NCERT Exemplar for Class 12 Maths - Vector Algebra - Free PDF Download

Solved Examples 

Short Answer (S.A.)

1. Find the unit vector in the direction of the sum of the vectors $\overrightarrow{a}=2\hat{i}-\hat{j}+2\hat{k}$ and $\overrightarrow{b}=-\hat{i}+\hat{j}+3\hat{k}$.

Ans: Let $\overrightarrow{c}$ be the sum of given vectors i.e $\overrightarrow{a}$ and $\overrightarrow{b}$

$\overrightarrow{c}$ = $\overrightarrow{a}$ + $\overrightarrow{b}$ 

$\overrightarrow{c}$ = $2\hat{i}-\hat{j}+2\hat{k}$ $-\hat{i}+\hat{j}+3\hat{k}$

$\overrightarrow{c}$ = $\hat{i}+5\hat{k}$

|$\overrightarrow{c}$| = $\sqrt{1^2+5^2}$ = $\sqrt{26}$

Now, the unit vector is given by 

$\hat{c}={\displaystyle \frac{\overrightarrow{c}}{\left|\overrightarrow{c}\right|}}={\displaystyle \frac{\hat{i}+5\hat{k}}{\sqrt{26}}}={\displaystyle \frac{1}{\sqrt{26}}}\hat{i}+{\displaystyle \frac{5}{\sqrt{26}}}\hat{k}$ 

2. Find a vector of magnitude 11 in the direction opposite to that of $\overrightarrow{PQ}$, where P and Q are the points (1, 3, 2) and (–1, 0, 8), respectively.

Ans: The vector with initial point P (1, 3, 2) and terminal point Q (–1, 0, 8) is given by

$\overrightarrow{\text{PQ}}$ = $(-1-1)\hat{i}+(0-3)\hat{j}+(8-2)\hat{k}$

$\overrightarrow{\text{PQ}}$ = $-2\hat{i}-3\hat{j}+6\hat{k}$

We know , $\overrightarrow{\text{QP}}$  = -$\overrightarrow{\text{PQ}}$

$\overrightarrow{\text{QP}}$  = $$-(-2\hat{i}-3\hat{j}+6k)$$

$\overrightarrow{\text{QP}}$  = $$-2\hat{i}-3\hat{j}+6k$$

Now , the unit vector in the direction of $\overrightarrow{\text{QP}}$ is given by 

$\widehat{QP}={\displaystyle \frac{\overrightarrow{QP}}{\left|\overrightarrow{QP}\right|}}={\displaystyle \frac{2\hat{i}+3\hat{j}-6\hat{k}}{\sqrt{2^2+3^2+{(-6)}^2}}}={\displaystyle \frac{2\hat{i}+3\hat{j}-6\hat{k}}{\sqrt{49}}}={\displaystyle \frac{2}{7}}\hat{i}+{\displaystyle \frac{3}{7}}\hat{j}-{\displaystyle \frac{6}{7}}\hat{k}$

Therefore , the  vector of magnitude 11 in direction of $\overrightarrow{\text{QP}}$ is given by 

11$\widehat{QP}$ = 11$({\displaystyle \frac{2}{7}}\hat{i}+{\displaystyle \frac{3}{7}}\hat{j}-{\displaystyle \frac{6}{7}}\hat{k})$=${\displaystyle \frac{22}{7}}\hat{i}+{\displaystyle \frac{33}{7}}\hat{j}-{\displaystyle \frac{66}{7}}\hat{k}$

3. Find the position vector of a point R which divides the line joining the two points P and Q with position vectors   $\overrightarrow{OP}$  = $2\hat{a}+\hat{b}$  and  $\overrightarrow{OQ}$  = $\hat{a}-2\hat{b}$ , respectively, in the ratio 1:2, 

(i) internally and (ii) externally.

Ans: (i) The position vector of the point R which is dividing the join of P and Q internally in the ratio 1:2 is given by

$\overrightarrow{OR}$ =${\displaystyle \frac{2(2\hat{a}+\hat{b})+1(\hat{a}-2\hat{b})}{1+2}}$ =${\displaystyle \frac{5\hat{a}}{3}}$

(ii) The position vector of the point R′  which is dividing the join of P and Q in the ratio 1 : 2 externally is given by

$\overrightarrow{O{R}^{\prime }}$ =${\displaystyle \frac{2(2\hat{a}+\hat{b})-1(\hat{a}-2\hat{b})}{2-1}}$  = $3\hat{a}+4\hat{b}$

4. If the points (–1, –1, 2), (2, m, 5) and (3,11, 6) are collinear, find the value of m.

Ans: Given points are :  A (–1, –1, 2), B (2, m, 5) and C (3, 11, 6). 

