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 $\vec{a}=2\hat{i}-\hat{j}+2\hat{k}$ and $\vec{b}=-\hat{i}~+\hat{j}~+3\hat{k}$.

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

$\vec{c}$ = $\vec{a}$ + $\vec{b}$ 

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

$\vec{c}$ = $~\hat{i}+5\hat{k}$

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

Now, the unit vector is given by 

$\hat{c}=\dfrac{{\vec{c}}}{\left|{\vec{c}}\right|}=\dfrac{\hat{i}+5\hat{k}~}{\sqrt{26}}=\dfrac{1}{\sqrt{26}}\hat{i}+\dfrac{5}{\sqrt{26}}\hat{k}$ 

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

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

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

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

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

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

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

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

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

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

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

$\overrightarrow{OR}$ =$\dfrac{2\left( 2\hat{a}~+\hat{b}~ \right)~+~1\left( \hat{a}~-2\hat{b}~~ \right)}{1+2}$ =$\dfrac{5\hat{a}}{3}$

(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

$\overrightarrow{OR'}$ =$\dfrac{2\left( 2\hat{a}~+\hat{b}~ \right)~-~1\left( \hat{a}~-2\hat{b}~~ \right)}{2-1}$  = $3\hat{a}~+4\hat{b}~$

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 

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

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

Also, 

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

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

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

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

$3\hat{i}~+\left( m+1 \right)\hat{j}~+3\hat{k}$ = 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 $\vec{r}$ of magnitude 3$\sqrt{\text{2}}$ units which makes an angle of $\dfrac{\pi }{4}$ and $\dfrac{\pi }{2}$ with y and z - axes, respectively.

Ans: Here we have been given m and n as 

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

and n = cos$\dfrac{\pi }{2}$ = 0

we know that 

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

$\Rightarrow {{l}^{2~}}~+\dfrac{1}{2}+0~~=~1$

$\Rightarrow {{l}^{~}}~=\pm \dfrac{1}{\sqrt{2}}~$

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

$\vec{r}$=3$\sqrt{2}$($\pm\dfrac{1}{\sqrt{2}}\hat{i}~+\dfrac{1}{\sqrt{2}}\hat{j}~+0\hat{k}$)

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

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

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

Ans: We have

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

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

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

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

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

$\Rightarrow(2\hat{i}~-\hat{j}~+\hat{k}$).λ$\left[\left(\hat{i}~+\hat{j}~-2\hat{k}\right)+\left(\hat{i}~+3\hat{j}~-\hat{k}\right)\right]$ = 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   -$\hat{i}~+\hat{3j}~+4\hat{k}$.

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

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

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

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

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

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

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

$\dfrac{\vec{a}\times\vec{b}~~}{\left| \vec{a}\times\vec{b}~ \right|}~$=  $\dfrac{\hat{5i}~-5\hat{j}~+5\hat{k}~~}{5\sqrt{3}}~$

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

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

In terms of components, we have

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

= 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,  $\dfrac{sinA}{a}~=~\dfrac{sinB}{b}~=~\dfrac{sinC}{c}~~$  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 $\vec{a}$ , $\vec{b}$ , and $\vec{c}$ respectively 

Now we have , 

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

⇒$\vec{a}$ + $\vec{b}$  =  -$\vec{c}$

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

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

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

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

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

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

ab sin C = bc sinA = ca sinB

Dividing by abc, we get

$\dfrac{sinC}{c}~ = ~\dfrac{sinA}{a}~=~\dfrac{sinB}{b}~~$

$\Rightarrow \dfrac{sinA}{a}~=~\dfrac{sinB}{b}~=~\dfrac{sinC}{c}~~$

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 $\vec{a}$ = 6$\hat{i}~+2\hat{j}~+3\hat{k}$  

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

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

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

Hence the correct option is (B)

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

(A) $\dfrac{\overrightarrow{3a}~+~2\vec{b}}{3}$

(B)  $\vec{a}$

(C) $\dfrac{\overrightarrow{5a}~-\vec{b}}{3}~~~~$

(D) $\dfrac{\overrightarrow{4a}~+~\vec{b}}{3}$

Ans: The correct answer is option (D).

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

$\dfrac{2\left(\vec{a}~+~\vec{b}\right)+1\left(2\vec{a}-\vec{b}\right)}{2+1}$ = $\dfrac{\overrightarrow{4a}~+~\vec{b}}{3}$

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) $\dfrac{\pi }{3}$

(B) $\dfrac{2\pi }{3}$

(C) $\dfrac{-\pi }{3}$ 

(D) $\dfrac{5\pi }{6}$

Ans: Option (B) is the correct answer.

The angle between two vectors is given by 

$\cos\theta$ = $\dfrac{\vec{a}~.~\vec{b}}{\left| \vec{a}~\left| . \right|\vec{b} \right|}$ = \[\dfrac{(\hat{i}~-\hat{j})~.(\hat{j}-\hat{k})}{|(\hat{i}~-\hat{j})|~.|(\hat{j}~~-\hat{k})|}\]  = $\dfrac{-1}{\sqrt{{{1}^{2}}+{{1}^{2~}}~}.~\sqrt{{{1}^{2}}+{{1}^{2~}}}}$ = $\dfrac{-1}{2}$ 

θ  = $\dfrac{2\pi }{3}$

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 

$\left( 2\hat{i}~~-\hat{j}~+2\hat{k} \right)$.$\left( 3\hat{i}~+\lambda \hat{j}~+\hat{k} \right)$ = 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  $\vec{a}$ and $\vec{b}$   is given by |a- x$\vec{b}$ | 

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

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

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

Hence the correct option is (B)

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

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

⇒12 = 8.3.|sin$~\theta ~$|

⇒|sin$~\theta ~$| = $\dfrac{1}{2}$

⇒$\theta $ = ±$\dfrac{\pi }{6}$

Now we have,

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

$\Rightarrow \vec{a}$.$\vec{b}$  = |$\vec{a}$| .|$\vec{b}$| |cos$~\theta ~$| = 8 x 3 x$~\dfrac{\sqrt{3}}{2}$ = 12$\sqrt{3}$

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

Hence the correct option is (C)

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

(A) $\dfrac{\sqrt{\text{34}}}{\text{2}}$

(B) $\dfrac{\sqrt{\text{48}}}{\text{2}}$

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

(D) None of these

Ans: The correct answer is  option (A).

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

$|\overrightarrow{AD~}$| =  $\dfrac{1}{2}$| $3\hat{i}~~-\hat{j}~+\hat{4k}$|

$\Rightarrow |\overrightarrow{AD~}$| =  $\dfrac{\sqrt{34}}{2}$

Hence the correct option is (A).

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

(A) $\dfrac{\text{2}}{\text{3}}$

(B) $\dfrac{\text{1}}{\text{3}}$

(C) 2

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

Ans: Option (A) is the correct answer.

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

$\dfrac{\vec{a}~.\vec{b}~}{\left| {\vec{b}} \right|}$ = $\dfrac{\left( 2\hat{i}~-\hat{j}~+\hat{k} \right).\left( \hat{i}~+2\hat{j}~+2\hat{k}~ \right)}{(|2\hat{i}~-\hat{j}~+\hat{k})|.(|\hat{i}~+2\hat{j}~+2\hat{k|}~)}$ = $\dfrac{2}{3}$

Hence the correct option is (A)

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

(A) 30°

(B) 45° 

(C) 60°

(D) 90°

Ans: Option (A) is the correct answer.

