $$|\vec{a}|$$ = 5, $$|\vec{b}|$$ = 12, $$|\vec{c}|$$ = 13 and $$\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$$ then find the value of .
Solution:
(i) Given $$\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$$ ;
on squaring, we ahve

(ii) Given $$|\vec{a}|$$ = 5 ;
$$|\vec{b}|$$ = 12 ;
$$|\vec{c}|$$ = 13
and $$\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$$ ;
on squaring we have

Question 13.
(i) We are given two vectors $$\vec{a} \text { and } \vec{b}$$ such that $$(\vec{a})^2=(\vec{b})^2$$. Is it necessary that $$\vec{a}=\vec{b}$$ ? Justify your answer.
(ii) For two non-zero vectors $$\vec{a} \text { and } \vec{b}$$, state when $$|\vec{a}+\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2$$ holds.
(iii) If $$\vec{a}=\overrightarrow{0}$$ or $$\vec{b}=\overrightarrow{0}$$, then $$\vec{a} \cdot \vec{b}$$ = 0. Is the converse true ? Justify your answer by an example.
(iv) If $$\vec{a}, \vec{b}$$ are non – zero vectors and $$\vec{a} \cdot \vec{b}$$ ≥ 0, then what can you say about the angle θ between the vectors $$\vec{a} \text { and } \vec{b}$$.
Solution:
(i) Let $$\vec{a}=\hat{i}+2 \hat{j}$$
and $$\vec{b}=2 \hat{i}-\hat{j}$$

Here $$\vec{a}^2=|\vec{a}|^2$$
= $$\left(\sqrt{1^2+2^2}\right)^2$$
= $$(\sqrt{5})^2$$ = 5

and $$\vec{b}^2=|\vec{b}|^2$$
= $$\left(\sqrt{2^2+(-1)^2}\right)^2$$
= $$(\sqrt{5})^2$$ = 5
i.e. $$\vec{a}^2=\vec{b}^2$$ but $$\vec{a} \neq \vec{b}$$.

(ii) $$|\vec{a}+\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2$$ holds

(iii) $$\vec{a} \cdot \vec{b}=\overrightarrow{0} \cdot \vec{b}$$ = 0
if $$\vec{a}$$ = 0
When $$\vec{b}$$ = 0 ;
$$\vec{a} \cdot \vec{b}=\vec{a} \cdot \overrightarrow{0}$$ = 0
but converse need not be true.
e.g. $$\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$$
and $$\vec{b}=\hat{i}+\hat{j}-\hat{k}$$
but $$\vec{a} \cdot \vec{b}=(2 \hat{i}-\hat{j}+\hat{k}) \cdot(\hat{i}+\hat{j}-\hat{k})$$
= 2 (1) – 1 (1) + 1 (- 1)
= 2 – 2 = 0
Here $$\vec{a} \cdot \vec{b}$$ = 0 but neither $$\vec{a}=\overrightarrow{0}$$ nor $$\vec{b}=\overrightarrow{0}$$.

(iv) Since $$\vec{a} \text { and } \vec{b}$$ are non-zero vectors.
∴ $$|\vec{a}| \cdot \mid \vec{b}$$ > 0
also $$\vec{a} \cdot \vec{b}$$ ≥ 0
Thus cos θ = $$\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}$$ ≥ 0
∴ 0 ≤ θ ≤ $$\frac{\pi}{2}$$

Question 14.
If $$\vec{a} \cdot \vec{a}$$ = 0 and $$\vec{a} \cdot \vec{b}$$ = 0, then what can you say about the vector $$\vec{b}$$ ?
Solution:
Given $$\vec{a} \cdot \vec{a}$$ = 0
and $$\vec{a} \cdot \vec{b}$$ = 0
⇒ $$|\vec{a}|^2$$ = 0 i.e. $$\vec{a}=\overrightarrow{0}$$
and ($$\vec{a} \perp \vec{b}$$ or $$\vec{a}=\overrightarrow{0}$$ or $$\vec{b}=\overrightarrow{0}$$)
i.e. $$\vec{a}=\overrightarrow{0}$$ and $$\vec{b}$$ can be any vector.