It is given that the points are collinear 

$\overrightarrow{AB}$  = $(2+1)\hat{i}+(m+1)\hat{j}+(5-2)\hat{k}$

$\overrightarrow{AB}$  = $3\hat{i}+(m+1)\hat{j}+3\hat{k}$

Also, 

$\overrightarrow{AC}$  = $(3+1)\hat{i}+(11+1)\hat{j}+(6-2)\hat{k}$

$\overrightarrow{AC}$  = $4\hat{i}+12\hat{j}+4\hat{k}$

Since A, B, C, are collinear, we have

$\overrightarrow{AB}$  = λ$\overrightarrow{AC}$ 

$3\hat{i}+(m+1)\hat{j}+3\hat{k}$ = λ($4\hat{i}+12\hat{j}+4\hat{k}$)

⇒ 3 = 4 λ and m + 1 = 12 λ

Hence , m = 8

5. Find a vector $\overrightarrow{r}$ of magnitude 3$\sqrt{\text{2}}$ units which makes an angle of ${\displaystyle \frac{\pi }{4}}$ and ${\displaystyle \frac{\pi }{2}}$ with y and z - axes, respectively.

Ans: Here we have been given m and n as 

m = cos${\displaystyle \frac{\pi }{4}}$ = ${\displaystyle \frac{1}{\sqrt{2}}}$ 

and n = cos${\displaystyle \frac{\pi }{2}}$ = 0

we know that 

$l^2+m{}^2+n^2=1$

$\Rightarrow l^2+{\displaystyle \frac{1}{2}}+0=1$

$\Rightarrow l^{}=\pm {\displaystyle \frac{1}{\sqrt{2}}}$

Therefore, the required vector $\overrightarrow{r}$  =3$\sqrt{2}$( $l\hat{i}+m\hat{j}+n\hat{k}$)

$\overrightarrow{r}$=3$\sqrt{2}$($\pm {\displaystyle \frac{1}{\sqrt{2}}}\hat{i}+{\displaystyle \frac{1}{\sqrt{2}}}\hat{j}+0\hat{k}$)

$\Rightarrow \overrightarrow{r}$  = $\pm 3\hat{i}+3\hat{j}$

Hence ,  $\overrightarrow{r}$  = $\pm 3\hat{i}+3\hat{j}$

6. If  $\overrightarrow{a}$  = $2\hat{i}-\hat{j}+\hat{k}$ , $\overrightarrow{b}$  = $\hat{i}+\hat{j}-2\hat{k}$ and $\overrightarrow{c}$  = $\hat{i}+3\hat{j}-\hat{k}$ , find λ such that $\overrightarrow{a}$  is perpendicular to λ$\overrightarrow{b}$ + $\overrightarrow{c}$ .

Ans: We have

λ$\overrightarrow{b}$ + $\overrightarrow{c}$  = λ($\hat{i}+\hat{j}-2\hat{k}$)  + ($\hat{i}+3\hat{j}-\hat{k}$)

⇒λ$\overrightarrow{b}$ + $\overrightarrow{c}$  = (λ+1)$\hat{i}+(\lambda +3)\hat{j}-(2\lambda +1)\hat{k}$)  

Now since $\overrightarrow{a}$ is perpendicular to λ$\overrightarrow{b}$ + $\overrightarrow{c}$

$\overrightarrow{a}$.(λ$\overrightarrow{b}$ + $\overrightarrow{c}$ ) = 0

$\Rightarrow (2\hat{i}-\hat{j}+\hat{k}$).(λ$\overrightarrow{b}$ + $\overrightarrow{c}$ )  = 0

$\Rightarrow (2\hat{i}-\hat{j}+\hat{k}$).λ$[(\hat{i}+\hat{j}-2\hat{k})+(\hat{i}+3\hat{j}-\hat{k})]$ = 0

⇒2 (λ + 1) – (λ + 3) – (2λ + 1) =0

⇒λ = -2

Hence the value of λ = -2

7. Find all vectors of magnitude 10$\sqrt{3}$that are perpendicular to the plane of  $\hat{i}+2\hat{j}+\hat{k}$  and   -$\hat{i}+\widehat{3j}+4\hat{k}$.