Here we have 

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

$\Rightarrow \vec{a}$.$\vec{b}$ = $\dfrac{\sqrt{3}}{2}$

⇒cosθ =$\dfrac{\sqrt{3}}{2}$

⇒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) \[\dfrac{{\hat{i}-\hat{j}}}{\sqrt{{2}}}\]

(D) \[\dfrac{{\hat{i}+\hat{j}}}{\sqrt{{2}}}\]

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  

$\dfrac{(\hat{i}~-\hat{j}~\times\left( \hat{i}~+\hat{j}~~ \right)}{\left| \left( \hat{i}~-\hat{j} \right)\times\left( \hat{i}~+\hat{j}~ \right) \right|}$  =  $\hat{k}$

Hence the correct option is (A)

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

(A) [0, 6] 

(B) [– 3, 6]

(C) [ 3, 6] 

(D) [1, 2]

Ans: Option (A) is the correct answer.

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

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

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

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

|k$\vec{a}$| $\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 $\vec{c}$ denote the sum of $\vec{a}$ and $\vec{b}$. So, 

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

Unit vector along vector c will be

$\hat{c}=\dfrac{{\vec{c}}}{\left|{\vec{c}}\right|}=\dfrac{2\hat{i}+\hat{j}+2\hat{k}~}{3}=\dfrac{2}{3}\hat{i}+\dfrac{1}{3}\hat{j}+\dfrac{2}{3}\hat{k}$ 

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 $\vec{a}=\hat{i}+\hat{j}+2\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}-2\hat{k}$

(i) We have to find unit vector along

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

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

$\widehat{b}=\dfrac{{\vec{b}}}{\left|{\vec{b}}\right|}=\dfrac{2\hat{i}+\hat{j}-2\hat{k}}{\sqrt{{{2}^{2}}+{{1}^{2}}+{{2}^{2}}}}=\dfrac{1}{\sqrt{9}}\left( 2\hat{i}+\hat{j}-2\hat{k} \right)$ 

$\widehat{b}=\dfrac{2}{3}\hat{i}+\dfrac{1}{3}\hat{j}-\dfrac{2}{3}\hat{k}$ 

(ii) \[2\overrightarrow{a}-\overrightarrow{b}\]

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

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

$=\hat{j}+6\hat{k}$ 

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

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).

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

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

$\Rightarrow \widehat{PQ}=\dfrac{1}{7}\left( -2\hat{i}+3\hat{j}-6\hat{k}~ \right)$ 

4. If \[\vec{a}\] and \[\vec{b}\] 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 $OA=\vec{a}$ and $OB=\vec{b}$

$BA=OA-OB=\vec{a}-\vec{b}$ 

$\Rightarrow BC=1.5BA=1.5\left( \vec{a}-\vec{b}~ \right)$.  

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

$\Rightarrow 1.5\left( \vec{a}-\vec{b}~ \right)=\vec{c}-\vec{b}$ 

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

$\Rightarrow OC=\vec{c}=\dfrac{1}{2}\left( 3\vec{a}-\vec{b} \right)$ 

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

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

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

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

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  \[\vec{r}\]  is inclined at equal angles to the three axes. If the magnitude of  \[\vec{r}\]  is  ${2}\sqrt{{3}}$ units, find \[\vec{r}\].

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

$l=m=n$

${{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1\Rightarrow 3{{l}^{2}}=1$ 

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

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

$\vec{r}=2\sqrt{3}\left( \pm \dfrac{1}{\sqrt{3}}\hat{i}\pm \dfrac{1}{\sqrt{3}}\hat{j}\pm \dfrac{1}{\sqrt{3}}\hat{k} \right)=\pm 2\left( \hat{i}+\hat{j}+\hat{k} \right)$ 

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

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

Let direction cosines are $l,~m~and~n$

$\therefore l=\dfrac{2x}{14}=\dfrac{x}{7}$ 

$\therefore m=\dfrac{3x}{14}$ 

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

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

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

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

$\Rightarrow \left( 4+9+36 \right){{x}^{2}}=196$ 

$\Rightarrow 49{{x}^{2}}=4\times 49$ 

$\Rightarrow {{x}^{2}}=4$ 

$\therefore x=\pm 2$ 

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

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

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

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

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 $\vec{a}=2\hat{i}-\hat{j}+2\hat{k}$ and $\vec{b}=4\hat{i}-\hat{j}+3\hat{k}$ 

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

$\vec{c}=\vec{a}\times \vec{b}=\left| \hat{i}~\hat{j}~\hat{k}~2~-1~2~4~-1~3~ \right|=\left( -3+2 \right)\hat{i}-\left( 6-8 \right)\hat{j}+\left( -2+4 \right)\hat{k}$ 

$\vec{c}=-\hat{i}+2\hat{j}+2\hat{k}$ 

Required vector should have magnitude 6 then 

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

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

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  $\vec{a}=2\hat{i}-\hat{j}+\hat{k}$  and  $\vec{b}=3\hat{i}+4\hat{j}-\hat{k}$

Angle between vectors be $\theta $ then

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

$\Rightarrow cos~\theta ~=\dfrac{1}{\sqrt{6}\times \sqrt{26}~}\left( 6-4-1 \right)$ 

$\Rightarrow cos~\theta ~=\dfrac{1}{2\sqrt{39}~}$ 

$\Rightarrow \theta =\left( \dfrac{1}{2\sqrt{39}} \right)~$ 

10. If \[\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0\], show that \[\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{c}\times \overrightarrow{a}\]. Interpret the result geometrically?

Ans: Given $\vec{a}+\vec{b}+\vec{c}=0$ 

$\Rightarrow \vec{b}=-\left( \vec{a}+\vec{c} \right)$

$\vec{a}\times \vec{b}=\vec{a}\times \left\{ -\left( \vec{a}+\vec{c} \right) \right\}$ 

$=-\vec{a}\times \vec{a}-\vec{a}\times \vec{c}$ 

$=0+\vec{c}\times \vec{a}$ 

$\vec{a}\times \vec{b}=\vec{c}\times \vec{a}$ 

Again, $\vec{b}\times \vec{c}=\left\{ -\left( \vec{a}+\vec{c} \right) \right\}\times \vec{c}$

$=-\vec{a}\times \vec{c}-\vec{c}\times \vec{c}$ 

$\vec{b}\times \vec{c}=\vec{c}\times \vec{a}$ 

Hence, $\vec{a}\times \vec{b}=\vec{b}\times \vec{c}=\vec{c}\times \vec{a}$. 

Proved. 

Geometrical interpretation of the result.

If ABCD is parallelogram such that $AB=\vec{a}$ and $AD=\vec{b}$ and these

adjacent sides are making angle $\theta $ between each other, then

Area of parallelogram ABCD $=\left| {\vec{a}} \right|\left| {\vec{b}} \right|sin~\theta ~=\left| \vec{a}\times \vec{b} \right|$ 

Since, parallelogram on the same base and between the same parallels are 

equal in area.