Question 15.
Find the angle between the vectors $$\vec{a}+\vec{b}$$ and $$\vec{a}-\vec{b}$$ if $$\vec{a}=2 \hat{i}-\hat{j}+3 \hat{k}$$ and $$\vec{b}=3 \hat{i}+\hat{j}-2 \hat{k}$$.
Solution:
Given $$\vec{a}=2 \hat{i}-\hat{j}+3 \hat{k}$$
and $$\vec{b}=3 \hat{i}+\hat{j}-2 \hat{k}$$

Question 16.
Find the projection of $$\vec{b}+\vec{c}$$ on $$\vec{a}$$ where $$\vec{a}=2 \hat{i}-2 \hat{j}+\hat{k}$$, $$\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}$$ and $$\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}$$.
Solution:
Given $$\vec{a}=2 \hat{i}-2 \hat{j}+\hat{k}$$,
$$\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}$$
and $$\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}$$

Here $$\vec{b} \cdot \vec{a}=(2 \hat{i}-2 \hat{j}+\hat{k}) \cdot(\hat{i}+2 \hat{j}-2 \hat{k})$$
= 2 (1) – 2 (2) + 1 (- 2)
= 2 – 4 – 2 = – 4

$$\vec{c} \cdot \vec{a}=(2 \hat{i}-\hat{j}+4 \hat{k}) \cdot(2 \hat{i}-2 \hat{j}+\hat{k})$$
= 2 (2) – 1 (- 2) + 4 (1)
= 4 + 2+ 4 = 10

∴ required projection of $$\vec{b}+\vec{c} \text { on } \vec{a}$$
= $$\frac{(\vec{b}+\vec{c}) \cdot \vec{a}}{|\vec{a}|}$$
= = $$\frac{(\vec{b} \cdot \vec{a})+(\vec{c} \cdot \vec{a})}{|\vec{a}|}$$
= $$\frac{-4+10}{\sqrt{2^2+(-2)^2+1^2}}$$
= $$\frac{6}{3}$$
= 2

Question 17.
(i) If $$\vec{a}=\hat{i}-\hat{j}+7 \hat{k}$$ and $$\vec{b}=5 \hat{i}-\hat{j}+\lambda \hat{k}$$, then find λ such that $$\vec{a}-\vec{b}$$ and $$\vec{a}+\vec{b}$$ are perpendicular to each other.
(ii) For any (non-zero) vectors $$\vec{a}$$, and $$\vec{b}$$ prove that the vectors $$|\vec{a}| \vec{b}+|\vec{b}| \vec{a}$$ and $$|\vec{a}| \vec{b}-|\vec{b}| \vec{a}$$ are perpendicular to each other. (NCERT)
Solution:
(i) Given $$\vec{a}=\hat{i}-\hat{j}+7 \hat{k}$$
and $$\vec{b}=5 \hat{i}-\hat{j}+\lambda \hat{k}$$
Since $$\vec{a}-\vec{b}$$ and $$\vec{a}-\vec{b}$$ are ⊥ to each other.
∴ $$(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})$$ = 0
⇒ $$\vec{a} \cdot \vec{a}-\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}-\vec{b} \cdot \vec{b}$$ = 0
⇒ $$|\vec{a}|^2-|\vec{b}|^2$$ = 0
[∵$$\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}$$]
⇒ {12 + (- 1)2 + 72} – {52 + (- 1)2 + λ2} = 0
⇒ 51 – (26 + λ2) = 0
⇒ 25 – λ2 = 0
⇒ λ = ± 5

(ii) Now,

Question 18.
If $$\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}$$, $$\vec{b}=-\hat{i}+2 \hat{j}+\hat{k}$$ and $$\vec{c}=3 \hat{i}+\hat{j}$$ are such that $$\vec{a}+\lambda \vec{b}$$ is perpendiculai to $$\vec{c}$$, then find the value of λ ?
Solution:
Given $$\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}$$,
$$\vec{b}=-\hat{i}+2 \hat{j}+\hat{k}$$
and $$\vec{c}=3 \hat{i}+\hat{j}$$