Ans:  Let us assume that $\overrightarrow{a}$ = $\hat{i}+2\hat{j}+\hat{k}$  and $\overrightarrow{b}$ =  -$\hat{i}+\widehat{3j}+4\hat{k}$

Now, $\overrightarrow{a}\times \overrightarrow{b}$=($\hat{i}+2\hat{j}+\hat{k}$)x(-$\hat{i}+\widehat{3j}+4\hat{k}$)

$\Rightarrow \overrightarrow{a}\times \overrightarrow{b}$  = ( $\hat{i}+2\hat{j}+\hat{k}$)x(-$\hat{i}+\widehat{3j}+4\hat{k}$)

$\Rightarrow \overrightarrow{a}\times \overrightarrow{b}$  = $\hat{i}(8-3)-\hat{j}(4+1)+\hat{k}$(3+2)

$\Rightarrow \overrightarrow{a}\times \overrightarrow{b}$    =  $\widehat{5i}-5\hat{j}+5\hat{k}$

⇒|$\overrightarrow{a}$×$\overrightarrow{b}$ | = $\sqrt{5^2{+}_{}{(-5)}^2+5^2}$ =  5$\sqrt{3}$

 Unit vector perpendicular to the plane of a  and b  is given by

${\displaystyle \frac{\overrightarrow{a}\times \overrightarrow{b}}{|\overrightarrow{a}\times \overrightarrow{b}|}}$=  ${\displaystyle \frac{\widehat{5i}-5\hat{j}+5\hat{k}}{5\sqrt{3}}}$

Therefore ,  vectors of magnitude of 10$\sqrt{3}$  that are perpendicular to plane of a  and b  are  ±10$\sqrt{3}$ (${\displaystyle \frac{\widehat{5i}-5\hat{j}+5\hat{k}}{5\sqrt{3}}}$)  =  ±10(   $\hat{i}-\hat{j}+\hat{k}$)

Long Answer (L.A)

8. Using vectors, prove that cos (A – B) = cosA cosB + sinA sinB.

Ans: Let $\widehat{OP}$ and $\widehat{OQ}$ be unit vectors making angles A and B, respectively, with positive direction of x-axis. Then ∠QOP = A – B

Now, We know $\widehat{OP}$=$\widehat{OM}$ + $\widehat{MP}$ =   $\hat{i}cosA+\hat{j}sinA$

and $\widehat{OQ}$=$\widehat{ON}$ + $\widehat{NQ}$ = $cosB\hat{i}+sinB\hat{j}$

By definition , we have 

$\widehat{OP}$ .$\widehat{OQ}$  = |$\widehat{OP}$||$\widehat{OQ}$| cos A-B =  cos (A – B) ……..eqn (1)

In terms of components, we have

$\widehat{OP}$.$\widehat{OQ}$=($\hat{i}cosA+\hat{j}sinA)$.$\widehat{(i}cosB+\hat{j}sinB)$ 

= cosA cosB + sinA sinB                                        ……….eqn(2)

From (1) and (2), we have 

cos (A – B) = cosA cosB + sinA sinB.

Hence Proved 

9. Prove that in a ∆ ABC,  ${\displaystyle \frac{sinA}{a}}={\displaystyle \frac{sinB}{b}}={\displaystyle \frac{sinC}{c}}$  where a, b, c represent the magnitudes of the sides opposite to vertices A, B, C, respectively.

Ans: Let the three sides of the triangle BC, CA and AB be represented by $\overrightarrow{a}$ , $\overrightarrow{b}$ , and $\overrightarrow{c}$ respectively 

Now we have , 

$\overrightarrow{a}$ + $\overrightarrow{b}$ + $\overrightarrow{c}$ = 0 

⇒$\overrightarrow{a}$ + $\overrightarrow{b}$  =  -$\overrightarrow{c}$

On pre cross multiplying by $\overrightarrow{a}$  , and post cross multiplying by $\overrightarrow{b}$ , we get 

$\overrightarrow{a}$ × $\overrightarrow{b}$  =  $\overrightarrow{c}$ × $\overrightarrow{a}$ 

and $\overrightarrow{a}$ × $\overrightarrow{b}$  = $\overrightarrow{b}$ × $\overrightarrow{c}$ 

Therefore , $\overrightarrow{a}$ × $\overrightarrow{b}$  =  $\overrightarrow{b}$ × $\overrightarrow{c}$  =  $\overrightarrow{c}$ × $\overrightarrow{a}$ 

⇒|$\overrightarrow{a}$ × $\overrightarrow{b}$ | =  |$\overrightarrow{b}$ × $\overrightarrow{c}|$  = | $\overrightarrow{c}$ × $\overrightarrow{a}$ |

⇒|$\overrightarrow{a}$||$\overrightarrow{b}$|sin(π–C)=|$\overrightarrow{b}$||$\overrightarrow{c}$|sin(π–A)=|$\overrightarrow{c}$||$\overrightarrow{a}$|sin(π-B)

ab sin C = bc sinA = ca sinB

Dividing by abc, we get

${\displaystyle \frac{sinC}{c}}={\displaystyle \frac{sinA}{a}}={\displaystyle \frac{sinB}{b}}$

$\Rightarrow {\displaystyle \frac{sinA}{a}}={\displaystyle \frac{sinB}{b}}={\displaystyle \frac{sinC}{c}}$

Hence Proved 

Objective Type Questions 

10. The magnitude of the vector 6$\hat{i}+2\hat{j}+3\hat{k}$  is

(A) 5 

(B) 7

(C) 12

(D) 1

Ans: The correct answer is option (B).