So, $\left| \vec{a}\times \vec{b} \right|=\left| \vec{b}\times \vec{c} \right|=\left| \vec{c}\times \vec{a} \right|$ 

This also implies that $\vec{a}\times \vec{b}=\vec{b}\times \vec{c}=\vec{c}\times \vec{a}$ 

So, area of the parallelogram formed by taking any two sides represented

By  $\vec{a},~~\vec{b}~and~\vec{c}$  as adjacent are equal.

11. Find the sine of the angle between the vectors and $\overrightarrow{a}=3\hat{i}+\hat{j}+2\hat{k}$ and $\overrightarrow{b}=2\hat{i}+2\hat{j}+4\hat{k}$

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

Let the angle between vectors be $\theta $ then

$sin~\theta ~=\dfrac{\left| \vec{a}\times \vec{b} \right|}{\left| {\vec{a}} \right|\left| {\vec{b}} \right|}$ 

$\vec{a}\times \vec{b}$=$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 2 & -2 & 4 \\ \end{vmatrix}$ = $\left( 4+4 \right)\hat{i}-\left( 12-4 \right)\hat{j}+\left( -6-2 \right)\hat{k}$ 

$\vec{a}\times \vec{b}=8\hat{i}-8\hat{j}-8\hat{k}\Rightarrow \left| \vec{a}\times \vec{b} \right|=8\sqrt{1+1+1}=8\sqrt{3}$ 

$\left| ~\vec{a} \right|=\sqrt{9+1+4}=\sqrt{14}$ 

$\left| ~\vec{b} \right|=\sqrt{4+4+16}=\sqrt{24}$ 

$sin~\theta ~=\dfrac{\left| \vec{a}\times \vec{b} \right|}{\left| {\vec{a}} \right|\left| {\vec{b}}\right|}=\dfrac{8\sqrt{3}}{\sqrt{14}\sqrt{24}}=\dfrac{2\times 4\sqrt{3}}{4\sqrt{3}\sqrt{7}}$ 

$=\dfrac{2}{\sqrt{7}}$.

12. If A, B, C, D are the points with position vectors \[\hat{i}+\hat{j}+\hat{k}\], \[2\hat{i}+\hat{j}+3\hat{k}\], \[2\hat{i}-3\hat{k}\], and \[3\hat{i}-2\hat{j}+\hat{k}\], respectively, find the projection of AB along CD. 

Ans: Given, $OA=\hat{i}+\hat{j}-\hat{k}$, $OB=2\hat{i}-\hat{j}+3\hat{k}$, 

$OC=2\hat{i}-3\hat{k}$, and $OD=3\hat{i}-2\hat{j}+\hat{k}$

$AB=OB-OA=\left(2\hat{i}-\hat{j}+3\hat{k}\right)-\left(\hat{i}+\hat{j}-\hat{k}\right)=\hat{i}-2\hat{j}+4\hat{k}$ 

$CD=OD-OC=\left(3\hat{i}-2\hat{j}+\hat{k}\right)-\left(2\hat{i}-3\hat{k}\right)=\hat{i}-2\hat{j}+4\hat{k}$ 

Now, projection of AB along CD $=\dfrac{AB.CD}{\left| CD \right|}$ 

$=\dfrac{1}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{4}^{2}}}}\left(\hat{i}-2\hat{j}+4\hat{k} \right).\left( \hat{i}-2\hat{j}+4\hat{k}~ \right)$ 

$=\dfrac{1}{\sqrt{21}}\left( 1+4+16 \right)$ 

$=\dfrac{21}{\sqrt{21}}$ 

$=\sqrt{21}$ 

13. Using vectors, find the area of the triangle ABC with vertices 

A (1, 2, 3), B (2, – 1, 4) and C (4, 5, – 1). 

Ans: Given A (1, 2, 3), B (2, – 1, 4) and C (4, 5, – 1). 

$AB=\hat{i}-3\hat{j}+\hat{k}$, $AC=3\hat{i}+3\hat{j}-4\hat{k}$

Area of $\Delta ABC=\dfrac{1}{2}\left| AB\times AC \right|$ 

$=\dfrac{1}{2}\left|\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 3 & 3 & -4 \\ \end{vmatrix} \right|$

$=\dfrac{1}{2}\left| \left( 9\hat{i}+7\hat{j}+12\hat{k} \right) \right|$ 

$=\dfrac{1}{2}\sqrt{{{9}^{2}}+{{7}^{2}}+{{12}^{2}}}$ 

$=\dfrac{1}{2}\sqrt{81+49+144}$ 

$=\dfrac{1}{2}\sqrt{274}$ sq. units

14. Using vectors, prove that the parallelogram on the same base and between the same parallels are equal in area. 

Ans: Let ABCD and ABFE are parallelograms on the same base AB and

between the same parallel line AB and DF.

Let $AB=\vec{a}$ and $AD=\vec{b}$ 

Area of parallelogram $ABCD=~\vec{a}\times \vec{b}$ 

Again, area of parallelogram $ABFE=AB\times AE$ 

$=AB\times \left( AD+DE \right)$ 

$=\vec{a}\times \left( \vec{b}+k\vec{a} \right)$;      $\left[\because DE=k\vec{a}\right]$ 

$=\vec{a}\times \vec{b}+0$ 

$=\vec{a}\times \vec{b}$ 

area of parallelogram $ABFE=$ area of parallelogram $ABCD$

Hence, proved.

Long Answer (L.A.)

15. Prove that in any triangle ABC, $\cos A = \dfrac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}$, where a, b, c are the magnitudes of the sides opposite to the vertices A, B, C, respectively. 

Ans:

Here components of c are $ccoscos~A~$ and $csinsin~A~$ are drawn. 

Since, $CD=b-C cos~A~$

In $\Delta BDC$, using Pythagoras theorem,

${{a}^{2}}={{\left( b-C cos~A~ \right)}^{2}}+{{\left( c~sin~A~ \right)}^{2}}$ 

$\Rightarrow {{a}^{2}}={{b}^{2}}+{{c}^{2}}cos^{2}A~-2bc~cos~A~+{{c}^{2}}cos^{2}A~$ 

$\Rightarrow {{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc~cos~A~$

$\Rightarrow 2bcos~A~={{b}^{2}}+{{c}^{2}}-{{a}^{2}}$ 

$\Rightarrow cos~A~=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}$ 

Hence proved.

16. If \[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\] determine the vertices of a triangle, show that \[\dfrac{1}{2}\left[ \overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}+\overrightarrow{a}\times \overrightarrow{c} \right]\] gives the vector area of the triangle. Hence deduce the condition that the three points \[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\] are collinear. Also find the unit vector normal to the plane of the triangle.

Ans: Given that $\vec{a},~\vec{b},~\vec{c}$ are the vertices of the $\Delta ABC$ as

Area of $\Delta ABC=\dfrac{1}{2}\left| AB\times AC \right|$

$=\dfrac{1}{2}\left| \left( \vec{b}-\vec{a} \right)\times \left( \vec{c}-\vec{a} \right) \right|$ 

$=\dfrac{1}{2}\left| \vec{b}\times \vec{c}-\vec{b}\times \vec{a}-\vec{a}\times \vec{c}+\vec{a}\times \vec{a} \right|$ 

$=\dfrac{1}{2}\left| \vec{b}\times \vec{c}+\vec{a}\times \vec{b}+\vec{c}\times \vec{a} \right|$ 

Hence proved.