Now $$\vec{a}+\lambda \vec{b}$$
= $$(2 \hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-\hat{i}+2 \hat{j}+\hat{k})$$
= $$(2-\lambda) \hat{i}+(2+2 \lambda) \hat{j}+(3+\lambda) \hat{k}$$

Now $$\vec{a}+\lambda \vec{b}$$ is ⊥ to $$\vec{c}$$
∴ $$\vec{a}+\lambda \vec{b}$$ . $$\vec{c}$$ = 0
i.e.  = 0
⇒ 3 (2 – λ) + (2 + 2λ) 1 = 0
⇒ 8 – λ = 0
⇒ λ = 8

Question 19.
(i) If $$\vec{c}$$ is perpendicular to $$\vec{a} \text { and } \vec{b}$$ both, then show that it is perpendicular to $$\vec{a}+\vec{b}$$ as well as $$\vec{a}-\vec{b}$$.
(ii) Show that the vectors $$\frac{1}{7}(2 \hat{i}+3 \hat{j}+6 \hat{k})$$, $$\frac{1}{7}(3 \hat{i}-6 \hat{j}+2 \hat{k})$$ and $$\frac{1}{7}(6 \hat{i}+2 \hat{j}-3 \hat{k})$$ are mutually perpendicular unit vccotrs. (NCERT)
(iii) If $$\vec{a} \text { and } \vec{b}$$ are unit vectors and θ is the angle between them, then prove that cos $$\frac{\theta}{2}=\frac{1}{2}|\vec{a}+\vec{b}|$$.
Solution:
(i) Since $$\vec{c}$$ is ⊥ to $$\vec{a} \text { and } \vec{b}$$
∴ $$\vec{c} \cdot \vec{a}$$ = $$\vec{c} \cdot \vec{b}$$
Now $$\vec{c} \cdot(\vec{a}+\vec{b})=\vec{c} \cdot \vec{a}+\vec{c} \cdot \vec{b}$$
= 0 + 0 = 0
and $$\vec{c} \cdot(\vec{a}-\vec{b})=\vec{c} \cdot \vec{a}-(\vec{c} \cdot \vec{b})$$
= 0 – 0 = 0
Thus $$\vec{c}$$ is ⊥ to both $$\vec{a}+\vec{b} \text { and } \vec{a}-\vec{b}$$.

(ii) Given $$\vec{a}=\frac{1}{7}(2 \hat{i}+3 \hat{j}+6 \hat{k})$$,
$$\vec{b}=\frac{1}{7}(3 \hat{i}-6 \hat{j}+2 \hat{k})$$
and $$\vec{c}=\frac{1}{7}(6 \hat{i}+2 \hat{j}-3 \hat{k})$$

(iii) Given $$\vec{a} \text { and } \vec{b}$$ s.t. $$|\hat{a}|=|\hat{b}|$$ = 1
and given θ be the angle between them
Now, $$|\hat{a}+\hat{b}|^2=(\hat{a}+\hat{b}) \cdot(\hat{a}+\hat{b})$$
= $$|\hat{a}|^2+|\hat{b}|^2+2 \hat{a} \cdot \hat{b}$$
[∵ $$\hat{a}^2=|\vec{a}|^2, \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}$$]
= 1 + 1 + 2 $$|\hat{a}|=|\hat{b}|$$ cos θ
= 2 + 2 cos θ
= 2 (1 + cos θ)
= 2 × 2 cos2 $$\frac{\theta}{2}$$
= 4 cos2 $$\frac{\theta}{2}$$
⇒ $$|\hat{a}+\hat{b}|=2 \cos \frac{\theta}{2}$$
$$\cos \frac{\theta}{2}=\frac{1}{2}|\hat{a}+\hat{b}|$$

Question 20.
Find the angles which the vector $$\vec{a}=3 \hat{i}-6 \hat{j}+2 \hat{k}$$ makes with the co-ordinate axes.
Solution:
Given $$\vec{a}=3 \hat{i}-6 \hat{j}+2 \hat{k}$$
Let α, β and γ are the angle made by the vector with the coordinate axes.