Let $\overrightarrow{a}$ = 6$\hat{i}+2\hat{j}+3\hat{k}$  

Now magnitude of $\overrightarrow{a}$ is given by 

|$\overrightarrow{a}$| = $\sqrt{6^2+2^2+3^2}$ = $\sqrt{36+4+9}$  = $\sqrt{49}$   = 7.

Therefore magnitude of the vector $\overrightarrow{a}$ is 7units

Hence the correct option is (B)

11. The position vector of the point which divides the join of points with position vectors $\overrightarrow{a}$ + $\overrightarrow{b}$  and  2$\overrightarrow{a}$ - $\overrightarrow{b}$ in the ratio 1 : 2 is

(A) ${\displaystyle \frac{\overrightarrow{3a}+2\overrightarrow{b}}{3}}$

(B)  $\overrightarrow{a}$

(C) ${\displaystyle \frac{\overrightarrow{5a}-\overrightarrow{b}}{3}}$

(D) ${\displaystyle \frac{\overrightarrow{4a}+\overrightarrow{b}}{3}}$

Ans: The correct answer is option (D).

The position vector of required point by using section formula is calculated as 

${\displaystyle \frac{2(\overrightarrow{a}+\overrightarrow{b})+1(2\overrightarrow{a}-\overrightarrow{b})}{2+1}}$ = ${\displaystyle \frac{\overrightarrow{4a}+\overrightarrow{b}}{3}}$

Hence the correct option is (D)

12. The vector with initial point P (2, –3, 5) and terminal point Q(3, –4, 7) is

(A) $\hat{i}-\hat{j}+2\hat{k}$  

(B)  5$\hat{i}-7\hat{j}+12\hat{k}$  

(C) -$\hat{i}+\hat{j}-2\hat{k}$  

(D) 0

Ans: Option (A) is the correct answer.

Initial point P (2, –3, 5)

Terminal point Q(3, –4, 7)

The vector is given by 

(3-2)$\hat{i}$ + (-4-(-3))$\hat{j}$ + (7-5)$\hat{k}$  = $\hat{i}-\hat{j}+2\hat{k}$  

Hence the correct option is (A).

13. The angle between the vectors  $\hat{i}-\hat{j}$ and $\hat{j}-\hat{k}$ is 

(A) ${\displaystyle \frac{\pi }{3}}$

(B) ${\displaystyle \frac{2\pi }{3}}$

(C) ${\displaystyle \frac{-\pi }{3}}$ 

(D) ${\displaystyle \frac{5\pi }{6}}$

Ans: Option (B) is the correct answer.

The angle between two vectors is given by 

$\cos \theta$ = ${\displaystyle \frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}|.|\overrightarrow{b}\right|}}$ = $${\displaystyle \frac{(\hat{i}-\hat{j}).(\hat{j}-\hat{k})}{|(\hat{i}-\hat{j})|.|(\hat{j}-\hat{k})|}}$$  = ${\displaystyle \frac{-1}{\sqrt{1^2+1^2}.\sqrt{1^2+1^2}}}$ = ${\displaystyle \frac{-1}{2}}$ 

θ  = ${\displaystyle \frac{2\pi }{3}}$

Hence the correct option is (B)

14. The value of λ for which the two vectors $2\hat{i}-\hat{j}+2\hat{k}$ and $3\hat{i}+\lambda \hat{j}+\hat{k}$ are perpendicular is

(A) 2 

(B) 4 

(C) 6 

(D) 8

Ans: Option (D) is the correct answer.