For three points to be collinear $ar\left( \Delta ABC \right)$ will be zero. So,

$\dfrac{1}{2}\left| \vec{b}\times \vec{c}+\vec{a}\times \vec{b}+\vec{c}\times \vec{a} \right|=0$

$\vec{b}\times \vec{c}+\vec{a}\times \vec{b}+\vec{c}\times \vec{a}=0$ 

This is the required condition for collinearity of three points $\vec{a},~\vec{b},~\vec{c}$.

Let $\hat{n}$ be the unit vector normal to the plane of the $\Delta ABC$

$\therefore ~~\hat{n}=\dfrac{AB\times AC}{\left| AB\times AC \right|}$

$=\dfrac{\vec{b}\times \vec{c}+\vec{a}\times \vec{b}+\vec{c}\times \vec{a}}{\left| \vec{b}\times \vec{c}+\vec{a}\times \vec{b}+\vec{c}\times \vec{a} \right|}$ 

17. Show that area of the parallelogram whose diagonals are given by \[\overrightarrow{a}\] and \[\overrightarrow{b}\] is \[\dfrac{\left| \overrightarrow{a}\times \overrightarrow{b} \right|}{2}\]. Also find the area of the parallelogram whose diagonals are \[2\hat{i}-\hat{j}+\hat{k}\] and \[\hat{i}+3\hat{j}+\hat{k}\]. 

Ans: ABCD be a parallelogram such that

$AB=\vec{p}=DC,~~AD=\vec{q}=BC$ 

By triangle law of addition, we get

$AC=\vec{p}+\vec{q}=\vec{a}$ 

$BD=-\vec{p}+\vec{q}=\vec{b}$ 

$AC+BD=2\vec{q}=\vec{a}+\vec{b}$ 

$\vec{q}=\dfrac{1}{2}\left( \vec{a}+\vec{b} \right)$ 

And $AC-BD=2\vec{p}=\vec{a}-\vec{b}$

$\vec{p}=\dfrac{1}{2}\left( \vec{a}-\vec{b} \right)$ 

Now, $\vec{p}\times \vec{q}=\dfrac{1}{4}\left( \vec{a}-\vec{b} \right)\times \left( \vec{a}+\vec{b} \right)$ 

$=\dfrac{1}{4}\left( \vec{a}\times \vec{a}+\vec{a}\times \vec{b}-\vec{b}\times \vec{a}-\vec{b}\times \vec{b} \right)$ 

$=\dfrac{1}{4}\left( 2\vec{a}\times \vec{b} \right)$ 

$=\dfrac{1}{2}\left( \vec{a}\times \vec{b} \right)$ 

Hence, area of parallelogram ABCD $=\vec{p}\times \vec{q}=\dfrac{1}{2}\left( \vec{a}\times \vec{b} \right)$ 

Area of parallelogram, whose diagonals are $2\hat{i}-\hat{j}+\hat{k}$ and $\hat{i}+3\hat{j}-\hat{k}$ 

$=\dfrac{1}{2}\left( 2\hat{i}-\hat{j}+\hat{k} \right)\times \left( \hat{i}+3\hat{j}-\hat{k} \right)$ 

$=\dfrac{1}{2}\left|\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 3 & -1 \\ \end{vmatrix}\right|$

$=\dfrac{1}{2}\left| \left( -2\hat{i}+3\hat{j}+7\hat{k} \right) \right|$  

$=\dfrac{1}{2}\sqrt{{{2}^{2}}+{{3}^{2}}+{{7}^{2}}}$  

$=\dfrac{1}{2}\sqrt{4+9+49}$  

$=\dfrac{1}{2}\sqrt{62}$ sq. units

18.If \[\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}\] and \[\overrightarrow{b}=\hat{j}-\hat{k}\], find a vector \[\overrightarrow{c}\] such that \[\overrightarrow{a}\times \overrightarrow{c}=\overrightarrow{b}\] and \[\overrightarrow{a}\overrightarrow{c}=3\].

Ans: Given, $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{j}-\hat{k}$ 

Let, $\vec{c}=x\hat{i}+y\hat{j}+z\hat{k}$ 

For, $\vec{a}\times \vec{c}=\vec{b}$, 

$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z\\ \end{vmatrix}$ = $\hat{j}-\hat{k}$ 

$\left( z-y \right)\hat{i}-\left( z-x \right)\hat{j}+\left( y-x \right)\hat{k}=\hat{j}-\hat{k}$ 

By comparing we have

$z-y=0\Rightarrow y=z$ 

$z-x=-1\Rightarrow x-z=1$ 

$y-x=-1\Rightarrow x-y=1$ 

Also, $\vec{a}\centerdot \vec{c}=3$ 

$\left( \hat{i}+\hat{j}+\hat{k} \right)\centerdot \left( x\hat{i}+y\hat{j}+z\hat{k} \right)=3$ 

$x+y+z=3$ 

$x+2y=3$ 

By solving $x-y=1~and~x+2y=3$ 

So, $x=\dfrac{5}{3},~y=\dfrac{2}{3}=z$ 

$~\vec{c}=\dfrac{1}{3}\left( 5\hat{i}+2\hat{j}+2\hat{k} \right)$ 

Objective Type Questions

Choose the correct answer from the given four options in each of the Exercises from 19 to 33 (M.C.Q) 

19. The vector in the direction of the vector \[\hat{i}+2\hat{j}+2\hat{k}\] that has magnitude 9 is 

(A) \[\hat{i}-2\hat{j}+2\hat{k}\]

(B) \[\dfrac{1}{3}\left( \hat{i}-2\hat{j}+2\hat{k} \right)\]

(C) \[3\left( \hat{i}+2\hat{j}+2\hat{k} \right)\]

(D) \[9\left( \hat{i}+2\hat{j}+2\hat{k} \right)\]

Ans: Let $\vec{a}=\hat{i}-2\hat{j}+2\hat{k}$ Unit vector, $\hat{a}=\dfrac{{\vec{a}}}{\left| {\vec{a}}\right|}=\dfrac{\hat{i}-2\hat{j}+2\hat{k}}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{2}^{2}}}}=\dfrac{1}{3}\left( \hat{i}-2\hat{j}+2\hat{k} \right)$ 

Required vector, 

$9\dfrac{{\vec{a}}}{\left| {\vec{a}} \right|}=\dfrac{9}{3}\left( \hat{i}-2\hat{j}+2\hat{k} \right)=3\left( \hat{i}-2\hat{j}+2\hat{k} \right)$ 

Option (c) is correct. 

20. The position vector of the point which divides the join of points $2\vec{a}-3\vec{b}$ and $\vec{a}+\vec{b}$ in the ratio 3:1 is 

(A) $\dfrac{1}{2}\left(3\vec{a}-2\vec{b}\right)$

(B) $\dfrac{1}{4}\left(7\vec{a}-8\vec{b}\right)$

(C) $\dfrac{1}{4}\left( 3\vec{a} \right)$

(D) $\dfrac{1}{4}\left( 5\vec{a} \right)$

Ans: Let point P divides the join of points $2\vec{a}-3\vec{b}$ and $\vec{a}+\vec{b}$ in the ratio $3:1$. So, 

Position vector of P 

$=\dfrac{1}{3+1}\left[\left(2\vec{a}-3\vec{b}\right)+3\left(\vec{a}+\vec{b} \right) \right]$ 

$=\dfrac{1}{4}\left( 5\vec{a} \right)$ 

Option (D) is correct.