Question 21.
(i) Dot products of a vector with vectors ,  and  are respectively – 1, 6 and 5. Find the vector.
(ii) The dot product of a vector with the vectors , and  are 0, 5 and 8 respectively. Find the vector.
Solution:
(i) Let the required vector be
$$\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$$ …………………..(1)
Now $$\vec{a} \cdot(3 \hat{i}-5 \hat{k})$$ = – 1
⇒ 3a1 – 5a3 = – 1 ………………………..(2)
also $$\vec{a} \cdot(2 \hat{i}+7 \hat{j})$$ = 6
⇒ 2a1 + 7a2 = 6 ……………………….(3)
and $$\vec{a} \cdot(\hat{i}+\hat{j}+\hat{k})$$ = 5
⇒ a1 + a2 + a3 = 5 ………………(4)
eqn. (3) – 7 × eqn. (4) ; we have
– 5a1 – 7a3 = – 29
⇒ 5a1 + 7a3 = 29 ………………………(5)
On solving (2) and (5) we get
46a1 = 145 – 7 = 138
⇒ a1 = 3 ;
a3 = 2
∴ from (4) ;
3 + 2 + a2 = 5
⇒ a2 = 0

(ii) Let the required vector be
$$\vec{a}=x \hat{i}+y \hat{j}+z \hat{k}$$ ……………..(1)
given, $$\vec{a} \cdot(\hat{i}+\hat{j}-3 \hat{k})$$ = 0

⇒ $$(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+\hat{j}-3 \hat{k})$$ = 0
⇒ x + y – 3z = 0 ………………..(2)
also, $$\vec{a} \cdot(\hat{i}+3 \hat{j}-2 \hat{k})$$ = 5

⇒ $$(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+3 \hat{j}-2 \hat{k})$$ = 5
⇒ x + 3y – 2z = 5 ……………….(3)
also, $$\vec{a} \cdot(2 \hat{i}+\hat{j}+4 \hat{k})$$ = 8

⇒ $$(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(2 \hat{i}+\hat{j}+4 \hat{k})$$ = 8
⇒ 2x + y + 4z = 8 ………………(4)

eqn. (3) – eqn. (2) gives ;
2y + z = 5 ………………….(5)
eqn. (4) – 2 × eqn. (3) gives ;
– 5y + 8z = – 2 ……………………(6)
eqn. (6) – 8 × eqn. (5) gives ;
– 5y – 16y = – 2 – 40
⇒ – 21 y = – 42
⇒ y = 2
∴ from(5) ;
4 + z = 5
⇒ z = 1
∴ from (2) ;
x + 2 – 3 = 0
⇒ x = 1
∴ required vector be $$\vec{a}=\hat{i}+2 \hat{j}+\hat{k}$$.

Question 22.
Show that the vectors $$\vec{a}=\hat{i}-3 \hat{j}-5 \hat{k}$$, $$\vec{b}=2 \hat{i}-\hat{j}+\hat{k}$$ and $$\vec{c}=\hat{i}+2 \hat{j}+6 \hat{k}$$ form a right-angled triangle.
Solution:
Given $$\vec{a}=\hat{i}-3 \hat{j}-5 \hat{k}$$,
$$\vec{b}=2 \hat{i}-\hat{j}+\hat{k}$$
and $$\vec{c}=\hat{i}+2 \hat{j}+6 \hat{k}$$

Now $$\vec{a}+\vec{c}=(\hat{i}-3 \hat{j}-5 \hat{k})+(\hat{i}+2 \hat{j}+6 \hat{k})$$
= $$2 \hat{i}-\hat{j}+\hat{k}=\vec{b}$$

∴ $$\vec{a}, \vec{b} \text { and } \vec{c}$$ are coplanar.
also no two of these vectors are parallel.
∴ gives vectors forms a triangle.

Also $$\vec{a} \cdot \vec{b}=(\hat{i}-3 \hat{j}-5 \hat{k}) \cdot(2 \hat{i}-\hat{j}+\hat{k})$$
= 1 (2) – 3 (- 1) – 5 (1) = 0
These dot product of two non-zero vectors is zero.
Therefore these vectors $$\vec{a} \text { and } \vec{b}$$ are ⊥ to each other.
Hence the sides represented by these vectors are perpendicular.
Thus, the given vectors forms a right angled triangle.