Since the vectors are perpendicular to each other 

$(2\hat{i}-\hat{j}+2\hat{k})$.$(3\hat{i}+\lambda \hat{j}+\hat{k})$ = 0                               

$\Rightarrow 6-\lambda +2$ = 0

$\Rightarrow \lambda$ = 8

Hence the correct option is (D)

15. The area of the parallelogram whose adjacent sides are  $\hat{i}+\hat{k}$  and $2\hat{i}+\hat{j}+\hat{k}$  is 

(A) $\sqrt{2}$

(B) $\sqrt{3}$  

(C) 3

(D) 4 

Ans: Area of parallelogram whose adjacent sides are  $\overrightarrow{a}$ and $\overrightarrow{b}$   is given by |a- x$\overrightarrow{b}$ | 

$\overrightarrow{a}$×$\overrightarrow{b}$   = ($\hat{i}+\hat{k}$) ×$(2\hat{i}+\hat{j}+\hat{k}$) 

$\Rightarrow \overrightarrow{a}$×$\overrightarrow{b}$ = $-\hat{i}+\hat{j}+\hat{k}$

⇒|$\overrightarrow{a}$×$\overrightarrow{b}$ | = $\sqrt{{(-1)}^2+{(1)}^2+{(1)}^2}$ = $\sqrt{3}$  

Hence the correct option is (B)

16. If  |$\overrightarrow{a}$|=8 , |$\overrightarrow{b}$| = 3 and | $\overrightarrow{a}\times \overrightarrow{b}$| = 12, then the value of $\overrightarrow{a}\cdot \overrightarrow{b}$ is

(A) 6$\sqrt{3}$

(B) 8$\sqrt{3}$

(C) 12$\sqrt{3}$ 

(D) None of these

Ans: Option (C) is the correct answer.

we know that

|$\overrightarrow{a}$ x$\overrightarrow{b}$| =  |$\overrightarrow{a}$| .|$\overrightarrow{b}$| |sin$\theta$|

⇒12 = 8.3.|sin$\theta$|

⇒|sin$\theta$| = ${\displaystyle \frac{1}{2}}$

⇒$\theta$ = ±${\displaystyle \frac{\pi }{6}}$

Now we have,

$\Rightarrow \overrightarrow{a}$ .$\overrightarrow{b}$  = |$\overrightarrow{a}$| .|$\overrightarrow{b}$| |cos$\theta$|

$\Rightarrow \overrightarrow{a}$.$\overrightarrow{b}$  = |$\overrightarrow{a}$| .|$\overrightarrow{b}$| |cos$\theta$| = 8 x 3 x${\displaystyle \frac{\sqrt{3}}{2}}$ = 12$\sqrt{3}$

Therefore, $\overrightarrow{a}$ .$\overrightarrow{b}$  = 12$\sqrt{3}$

Hence the correct option is (C)

17. The 2 vectors $\hat{j}+\hat{k}$ and $3\hat{i}-\hat{j}+\widehat{4k}$   represents the two sides AB and AC, respectively of a ∆ABC. The length of the median through A is

(A) ${\displaystyle \frac{\sqrt{\text{34}}}{\text{2}}}$

(B) ${\displaystyle \frac{\sqrt{\text{48}}}{\text{2}}}$

(C) $\sqrt{\text{18}}$

(D) None of these

Ans: The correct answer is  option (A).

We know that Median $\overrightarrow{AD}$ is given by 

$|\overrightarrow{AD}$| =  ${\displaystyle \frac{1}{2}}$| $3\hat{i}-\hat{j}+\widehat{4k}$|

$\Rightarrow |\overrightarrow{AD}$| =  ${\displaystyle \frac{\sqrt{34}}{2}}$

Hence the correct option is (A).

18. The projection of vector $\overrightarrow{a}$ = 2$\hat{i}-\hat{j}+\hat{k}$  along  $\overrightarrow{b}$ = $\hat{i}+2\hat{j}+2\hat{k}$  is

(A) ${\displaystyle \frac{\text{2}}{\text{3}}}$

(B) ${\displaystyle \frac{\text{1}}{\text{3}}}$

(C) 2

(D) $\sqrt{\text{6}}$

Ans: Option (A) is the correct answer.

Projection of a vector $\overrightarrow{a}$ on  $\overrightarrow{b}$ is given by 

${\displaystyle \frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{b}\right|}}$ = ${\displaystyle \frac{(2\hat{i}-\hat{j}+\hat{k}).(\hat{i}+2\hat{j}+2\hat{k})}{(|2\hat{i}-\hat{j}+\hat{k})|.(|\hat{i}+2\hat{j}+2\widehat{k|})}}$ = ${\displaystyle \frac{2}{3}}$

Hence the correct option is (A)

19. If $\overrightarrow{a}$ and $\overrightarrow{b}$  are unit vectors, then what is the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$  for $\sqrt{3}\overrightarrow{a}-\overrightarrow{b}$  to be a unit vector?

(A) 30°

(B) 45° 

(C) 60°

(D) 90°

Ans: Option (A) is the correct answer.