21. The vector having initial and terminal points as (2, 5, 0) and (–3, 7, 4), respectively is 

(A) \[-\hat{i}+12\hat{j}+4\hat{k}\]

(B) \[5\hat{i}+2\hat{j}-4\hat{k}\]

(C) \[-5\hat{i}+2\hat{j}+4\hat{k}\]

(D) \[\hat{i}+\hat{j}+\hat{k}\]

Ans: Required vector $=\left( -3-2 \right)\hat{i}+\left( 7-5 \right)\hat{j}+\left( 4-0 \right)\hat{k}$ 

$=-5\hat{i}+2\hat{j}+4\hat{k}$ 

Option (C) is correct. 

22. The angle between two vectors \[\overrightarrow{a}\] and \[\overrightarrow{b}\] with magnitudes $\sqrt{\text{3}}$ and 4, respectively, and $\vec{a}\centerdot \vec{b}$ is 

(A) $\dfrac{\text{ }\text{ }\!\!\pi\!\!\text{ }\text{ }}{\text{6}}$

(B) $\dfrac{\text{ }\text{ }\!\!\pi\!\!\text{ }\text{ }}{\text{3}}$

(C) $\dfrac{\text{ }\text{ }\!\!\pi\!\!\text{ }\text{ }}{\text{2}}$

(D) $\dfrac{\text{5 }\text{ }\!\!\pi\!\!\text{ }\text{ }}{\text{2}}$ 

Ans: Given $\left| {\vec{a}} \right|=\sqrt{3}$, $\left| {\vec{b}} \right|=4$ and $\vec{a}\centerdot \vec{b}=2\sqrt{3}$ 

$\Rightarrow \vec{a}\centerdot \vec{b}=2\sqrt{3}$ 

$\Rightarrow \left| {\vec{a}} \right|\left| {\vec{b}} \right|cos~\theta ~=2\sqrt{3}$ 

$\Rightarrow cos~\theta ~=\dfrac{2\sqrt{3}}{4\sqrt{3}}=\dfrac{1}{2}$ 

$\Rightarrow \theta =\dfrac{\pi }{3}$. 

Option (B) is correct.

23. Find the value of λ such that the vectors, \[\overrightarrow{a}=2\hat{i}+\lambda \hat{j}+\hat{k}\]  and \[\overrightarrow{b}=\hat{i}+2\hat{j}+3\hat{k}\] are orthogonal 

(A) 0

(B) 1

(C) $\dfrac{\text{3}}{\text{2}}$

(D) $\text{-}\dfrac{\text{5}}{\text{2}}$ 

Ans: Given nonzero vectors are $\vec{a}=2\hat{i}+\lambda \hat{j}+\hat{k}$  and $\vec{b}=\hat{i}+2\hat{j}+3\hat{k}$

If both the vectors are orthogonal then scalar product will be zero

$\Rightarrow \vec{a}\centerdot \vec{b}=\left( 2\hat{i}+\lambda \hat{j}+\hat{k} \right)\centerdot \left( \hat{i}+2\hat{j}+3\hat{k} \right)=0$ 

$\Rightarrow 2+2\lambda +3=0$ 

$\Rightarrow \lambda =-\dfrac{5}{2}$ 

Option (D) is correct.

24. The value of λ for which the vectors \[3\hat{i}-6\hat{j}+\hat{k}\] and \[2\hat{i}-4\hat{j}+\lambda \hat{k}\] are parallel is 

(A) $\dfrac{\text{2}}{\text{3}}$

(B) $\dfrac{\text{3}}{\text{2}}$

(C) $\dfrac{\text{5}}{\text{2}}$

(D) $\dfrac{\text{2}}{\text{5}}$

Ans: Given vectors are $3\hat{i}-6\hat{j}+\hat{k}$ and $2\hat{i}-4\hat{j}+\lambda \hat{k}$ are parallel then it may be written as 

$3\hat{i}-6\hat{j}+\hat{k}=p\left( 2\hat{i}-4\hat{j}+\lambda \hat{k} \right)$ 

Where p is a nonzero scalar quantity.

By comparing we have

$\Rightarrow 3=2p\Rightarrow p=\dfrac{3}{2}$ 

$\Rightarrow -6=-4p\Rightarrow p=\dfrac{3}{2}$ 

$\Rightarrow 1=\lambda p\Rightarrow \lambda =\dfrac{1}{p}=\dfrac{2}{3}$ 

Hence option (A) is correct. 

25. The vectors from origin to the points A and B are \[\overrightarrow{a}=2\hat{i}-3\hat{j}+2\hat{k}\] and \[\overrightarrow{b}=2\hat{i}+3\hat{j}+\hat{k}\], respectively, then the area of triangle OAB is

(A) 340

(B) 5

(C) $\sqrt{\text{229}}$

(D) $\dfrac{\text{1}}{\text{2}}\sqrt{\text{229}}$ 

Ans: Area of $\Delta ABC=\vec{a}\times \vec{b}=\dfrac{1}{2}\left| \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 2 \\ 2 & 3 & 1 \\ \end{vmatrix} \right|$ 

$=\dfrac{1}{2}\left| \left( -3-6 \right)\hat{i}-\left( 2-4 \right)\hat{j}+\left( 6+6 \right)\hat{k} \right|$ 

$=\dfrac{1}{2}\left| -9\hat{i}+2\hat{j}+12\hat{k} \right|$ 

$=\dfrac{1}{2}\sqrt{81+4+144}$ 

Area of  $\Delta ABC=\dfrac{1}{2}\sqrt{229}$ 

Option (D) is correct.

26. For any vector ${\vec{a}}$, the value of \[{{\left( \overrightarrow{a}\times \hat{i}\right)}^{2}}+{{\left(\overrightarrow{a}\times\hat{j}\right)}^{2}}+{{\left( \overrightarrow{a}\times \hat{k} \right)}^{2}}\] is equal to

(A) \[{{\overrightarrow{a}}^{2}}\]

(B) \[3{{\overrightarrow{a}}^{2}}\]

(C) \[4{{\overrightarrow{a}}^{2}}\]

(D) \[2{{\overrightarrow{a}}^{2}}\]

Ans: Let $\vec{a}=x\hat{i}+y\hat{j}+z\hat{k}$

$\vec{a}\times \hat{i}=\left( x\hat{i}+y\hat{j}+z\hat{k} \right)\times \hat{i}$ 

$=0-y\hat{k}+z\hat{i}$ 

$\vec{a}\times \hat{i}=-y\hat{k}+z\hat{i}$ 

${{\left( \vec{a}\times \hat{i} \right)}^{2}}={{\left( -y\hat{k}+z\hat{i} \right)}^{2}}={{y}^{2}}+{{z}^{2}}$ 

Similarly, ${{\left( \vec{a}\times \hat{j} \right)}^{2}}={{x}^{2}}+{{z}^{2}}$ and ${{\left( \vec{a}\times \hat{k} \right)}^{2}}={{x}^{2}}+{{y}^{2}}$ 

Now, by adding we have

${{\left( \vec{a}\times \hat{i} \right)}^{2}}+{{\left( \vec{a}\times \hat{j} \right)}^{2}}+{{\left( \vec{a}\times \hat{k} \right)}^{2}}=2\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)$ 

${{\vec{a}}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}$ 

${{\left( \vec{a}\times \hat{i} \right)}^{2}}+{{\left( \vec{a}\times \hat{j} \right)}^{2}}+{{\left( \vec{a}\times \hat{k} \right)}^{2}}=2{{\vec{a}}^{2}}$

Option (D) is correct. 