Question 23.
If A, B, C have position vectors $$\hat{j}+\hat{k}$$, $$3 \hat{i}+\hat{j}+5 \hat{k}$$ and $$3 \hat{j}+3 \hat{k}$$ respectively, then prove that ∆ABC is right angled at C.
Solution:
Let $$\vec{a}=\hat{j}+\hat{k}$$,
$$\vec{b}=3 \hat{i}+\hat{j}+5 \hat{k}$$
and $$\vec{c}=3 \hat{j}+3 \hat{k}$$
are the position vectors of points A, B, C respectively.

Question 24.
If the vertices of a AABC are A(- 1, 3, 2), B (2, 3, 5) and C (3, 5, – 2), then show that it is right angled at A. Also find the other two angles.
Solution:
The position vectors of the vertices A, B and C of a triangle are $$-\hat{i}+3 \hat{j}+2 \hat{k}$$, $$2 \hat{i}+3 \hat{j}+5 \hat{k}$$ and $$3 \hat{i}+5 \hat{j}-2 \hat{k}$$

Hence, the three sides of the triangle are represented by $$\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{BC}} \text { and } \overrightarrow{\mathrm{CA}}$$.

∴ $$\overrightarrow{\mathrm{AB}}$$ = P.V 0f B – P.V 0f A
= $$(2 \hat{i}+3 \hat{j}+5 \hat{k})-(-\hat{i}+3 \hat{j}+2 \hat{k})$$
= $$3 \hat{i}+3 \hat{k}$$

$$\overrightarrow{\mathrm{BC}}$$ = P.V 0f C – P.V 0f B
= $$(3 \hat{i}+5 \hat{j}-2 \hat{k})-(2 \hat{i}+3 \hat{j}+5 \hat{k})$$
= $$\hat{i}+2 \hat{j}-7 \hat{k}$$

$$\overrightarrow{\mathrm{CA}}$$ = P.V 0f A – P.V 0f C
= $$(-\hat{i}+3 \hat{j}+2 \hat{k})-(3 \hat{i}+5 \hat{j}-2 \hat{k})$$
= $$-4 \hat{i}-2 \hat{j}+4 \hat{k}$$

∴ $$\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{CA}}$$ = $$(3 \hat{i}+3 \hat{k}) \cdot(-4 \hat{i}-2 \hat{j}+4 \hat{k})$$
= 3 (- 4) + 0 (- 2) + 3 (4) = 0

Thus $$\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{CA}}$$ are ⊥to each other.
∴ ∠A = 90
Therefore, ∆ ABC is right-angled at A.

Question 25.
Prove that two proper vectors $$\vec{a} \text { and } \vec{b}$$ are at right angles iff $$(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^2+|\vec{b}|^2$$. (NCERT)
Solution:

Question 26.
If the coordinates of four points are A (2, 3, 4) , B (5, 4, – 1), C (3, 6, 2) and D (1, 2, 0), then show that $$\overrightarrow{\mathrm{AB}}$$ is perpendiculat to $$\overrightarrow{\mathrm{CD}}$$. (NCERT Exemplar)
Solution:
Given P.V. of A = $$2 \hat{i}+3 \hat{j}+4 \hat{k}$$ ;
P.V. of B = $$5 \hat{i}+4 \hat{j}-\hat{k}$$ ;
P.V. of C = $$3 \hat{i}+6 \hat{j}+2 \hat{k}$$
and P.V. of D = $$\hat{i}+2 \hat{j}+0 \hat{k}$$

∴ $$\overrightarrow{\mathrm{AB}}$$ = P.V. of B – P.V. of A
= $$(5 \hat{i}+4 \hat{j}-\hat{k})-(2 \hat{i}+3 \hat{j}+4 \hat{k})$$
= $$3 \hat{i}+\hat{j}-5 \hat{k}$$