Here we have 

${\left(\sqrt{3}\overrightarrow{a}-\overrightarrow{b}\right)}^2=3{\overrightarrow{a}}^2+{\overrightarrow{b}}^2-2\sqrt{3}\overrightarrow{a}.\overrightarrow{b}$

$\Rightarrow \overrightarrow{a}$.$\overrightarrow{b}$ = ${\displaystyle \frac{\sqrt{3}}{2}}$

⇒cosθ =${\displaystyle \frac{\sqrt{3}}{2}}$

⇒θ = 30°.

Hence the correct option is (A)

20. The unit vector perpendicular to the vectors $\hat{i}-\hat{j}$ and $\hat{i}+\hat{j}$ forming a right handed system is

(A) $\hat{k}$

(B) –$\hat{k}$

(C) $${\displaystyle \frac{\hat{i}-\hat{j}}{\sqrt{2}}}$$

(D) $${\displaystyle \frac{\hat{i}+\hat{j}}{\sqrt{2}}}$$

Ans: Option (A) is the correct answer.

Unit vector perpendicular to the vectors $\hat{i}-\hat{j}$ and $\hat{i}+\hat{j}$is given by  

${\displaystyle \frac{(\hat{i}-\hat{j}\times (\hat{i}+\hat{j})}{|(\hat{i}-\hat{j})\times (\hat{i}+\hat{j})|}}$  =  $\hat{k}$

Hence the correct option is (A)

21. If  |$\overrightarrow{a}$| = 3 and -1$\le k\le 2,$then |k$\overrightarrow{a}$| lies in the interval 

(A) [0, 6] 

(B) [– 3, 6]

(C) [ 3, 6] 

(D) [1, 2]

Ans: Option (A) is the correct answer.

Given  |$\overrightarrow{a}$| = 3 and -1$\le k\le 2,$

The smallest value of |k$\overrightarrow{a}$|  will exist at the numerically smallest value of k, i.e., at = 0, which gives 

|k$\overrightarrow{a}$|  = |k||$\overrightarrow{a}$|  = 0x3 = 0  

The numerically greatest value of k is 2 at which |k$\overrightarrow{a}$| =6  . 

|k$\overrightarrow{a}$| $\in$ [0, 6] 

Hence the correct option is (A)

Exercise – 10.3

Short Answer (S.A.) 

1. Find the unit vector in the direction of sum of vectors $$\overrightarrow{a}=2\hat{i}-\hat{j}+\hat{k}$$ and $$\overrightarrow{b}=2\hat{j}+\hat{k}$$.

Ans: Let $\overrightarrow{c}$ denote the sum of $\overrightarrow{a}$ and $\overrightarrow{b}$. So, 

$\overrightarrow{c}=\overrightarrow{a}+\overrightarrow{b}=2\hat{i}-\hat{j}+\hat{k}+(2\hat{j}+\hat{k})=2\hat{i}+\hat{j}+2\hat{k}$
$\left|\overrightarrow{c}\right|=\sqrt{2^2+1^2+2^2}=\sqrt{9}=3$  

Unit vector along vector c will be

$\hat{c}={\displaystyle \frac{\overrightarrow{c}}{\left|\overrightarrow{c}\right|}}={\displaystyle \frac{2\hat{i}+\hat{j}+2\hat{k}}{3}}={\displaystyle \frac{2}{3}}\hat{i}+{\displaystyle \frac{1}{3}}\hat{j}+{\displaystyle \frac{2}{3}}\hat{k}$ 

2.If $$\overrightarrow{a}=\hat{i}+\hat{j}+2\hat{k}$$ and $$\overrightarrow{b}=2\hat{i}+\hat{j}-2\hat{k}$$ find the unit vector in the direction of 

(i) $$6\overrightarrow{b}$$

Ans: Given that $\overrightarrow{a}=\hat{i}+\hat{j}+2\hat{k}$ and $\overrightarrow{b}=2\hat{i}+\hat{j}-2\hat{k}$

(i) We have to find unit vector along

$6\overrightarrow{b}=12\hat{i}+6\hat{j}-12\hat{k}$ 

We know that all parallel vectors have same unit vector so, unit vector along $6\overrightarrow{b}$ will be same as along $\overrightarrow{b}$ 

$\hat{b}={\displaystyle \frac{\overrightarrow{b}}{\left|\overrightarrow{b}\right|}}={\displaystyle \frac{2\hat{i}+\hat{j}-2\hat{k}}{\sqrt{2^2+1^2+2^2}}}={\displaystyle \frac{1}{\sqrt{9}}}(2\hat{i}+\hat{j}-2\hat{k})$ 

$\hat{b}={\displaystyle \frac{2}{3}}\hat{i}+{\displaystyle \frac{1}{3}}\hat{j}-{\displaystyle \frac{2}{3}}\hat{k}$ 