27. If $\left| {\vec{a}} \right|=10$, $\left| {\vec{b}} \right|=2$ and $\vec{a}\centerdot \vec{b}=12$ and  then value of $\vec{a}\centerdot \vec{b}=12$ is 

(A) 5

(B) 10

(C) 14

(D) 16

Ans: Given that  $\left| {\vec{a}} \right|=10$, $\left| {\vec{b}} \right|=2$ and $\vec{a}\centerdot \vec{b}=12$

$\vec{a}\centerdot \vec{b}=12$

$\Rightarrow \left| {\vec{a}} \right|\left| {\vec{b}} \right|cos~\theta ~=12$ 

$\Rightarrow 10\times 2cos~\theta ~=12$ 

$\Rightarrow cos~\theta ~=\dfrac{12}{20}=\dfrac{3}{5}$ 

$\Rightarrow sin~\theta ~=\sqrt{1-\theta ~}=\sqrt{1-{{\left( \dfrac{3}{5} \right)}^{2}}}$ 

$\Rightarrow sin~\theta ~=\sqrt{\dfrac{25-9}{25}}=\dfrac{4}{5}$ 

$\Rightarrow \left| \vec{a}\times \vec{b} \right|=\left| {\vec{a}} \right|\left| {\vec{b}} \right|sin~\theta ~$ 

$=10\times 2\times \dfrac{4}{5}$ 

$\Rightarrow \left| \vec{a}\times \vec{b} \right|=16$ 

Option (D) is correct.

28. The vectors \[\lambda \hat{i}+\hat{j}+2\hat{k}\], \[\hat{i}+\lambda \hat{j}-2\hat{k}\] and \[2\hat{i}-\hat{j}+\lambda \hat{k}\] are coplanar if 

(A) λ = –2

(B) λ = 0

(C) λ = 1

(D) λ = – 1

Ans: Given vectors $\lambda \hat{i}+\hat{j}+2\hat{k}$, $\hat{i}+\lambda \hat{j}-\hat{k}$ and $2\hat{i}-\hat{j}+\lambda \hat{k}$ are coplanar, so

$\begin{vmatrix} \lambda & 1 & 2 \\ 1 & \lambda & -2 \\ 2 & -1 & \lambda \\ \end{vmatrix} = 0 $

$\Rightarrow \lambda \left( {{\lambda }^{2}}-1 \right)-\left( \lambda +2 \right)+2\left( -1-2\lambda  \right)=0$ 

$\Rightarrow {{\lambda }^{3}}-\lambda -\lambda -2-2-4\lambda =0$ 

$\Rightarrow {{\lambda }^{3}}-6\lambda -4=0$ 

$\Rightarrow \left( \lambda +2 \right)\left( {{\lambda }^{2}}-2\lambda -2 \right)=0$ 

$\Rightarrow \lambda =-2,~\lambda =\dfrac{2\pm \sqrt{12}}{2}$ 

$\Rightarrow \lambda =-2,~\lambda =\dfrac{2\pm 2\sqrt{3}}{2}$

$\Rightarrow \lambda =-2,~\lambda =1\pm \sqrt{3}$ 

Option (A) is correct.

29. If \[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\] are unit vectors such that \[\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0\], then the value of \[\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}\] is 

(A) 1

(B) 3

(C) $\text{-}\dfrac{\text{3}}{\text{2}}$

(D) None of these 

Ans: Given that $\vec{a},~\vec{b},~\vec{c}$ are unit vectors such that $\vec{a}+\vec{b}+\vec{c}=0$ 

$\Rightarrow \left( \vec{a}+\vec{b}+\vec{c} \right)\centerdot \left( \vec{a}+\vec{b}+\vec{c} \right)=0$ 

$\Rightarrow \left( {{{\vec{a}}}^{2}}+{{{\vec{b}}}^{2}}+{{{\vec{c}}}^{2}}+2\left( \vec{a}\centerdot \vec{b}+\vec{b}\centerdot \vec{c}+\vec{c}\centerdot \vec{a} \right) \right)=0$ 

\[\Rightarrow 1+1+1+2\left( \vec{a}\centerdot \vec{b}+\vec{b}\centerdot \vec{c}+\vec{c}\centerdot \vec{a} \right)=0\] 

$\Rightarrow \vec{a}\centerdot\vec{b}+\vec{b}\centerdot \vec{c}+\vec{c}\centerdot \vec{a}=-\dfrac{3}{2}$ 

Option (C) is correct.

30. Projection vector of $\vec{a}$ on $\vec{b}$ is 

(A) $\dfrac{\vec{a}\centerdot \vec{b}}{{{\left| {\vec{b}} \right|}^{2}}}~\vec{b}$

(B) $\dfrac{\vec{a}\centerdot \vec{b}}{\left| {\vec{b}} \right|}$

(C) $\dfrac{\vec{a}\centerdot \vec{b}}{\left| {\vec{a}} \right|}$ 

(D) $\dfrac{\vec{a}\centerdot \vec{b}}{{{\left| {\vec{a}} \right|}^{2}}}~\vec{a}$

Ans:  Projection of $\vec{a}$ on $\vec{b}$ is $\dfrac{\vec{a}\centerdot \vec{b}}{\left| {\vec{b}} \right|}$ 

Projection vector of $\vec{a}$ on $\vec{b}$ is $\left( \dfrac{\vec{a}\centerdot \vec{b}}{\left| {\vec{b}} \right|} \right)\dfrac{{\vec{b}}}{\left| {\vec{b}} \right|}=\dfrac{\vec{a}\centerdot \vec{b}}{{{\left| {\vec{b}} \right|}^{2}}}~\vec{b}$ 

Option (A) is correct.

31. If  \[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\] are three vectors such that \[\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0\] and \[\left| \overrightarrow{a} \right|=2,\text{ }\left| \overrightarrow{b} \right|=3,\text{ }\left| \overrightarrow{c} \right|=5\], then value of \[\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{b}.\overrightarrow{a}\] is 

(A) 0

(B) 1

(C) – 19

(D) 38 

Ans: Given that $\vec{a},~\vec{b},~\vec{c}$ are vectors such that $\vec{a}+\vec{b}+\vec{c}=0$ and $\left| {\vec{a}} \right|=2,~\left| {\vec{b}} \right|=3$ and $\left| {\vec{c}} \right|=5$,

$\Rightarrow \left( \vec{a}+\vec{b}+\vec{c} \right)\centerdot \left( \vec{a}+\vec{b}+\vec{c} \right)=0$ 

$\Rightarrow \left( {{{\vec{a}}}^{2}}+{{{\vec{b}}}^{2}}+{{{\vec{c}}}^{2}}+2\left( \vec{a}\centerdot \vec{b}+\vec{b}\centerdot \vec{c}+\vec{c}\centerdot \vec{a} \right) \right)=0$ 

$\Rightarrow 4+9+25+2\left( \vec{a}\centerdot \vec{b}+\vec{b}\centerdot \vec{c}+\vec{c}\centerdot \vec{a} \right)=0$ 

$\Rightarrow \vec{a}\centerdot \vec{b}+\vec{b}\centerdot \vec{c}+\vec{c}\centerdot \vec{a}=-\dfrac{38}{2}=-19$  

Option (C) is correct.