∴ $$\overrightarrow{\mathrm{CD}}$$ = P.V. of D – P.V. of C
= $$(\hat{i}+2 \hat{j}+0 \hat{k})-(3 \hat{i}+6 \hat{j}+2 \hat{k})$$
= $$-2 \hat{i}-4 \hat{j}-2 \hat{k}$$

Here $$\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{CD}}$$
= $$(3 \hat{i}+\hat{j}-5 \hat{k}) \cdot(-2 \hat{i}-4 \hat{j}-2 \hat{k})$$
= 3 (- 2) + 1 (- 4) – 5 (- 2)
= – 6 – 4 + 10 = 0

Question 27.
If $$\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}$$ and $$\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}$$, then find the angle between $$2 \vec{a}+\vec{b}$$ and $$\vec{a}+2 \vec{b}$$.
Solution:
Given $$\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}$$
and $$\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}$$

Question 28.
If the points A, B and C with position vectors $$2 \hat{i}+\hat{j}+\hat{k}$$, $$\hat{i}-3 \hat{j}-5 \hat{k}$$ and $$\alpha \hat{i}-3 \hat{j}+\hat{k}$$ respectively are the vertices of a right-angled triangle at C, then find the value (s) of a.
Solution:
Since the points A, B and C with position vectors $$2 \hat{i}+\hat{j}+\hat{k}$$, $$\hat{i}-3 \hat{j}-5 \hat{k}$$ and $$\alpha \hat{i}-3 \hat{j}+\hat{k}$$ are the vertices of right angled triangle at C.

Question 29.
Let $$\vec{a} \text { and } \vec{b}$$ be unit vectors. If the vectors $$\vec{c}=\vec{a}+2 \vec{b}$$ and $$\vec{d}=5 \vec{a}-4 \vec{b}$$ are perpendicualr to each other, then find the angle between $$\vec{a} \text { and } \vec{b}$$.
Solution:
Since $$\vec{a} \text { and } \vec{b}$$ are unit vectors

Question 30.
Express the vector $$\vec{a}=5 \hat{i}-2 \hat{j}+5 \hat{k}$$ as the sum of two vectors such that one is parallel to the vector $$\vec{b}=3 \hat{i}+\hat{k}$$ and the other is perpendicular to .
Solution:
We want to find the vectors $$\vec{c} \text { and } \vec{d}$$ such that
$$\vec{a}=\vec{c}+\vec{d}$$ …………………………..(1)
given $$\vec{a}=5 \hat{i}-2 \hat{j}+5 \hat{k}$$
Since $$\vec{c}$$ is parallel to $$\vec{a}$$
∴ $$\vec{c}=\lambda \vec{b}$$ for some scalar λ
⇒ $$\vec{c}=\lambda(3 \hat{i}+\hat{k})$$
∴ From (1) ; we have
$$\vec{d}=\vec{a}-\vec{c}$$
= $$(5 \hat{i}-2 \hat{j}+5 \hat{k})-(3 \lambda \hat{i}+\lambda \hat{k})$$
⇒ $$\vec{d}=(5-3 \lambda) \hat{i}-2 \hat{j}+(5-\lambda) \hat{k}$$
also it is given that $$\vec{d} \perp \vec{b}$$
⇒ $$\vec{d} \cdot \vec{b}$$ = 0
$$[(5-3 \lambda) \hat{i}-2 \hat{j}+(5-\lambda) k] \cdot(3 \hat{i}+\hat{k})$$ = 0
3 (5 – 3λ) – 2 (0) + (5 – λ) . 1 = 0
15 – 9λ + 5 – λ = 0
⇒ 10λ = 20
⇒ λ = 2.
Thus $$\vec{c}=2(3 \hat{i}+\hat{k})=5 \hat{i}+2 \hat{k}$$
and $$\vec{d}=[(5-6) \hat{i}-2 \hat{j}+(5-2) \hat{k}]$$
= $$-\hat{i}-2 \hat{j}+3 \hat{k}$$