(ii) $$2\overrightarrow{a}-\overrightarrow{b}$$

Ans: We have to find unit vector along $\overrightarrow{c}=2\overrightarrow{a}-\overrightarrow{b}$ 

$\overrightarrow{c}=2\overrightarrow{a}-\overrightarrow{b}=2(\hat{i}+\hat{j}+2\hat{k})-(2\hat{i}+\hat{j}-2\hat{k})$ 

$=\hat{j}+6\hat{k}$ 

$\hat{c}={\displaystyle \frac{\overrightarrow{c}}{\left|\overrightarrow{c}\right|}}={\displaystyle \frac{\hat{j}+6\hat{k}}{\sqrt{1^2+6^2}}}={\displaystyle \frac{1}{\sqrt{37}}}(\hat{j}+6\hat{k})$ 

3. Find a unit vector in the direction of  $\overrightarrow{\text{PQ}}$  where P and Q have coordinates (5, 0, 8) and (3, 3, 2), respectively.

Ans: Given that P and Q have coordinates (5, 0, 8) and (3, 3, 2).

$\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}=(3\hat{i}+3\hat{j}+2\hat{k})-(5\hat{i}+8\hat{k})=-2\hat{i}+3\hat{j}-6\hat{k}$ 

$\Rightarrow \widehat{PQ}$=${\displaystyle \frac{\overrightarrow{PQ}}{\left|\overrightarrow{PQ}\right|}}$=${\displaystyle \frac{-2\hat{i}+3\hat{j}-6\hat{k}}{\sqrt{2^2+3^2+6^2}}}$=${\displaystyle \frac{1}{\sqrt{49}}}(-2\hat{i}+3\hat{j}-6\hat{k})$ 

$\Rightarrow \widehat{PQ}={\displaystyle \frac{1}{7}}(-2\hat{i}+3\hat{j}-6\hat{k})$ 

4. If $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$ are the position vectors of A and B, respectively, find the position vector of a point C in BA produced such that $BC=1.5BA$. 

Ans: Given that $OA=\overrightarrow{a}$ and $OB=\overrightarrow{b}$

$BA=OA-OB=\overrightarrow{a}-\overrightarrow{b}$ 

$\Rightarrow BC=1.5BA=1.5(\overrightarrow{a}-\overrightarrow{b})$.  

$\Rightarrow BC=OC-OB=\overrightarrow{c}-\overrightarrow{b}$ 

$\Rightarrow 1.5(\overrightarrow{a}-\overrightarrow{b})=\overrightarrow{c}-\overrightarrow{b}$ 

$\Rightarrow \overrightarrow{c}=1.5\overrightarrow{a}-1.5\overrightarrow{b}+\overrightarrow{b}$ 

$\Rightarrow OC=\overrightarrow{c}={\displaystyle \frac{1}{2}}(3\overrightarrow{a}-\overrightarrow{b})$ 

5. Using vectors, find the value of k such that the points A (k, – 10, 3), B (1, –1, 3) and C (3, 5, 3) are collinear.

Ans: Given points are A (k, – 10, 3), B (1, –1, 3) and C (3, 5, 3) Since points are collinear so

$AB=x\left(BC\right)\Rightarrow OB-OA=x(OC-OB)$

$\Rightarrow (\hat{i}-\hat{j}+3\hat{k})-(k\hat{i}-10\hat{j}+3\hat{k})=x[(3\hat{i}+5\hat{j}+3\hat{k})-(\hat{i}-\hat{j}+3\hat{k})]$ 

$\Rightarrow (1-k)\hat{i}+9\hat{j}=x(2\hat{i}+6\hat{j})$ 

By comparing the terms, we have

$1-k=2x\Rightarrow k=1-2x$ 

$\Rightarrow 9=6x\Rightarrow 2x=3$ 

$\Rightarrow k=1-3=-2$ 

Hence the value of $k=-2$. 

6. A vector  $$\overrightarrow{r}$$  is inclined at equal angles to the three axes. If the magnitude of  $$\overrightarrow{r}$$  is  $2\sqrt{3}$ units, find $$\overrightarrow{r}$$.