32. If $\left| {\vec{a}} \right|=4$ and −3 ≤ λ ≤ 2, then the range of $\left| \lambda \vec{a} \right|$ is 

(A)$\left[0, 8\right]$

(B) $\left[– 12, 8\right]$

(C) $\left[0, 12\right]$

(D) $\left[8, 12\right]$

Ans: Given $\left| {\vec{a}} \right|=4$ and −3 ≤ λ ≤ 2

$\Rightarrow \left| \lambda \vec{a} \right|=\left| \lambda  \right|\left| {\vec{a}} \right|=4\left| \lambda  \right|$ 

Now, $-3\le \lambda \le 2\Rightarrow 0\le \left| \lambda  \right|\le 3$ 

$\Rightarrow 0\le 4\left| \lambda  \right|\le 12$ 

$\Rightarrow \left| \lambda \vec{a} \right|\in \left[ 0,~12 \right]$ 

Option (C) is correct.

33. The number of vectors of unit length perpendicular to the vectors \[\overrightarrow{a}=2\hat{i}+\hat{j}+2\hat{k}\] and \[\overrightarrow{b}=\hat{j}+\hat{k}\] is 

(A) one

(B) two

(C) three

(D) infinite 

Ans: The number of vectors of unit length perpendicular to the vectors $\vec{a}=2\hat{i}+\hat{j}+2\hat{k}$ and $\vec{b}=\hat{j}+\hat{k}$ will be along the cross or vector product as

$\hat{c}=\pm \dfrac{\vec{a}\times \vec{b}}{\left| \vec{a}\times \vec{b} \right|}$ 

There are two-unit vectors. Option (B) is correct.

Fill in the blanks in each of the Exercises from 34 to 40. 

34. The vector \[\overrightarrow{a}+\overrightarrow{b}\] bisects the angle between the non-collinear vectors \[\overrightarrow{a}\] and \[\overrightarrow{b}\] if ________.

Ans: Given that vector $\vec{a}+\vec{b}$ bisects the angle between the non-collinear vectors $\vec{a}$ and $\vec{b}$ then let each angle is equal to $\theta $

Now, $\Rightarrow cos~\theta ~=\dfrac{\vec{a}\centerdot \left( \vec{a}+\vec{b}~ \right)}{\left| {\vec{a}} \right|\left| \left( \vec{a}+\vec{b}~ \right) \right|}=\dfrac{\vec{b}\centerdot \left( \vec{a}+\vec{b}~ \right)}{\left| {\vec{b}} \right|\left| \left( \vec{a}+\vec{b}~ \right) \right|}$ 

$\Rightarrow \dfrac{\vec{a}\centerdot \left( \vec{a}+\vec{b} \right)}{\left| {\vec{a}} \right|}=\dfrac{\vec{b}\centerdot \left( \vec{a}+\vec{b} \right)}{\left| {\vec{b}} \right|}$ 

$\Rightarrow \dfrac{\vec{a}\centerdot \left( \vec{a}+\vec{b} \right)}{\left| {\vec{a}} \right|}-\dfrac{\vec{b}\centerdot \left( \vec{a}+\vec{b} \right)}{\left| {\vec{b}} \right|}=0$ 

$\Rightarrow \left( \dfrac{{\vec{a}}}{\left| {\vec{a}} \right|}-\dfrac{{\vec{b}}}{\left| {\vec{b}} \right|} \right)\centerdot \left( \vec{a}+\vec{b} \right)=0$ 

35. If $\vec{r}\centerdot \vec{a}=0$, $\vec{r}\centerdot \vec{b}=0$ and $\vec{r}\centerdot \vec{c}=0$ for some non-zero vector $\vec{r}$, then the value of $\vec{a}\centerdot \left( \vec{b}\times \vec{c} \right)$ is ________ 

Ans:  Given that $\vec{r}\centerdot \vec{a}=0$, $\vec{r}\centerdot \vec{b}=0$ and $\vec{r}\centerdot \vec{c}=0$ also $\vec{r}$ is nonzero vector. So, $\vec{a},\vec{b}$ and $\vec{c}$ are coplanar vectors. We know that the scalar triple product of coplanar vectors is zero. 

$\therefore \vec{a}\centerdot \left( \vec{b}\times \vec{c} \right)=0$ 

36. The vectors \[\overrightarrow{a}=3\hat{i}-2\hat{j}+2\hat{k}\] and \[\overrightarrow{b}=-\hat{i}-2\hat{k}\] are the adjacent sides of a parallelogram. The acute angle between its diagonals is ________. 

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

Diagonals are 

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

$\vec{a}-\vec{b}=\left( 3\hat{i}-2\hat{j}+2\hat{k} \right)-\left( -\hat{i}-2\hat{k} \right)=4\hat{i}-2\hat{j}+4\hat{k}$ 

Let acute angle between diagonals be $\theta $ then

$cos\theta =\left| \dfrac{\left( \vec{a}+\vec{b} \right)\centerdot \left( \vec{a}-\vec{b} \right)}{\left| \vec{a}+\vec{b} \right|\left| \vec{a}-\vec{b} \right|} \right|~=\left| \dfrac{\left( 2\hat{i}-2\hat{j} \right)\centerdot \left( 4\hat{i}-2\hat{j}+4\hat{k} \right)}{\left| 2\hat{i}-2\hat{j} \right|\left| 4\hat{i}-2\hat{j}+4\hat{k} \right|} \right|~$ 

$=\left| \dfrac{8+4}{\sqrt{4+4}\sqrt{16+4+16}} \right|~=\left| \dfrac{12}{\sqrt{8}\sqrt{36}} \right|~$ 

$cos\theta=\left| \dfrac{1}{\sqrt{2}} \right|~$ 

$\theta =\dfrac{\pi }{4}$

37. The values of k for which $\left| k\vec{a} \right|<\left| {\vec{a}} \right|$ and $k\vec{a}+\dfrac{1}{2}\vec{a}$ is parallel to $\vec{a}$ holds true are _______. 

Ans: Given that $\left| k\vec{a} \right|<\left| {\vec{a}} \right|$ and $k\vec{a}+\dfrac{1}{2}\vec{a}$ 

$\Rightarrow \left| k \right|<1\Rightarrow -1<k<1$ 

Again, $k\vec{a}+\dfrac{1}{2}\vec{a}=\left( k+\dfrac{1}{2} \right)\vec{a}$ 

We know that null vector has no specific direction so $k\vec{a}+\dfrac{1}{2}\vec{a}$ will be parallel to $\vec{a}$ when  $k+\dfrac{1}{2}\ne 0\Rightarrow k=-\dfrac{1}{2}$ 

Hence $k\in \left( -1,~1 \right)-\left\{ -\dfrac{1}{2} \right\}$. 