Question 31.
If $$\vec{\alpha}=3 \hat{i}+4 \hat{j}+5 \hat{k}$$ and $$\vec{\beta}=2 \hat{i}+\hat{j}-4 \hat{k}$$ then express $$\vec{\beta}$$ in the form $$\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2$$. where $$\overrightarrow{\beta_1}$$ is parallel to $$\vec{\alpha} \text { and } \overrightarrow{\beta_2}$$ is perpendicualr to $$\vec{\alpha}$$.
Solution:
Given $$\vec{\alpha}=3 \hat{i}+4 \hat{j}+5 \hat{k}$$
and $$\vec{\beta}=2 \hat{i}+\hat{j}-4 \hat{k}$$
Given $$\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2$$ ……………….(1)
given $$\overrightarrow{\beta_1}$$ is parallel to $$\vec{\alpha}$$
∴ $$\overrightarrow{\beta_1}=\lambda \vec{\alpha}$$ for some scalar λ.

Question 32.
If the vectors $$\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$$, $$\vec{b}=2 \hat{i}+4 \hat{j}+\hat{k}$$ and $$\vec{c}=\lambda \hat{i}+\hat{j}+\mu \hat{k}$$ are mutually orthogonal, then find the values of λ and μ.
Solution:
Given $$\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$$
$$\vec{b}=2 \hat{i}+4 \hat{j}+\hat{k}$$
and $$\vec{c}=\lambda \hat{i}+\hat{j}+\mu \hat{k}$$
Since given vectors $$\vec{a}, \vec{b} \text { and } \vec{c}$$ are mutually orthogonal.
∴ $$\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}$$ = 0
Now $$\vec{a} \cdot \vec{c}$$ = 0
⇒ $$(\hat{i}-\hat{j}+2 \hat{k}) \cdot(\lambda \hat{i}+\hat{j}+\mu \hat{k})$$ = 0
⇒ λ – 1 + 2µ = 0
⇒ λ + 2µ = 1 ………………………………(1)
and $$\vec{b} \cdot \vec{c}$$ = 0
⇒ $$(2 \hat{i}+4 \hat{j}+\hat{k}) \cdot(\lambda \hat{i}+\hat{j}+\mu \hat{k})$$ = 0
⇒ 2λ + 4 + µ = 0
⇒ 2λ + µ = – 4 ………………..(2)
On solving (1) and (2) ; we have
λ = – 3 ; µ = 2

Question 33.
Find the values of λ and µ if the vectors $$\lambda \hat{i}-3 \hat{j}-6 \hat{k}$$ and $$3 \hat{i}-\mu \hat{j}-2 \hat{k}$$ are mutually perpendicular vectors of equal magnitude.
Solution:
Let $$\vec{a}=\lambda \hat{i}-3 \hat{j}-6 \hat{k}$$
and $$\vec{b}=3 \hat{i}-\mu \hat{j}-2 \hat{k}$$
Since $$\vec{a} \text { and } \vec{b}$$ are mutually ⊥ to each other.
∴ $$\vec{a} \cdot \vec{b}$$ = 0
⇒ $$(\lambda \hat{i}-3 \hat{j}-6 \hat{k}) \cdot(3 \hat{i}-\mu \hat{j}-2 \hat{k})$$ = 0
⇒ λ (3) – 3 (- µ) – 6 (- 2) = 0
⇒ 3λ + 3µ + 12 = 0
⇒ λ + µ + 4 = 0 ………………..(1)
also $$|\vec{a}|=|\vec{b}|$$
⇒ $$\sqrt{\lambda^2+(-3)^2+(-6)^2}=\sqrt{3^2+(-\mu)^2+(-2)^2}$$
⇒ $$\sqrt{\lambda^2+9+36}=\sqrt{9+\mu^2+4}$$
⇒ $$\sqrt{\lambda^2+45}=\sqrt{13+\mu^2}$$
On squaring ; we have
λ2 – µ2 = 13 – 45 = – 32
⇒ (λ – µ) (λ + µ) = – 32
⇒ (λ – µ) (- 4) = – 32 [using (1)]
⇒ λ – µ = 8 ……………………(2)
On adding (1) and (2) ; we have
2λ = 4
⇒ λ = 2
∴ from (1) ;
µ = – 6.