Ans: Given $\left|\overrightarrow{r}\right|=2\sqrt{3}$ and $\overrightarrow{r}$ is equally inclined to axes so

$l=m=n$

$l^2+m^2+n^2=1\Rightarrow 3l^2=1$ 

$\Rightarrow$ l = ±${\displaystyle \frac{1}{\sqrt{3}}}=m=n$

Now, $\overrightarrow{r}=\left|\overrightarrow{r}\right|(l\hat{i}+n\hat{j}+m\hat{k})$ 

$\overrightarrow{r}=2\sqrt{3}(\pm {\displaystyle \frac{1}{\sqrt{3}}}\hat{i}\pm {\displaystyle \frac{1}{\sqrt{3}}}\hat{j}\pm {\displaystyle \frac{1}{\sqrt{3}}}\hat{k})=\pm 2(\hat{i}+\hat{j}+\hat{k})$ 

7. A vector $$\overrightarrow{r}$$ has magnitude 14 and direction ratios 2, 3, – 6. Find the direction cosines and components of $$\overrightarrow{r}$$ , given that $$\overrightarrow{r}$$ makes an acute angle with x-axis.

Ans: Given that $\left|\overrightarrow{r}\right|=14$ and direction ratios 2, 3, -6. 

Let direction cosines are $l,mandn$

$\therefore l={\displaystyle \frac{2x}{14}}={\displaystyle \frac{x}{7}}$ 

$\therefore m={\displaystyle \frac{3x}{14}}$ 

$\therefore n=-{\displaystyle \frac{6x}{14}}=-{\displaystyle \frac{3x}{7}}$ 

Now, $l^2+m^2+n^2=1$ 

${\left({\displaystyle \frac{x}{7}}\right)}^2+{\left({\displaystyle \frac{3x}{14}}\right)}^2+{(-{\displaystyle \frac{3x}{7}})}^2=1$ 

$\Rightarrow {\displaystyle \frac{x^2}{49}}+{\displaystyle \frac{9x^2}{196}}+{\displaystyle \frac{9x^2}{49}}=1$ 

$\Rightarrow (4+9+36)x^2=196$ 

$\Rightarrow 49x^2=4\times 49$ 

$\Rightarrow x^2=4$ 

$\therefore x=\pm 2$ 

Since $\overrightarrow{r}$ makes acute angle with x-axis so value of $x=2$ 

Now, $l={\displaystyle \frac{2}{7}},m={\displaystyle \frac{3}{7}}$ and $n=-{\displaystyle \frac{6}{7}}$ 

$\overrightarrow{r}=\left|\overrightarrow{r}\right|(l\hat{i}+n\hat{j}+m\hat{k})=14({\displaystyle \frac{2}{7}}\hat{i}+{\displaystyle \frac{3}{7}}\hat{j}-{\displaystyle \frac{6}{7}}\hat{k})$ 

$\overrightarrow{r}=4\hat{i}+6\hat{j}-12\hat{k}$ 

8. Find a vector of magnitude 6, which is perpendicular to both the vectors $$2\hat{i}-\hat{j}+2\hat{k}$$ and $$4\hat{i}-\hat{j}+3\hat{k}$$.

Ans: Let $\overrightarrow{a}=2\hat{i}-\hat{j}+2\hat{k}$ and $\overrightarrow{b}=4\hat{i}-\hat{j}+3\hat{k}$ 

Vector product of the vectors is always perpendicular to both the vectors. So, 

$\overrightarrow{c}=\overrightarrow{a}\times \overrightarrow{b}=|\hat{i}\hat{j}\hat{k}2-124-13|=(-3+2)\hat{i}-(6-8)\hat{j}+(-2+4)\hat{k}$ 

$\overrightarrow{c}=-\hat{i}+2\hat{j}+2\hat{k}$ 

Required vector should have magnitude 6 then 

$\overrightarrow{r}=6\times {\displaystyle \frac{\overrightarrow{c}}{\left|\overrightarrow{c}\right|}}=6\times {\displaystyle \frac{-\hat{i}+2\hat{j}+2\hat{k}}{\sqrt{1^2+2^2+2^2}}}={\displaystyle \frac{6}{\sqrt{9}}}(-\hat{i}+2\hat{j}+2\hat{k})$ 

$\Rightarrow \overrightarrow{r}=-2\hat{i}+4\hat{j}+4\hat{k}$

9. Find the angle between the vectors $$2\hat{i}-\hat{j}+\hat{k}$$ and  $$3\hat{i}+4\hat{j}+\hat{k}$$. 

Ans: Given  $\overrightarrow{a}=2\hat{i}-\hat{j}+\hat{k}$  and  $\overrightarrow{b}=3\hat{i}+4\hat{j}-\hat{k}$

Angle between vectors be $\theta$ then

$cos\theta ={\displaystyle \frac{\overrightarrow{a}\bullet \overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}}={\displaystyle \frac{1}{\sqrt{2^2+1^2+1^2}\sqrt{3^2+4^2+1^2}}}(2\hat{i}-\hat{j}+\hat{k})\bullet (3\hat{i}+4\hat{j}-\hat{k})$