38. The value of the expression ${{\left| \vec{a}\times \vec{b} \right|}^{2}}+{{\left( \vec{a}\centerdot \vec{b} \right)}^{2}}$ is _______. 

Ans: ${{\left| \vec{a}\times \vec{b} \right|}^{2}}+{{\left( \vec{a}\centerdot \vec{b} \right)}^{2}}={{\left| {\vec{a}} \right|}^{2}}{{\left| {\vec{b}} \right|}^{2}}{sin^{2}\theta} ~+{{\left| {\vec{a}} \right|}^{2}}{{\left| {\vec{b}} \right|}^{2}} {cos^{2}\theta} ~$ 

$={{\left| {\vec{a}} \right|}^{2}}{{\left| {\vec{b}} \right|}^{2}}\left(sin^{2}\theta ~+cos^{2}\theta ~ \right)$ 

$={{\left| {\vec{a}} \right|}^{2}}{{\left| {\vec{b}} \right|}^{2}}$ 

39. If ${{\left| \vec{a}\times \vec{b} \right|}^{2}}+{{\left( \vec{a}\centerdot \vec{b} \right)}^{2}}=144$ and $\left| {\vec{a}} \right|=4$, then $\left| {\vec{b}} \right|$ is equal to _______. 

Ans: ${{\left| \vec{a}\times \vec{b} \right|}^{2}}+{{\left( \vec{a}\centerdot \vec{b} \right)}^{2}}={{\left| {\vec{a}} \right|}^{2}}{{\left| {\vec{b}} \right|}^{2}}sin^{2}\theta ~+{{\left| {\vec{a}} \right|}^{2}}{{\left| {\vec{b}} \right|}^{2}}cos^{2}\theta ~$ 

$={{4}^{2}}{{\left| {\vec{b}} \right|}^{2}}\left( sin^{2}\theta ~+cos^{2}\theta ~ \right)$ 

$\Rightarrow 144=16{{\left| {\vec{b}} \right|}^{2}}$ 

$\Rightarrow {{\left| {\vec{b}} \right|}^{2}}=9$ 

$\Rightarrow \left| {\vec{b}} \right|=3$

40. If $\left| {\vec{a}} \right|$ is any non-zero vector, then $\left( \vec{a}\centerdot \hat{i} \right)\hat{i}+\left( \vec{a}\centerdot j \right)\hat{j}+\left( \vec{a}\centerdot k \right)\hat{k}$ equals _______. 

Ans: Let $\vec{a}=x\hat{i}+y\hat{j}+z\hat{k}$ 

Now, $\left( \vec{a}\centerdot \hat{i} \right)\hat{i}=\left\{ \left( x\hat{i}+y\hat{j}+z\hat{k} \right)\centerdot \hat{i} \right\}\hat{i}=x\hat{i}$ 

Similarly, $\left( \vec{a}\centerdot j \right)\hat{j}=y\hat{j}~and~\left( \vec{a}\centerdot k \right)\hat{k}=z\hat{k}$ 

$\left( \vec{a}\centerdot \hat{i} \right)\hat{i}+\left( \vec{a}\centerdot j \right)\hat{j}+\left( \vec{a}\centerdot k \right)\hat{k}=x\hat{i}+y\hat{j}+z\hat{k}=\vec{a}$ 

State True or False in each of the following Exercises. 

41. If $\left| {\vec{a}} \right|=\left| {\vec{b}} \right|$, then necessarily it implies $\vec{a}=\pm \vec{b}$. 

Ans: Given that $\left| {\vec{a}} \right|=\left| {\vec{b}} \right|$ then it doesn’t imply that $\vec{a}=\pm \vec{b}$ because there are an infinite number of vectors with the same magnitude but different directions. 

So, the given statement is False.

42. Position vector of a point P is a vector whose initial point is the origin. 

Ans: The given statement is True. Because the initial point of all position vectors is always the origin.

43. If $\left| \vec{a}+\vec{b} \right|=\left| \vec{a}-\vec{b} \right|$, then the vectors $\vec{a}$ and $\vec{b}$ are orthogonal. 

Ans: Given that $\left| \vec{a}+\vec{b} \right|=\left| \vec{a}-\vec{b} \right|$ 

$\sqrt{{{a}^{2}}+{{b}^{2}}+2\vec{a}\centerdot \vec{b}}=\sqrt{{{a}^{2}}+{{b}^{2}}-2\vec{a}\centerdot \vec{b}}$ 

${{a}^{2}}+{{b}^{2}}+2\vec{a}\centerdot \vec{b}={{a}^{2}}+{{b}^{2}}-2\vec{a}\centerdot \vec{b}$ 

$\vec{a}\centerdot \vec{b}=0$, if vectors are non-zero then both will be orthogonal otherwise not.

Here nothing is about the vectors whether they are non-zero or not.

Hence the given statement is False.

44. The formula ${{\left( \vec{a}+\vec{b} \right)}^{2}}={{\vec{a}}^{2}}+{{\vec{b}}^{2}}+2\vec{a}\times \vec{b}$ is valid for non-zero vectors $\vec{a}$ and $\vec{b}$. 

Ans: ${{\left( \vec{a}+\vec{b} \right)}^{2}}=\left( \vec{a}+\vec{b} \right)\centerdot \left( \vec{a}+\vec{b} \right)$ 

\[={{\vec{a}}^{2}}+{{\vec{b}}^{2}}+\vec{a}\centerdot \vec{b}+\vec{a}\centerdot \vec{b}\] 

$={{\vec{a}}^{2}}+{{\vec{b}}^{2}}+2\vec{a}\centerdot \vec{b}$

Hence the given statement is False.

45. If $\vec{a}$ and $\vec{b}$ are adjacent sides of a rhombus, then $\vec{a}\centerdot \vec{b}=0$.

Ans: Given that $\vec{a}$ and $\vec{b}$ are adjacent sides of a rhombus, so angle between them never be right angle 

So, $\vec{a}\centerdot \vec{b}\ne 0$ 

Hence the given statement is False.

What is Good in Exemplar?

Students struggling with the concepts for the Chapter Vector Algebra must strictly solve the Exemplar. As Algebra is all about practice and concepts, Exemplars are the best source of information for the students. The Exemplar is designed in a methodical way which helps train students to get adjusted to varying levels of difficulties. So that the students can start with the easy questions and go ahead with the entire variety of them. Each question is available with a solution, so students also can check their results and work accordingly. Exemplar has notes that consist of formulas, these formulas are given separately for the students. Students can easily go concept by concept and clear their doubts. Vector Algebra is among the tough Chapters in Maths for Class 12. Students have to work hard for the Chapter to score better marks.

Chapter 10 Vector Algebra is based upon the superior equations of Algebra. This Chapter consists of an introduction to Vectors, types of Vectors, multiplication, and addition performed on Vectors, section formula, components of a Vector, projection of a Vector on a line, scalar product, and Vector product. This Chapter needs special attention, Exemplar sorts a life of a student with full information.

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