Question 34.
If $$\vec{a}=\hat{i}+\hat{j}-\hat{k}$$ and $$\vec{b}=\hat{i}-\hat{j}+\hat{k}$$, then find a unit vector which is perpendicular to $$\vec{a}$$ and is coplanar with $$\vec{a} \text { and } \vec{b}$$.
Solution:
Let $$\vec{r}$$ be any vector in the plane of $$\vec{a} \text { and } \vec{b}$$.
Then $$\vec{r}=\lambda \vec{a}+\mu \vec{b}$$,
where λ, µ are scalars.

Question 35.
Find a vector of magnitude 5 units which is coplanar with vectors $$3 \hat{i}-\hat{j}-\hat{k}$$ and $$\hat{i}+\hat{j}-2 \hat{k}$$ and is perpendicular to the vector $$2 \hat{i}+2 \hat{j}+\hat{k}$$.
Solution:
Let the required vector be $$\vec{d}=a \hat{i}+b \hat{j}+c \hat{k}$$
Since $$|\vec{d}|$$ = 5 units
⇒ a2 + b2 + c2 = 25
Let given vectors are $$\vec{\alpha}=3 \hat{i}-\hat{j}-\hat{k}$$
and $$\vec{\beta}=\hat{i}+\hat{j}-2 \hat{k}$$
Since $$\vec{d}$$ is coplanar with $$\vec{\alpha} \text { and } \vec{\beta}$$.

⇒ 2a + 2b + c = 0 ………………(5)
Using eqn. (2), (3) and (4) in eqn. (5) ; we have
2 (3λ + μ) + 2 (- λ + μ) + (- λ – 2μ) = 0
⇒ 3λ + 2μ = 0
⇒ μ = – $$\frac{3 \lambda}{2}$$ ………………(6)
∴ eqn. (2) ; (3) and (4) becomes ;

Question 36.
(i) If $$\vec{a}$$,$$\vec{b}$$ and $$\vec{c}$$ are vectors such that $$\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$$, then show taht the angle θ between $$\vec{b} \text { and } \vec{c}$$ is given that cos θ = $$=\frac{|\vec{a}|^2-|\vec{b}|^2-|\vec{c}|^2}{2|\vec{b}||\vec{c}|}$$.
(ii) If $$\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$$ and $$|\vec{a}|=5,|\vec{b}|=6 \quad \text { and } \quad|\vec{c}|=9$$, then find the angle between $$\vec{a} \text { and } \vec{b}$$.
Solution:
(i)

(ii) Given $$|\vec{a}|=5$$ ;
$$|\vec{b}|=6$$ ;
and $$|\vec{c}|=9$$
Let θ be the angle between $$\vec{a} \text { and } \vec{b}$$

Question 38.
In any ∆ABC, prove by vector method that cos B = $$\frac{c^2+a^2-b^2}{2 c a}$$. (ISC 2010).
Solution:
Let $$\overrightarrow{\mathrm{BC}}, \overrightarrow{\mathrm{CA}}, \overrightarrow{\mathrm{AB}}$$ represents the vectors $$\vec{a}, \vec{b} \text { and } \vec{c}$$ respectively
Now $$\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}+\overrightarrow{\mathrm{AB}}=\overrightarrow{0}$$

Question 39.
Prove by vector method that a diameter of a circle will subtend a right angle at a point on its circumference. (ISC 2003)
Solution:
Let O be the centre of circle with AB as diameter
and P be any point on the circumference of circle.
Take O as origin
let $$\overrightarrow{\mathrm{OP}}=\vec{b}$$ ;
$$\overrightarrow{\mathrm{OB}}=\vec{a}$$ ;
∴ $$\overrightarrow{\mathrm{OA}}=-\vec{a}$$
Now we want to prove that
∠APB = 90°

∴ ∠APB = 90°
Hence, the diameter of a circle will subtend a right angle at a point on its circumference.

Question 40.
Prove by vector method that the diagonals of a rectangle are perpendicular if and only if the rectangle is a square.
Solution:
Let ABCD is a rectangle.