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RD Sharma Class 12 Exercise VSA Scalar and dot product Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise VSA Scalar and dot product Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise VSA Scalar and dot product Solutions Maths - Download PDF Free Online

Updated on 27 Jan 2022, 05:02 PM IST

The class 12 students need to have proper guidance to complete their homework and prepare for their exams. But this is nearly impossible as a teacher or a tutor cannot standby a student for 24 x 7. Students must have the flexibility to check their answers and clarify their doubts at any time. Especially, the ones who are weak in mathematics and the concept of Scalar and Dot Product must find a proper solution book. RD Sharma Solutions The one that is most recommended for these students is the RD Sharma Class 12th Chapter 23 VSA.

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RD Sharma Class 12 Solutions Chapter23VSA Scaler and dot product - Other Exercise
Scalar or Dot Products Excercise: VSA
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter23VSA Scaler and dot product - Other Exercise

Scalar or Dot Products Excercise: VSA

Scalar or dot product exercise very short answer question 1

ANSWER: $\frac{\pi}{3}$
GIVEN: $|\vec{a}|=2 \quad \&|\vec{b}|=\sqrt{3} \quad \& \vec{a} \cdot \vec{b}=\sqrt{3}$
HINTS:Use the formula of cos θ
SOLUTION:Let θ be the angle between vectors $\vec{a} \& \vec{b}$ then,
$\begin{aligned} &\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \vec{b} \|} \\ & \end{aligned}$
$\Rightarrow \cos \theta=\frac{\sqrt{3}}{2 \sqrt{3}} \qquad\qquad\qquad[\because \vec{a} \cdot \vec{b}=\sqrt{3},|\vec{a}|=2 \&|\vec{b}|=\sqrt{3}] \\$
$\Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3} \qquad\qquad\qquad\left[\because \cos \frac{\pi}{3}=\frac{1}{2}\right]$

Scalar or dot product exercise very short answer question 2

ANSWER:$\frac{3}{2}$
GIVEN:$\vec{a} \cdot \vec{b}=6 \quad|\vec{a}|=3 \&|\vec{b}|=4$
HINTS:Use the formula of projection of $\vec{a} \text { on } \vec{b}$
SOLUTION:
$\\\text{Projection of } \vec{a} \; on\; \vec{b}=\frac{{\vec{a} \cdot \vec{b}}}{{|\vec{b}|}}$
$=\frac{6}{4}=\frac{3}{2}$

Scalar or dot product exercise very short answer question 3

ANSWER: $\frac{4}{\sqrt{51}}$
GIVEN: $\hat{4} \hat{i}-3 \hat{j}+3 \hat{k} \quad \& \quad 2 \hat{i}-\hat{j}-\hat{k}$
HINTS: you must know the formula of cosθ
SOLUTION:We know that
$\begin{aligned} &\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \\ & \end{aligned}$
$\Rightarrow \cos \theta=\frac{(4 \hat{i}-3 \hat{j}+3 \hat{k}) \cdot(2 \hat{i}-\hat{j}-\hat{k})}{\sqrt{16+9+9} \sqrt{4+1+1}}$
$\begin{aligned} &\Rightarrow \cos \theta=\frac{8+3-3}{\sqrt{34} \sqrt{6}}=\cos \theta=\frac{8}{2 \sqrt{51}} \\ & \end{aligned}$
$\Rightarrow \cos \theta=\frac{4}{\sqrt{51}}$

Scalar or dot product exercise very short answer question 4

ANSWER: $-2$
GIVEN:$3 \hat{i}+m \hat{j}+\hat{k} \quad \& \quad 2 \hat{i}-\hat{j}-8 \hat{k}$ are orthogonal
HINTS: you must know about the concept of perpendicular vector
SOLUTION:if vectors are perpendicular or orthogonal then,
$\begin{aligned} &\vec{a} \cdot \vec{b}=0 \\ & \end{aligned}$
$\Rightarrow(3 \hat{i}+m \hat{j}+\hat{k}) \cdot(2 \hat{i}-\hat{j}-8 \hat{k})=0 \\$
$\Rightarrow 6-m-8=0 \quad \Rightarrow-m-2=0 \\$
$\Rightarrow m=-2$

Scalar or dot product exercise very short answer question 5

ANSWER: $-24$
GIVEN: $3 \hat{i}-2 \hat{j}+4 \hat{k} \quad \& \quad 18 \hat{i}-12 \hat{j}-m \hat{k}$ are parallel
HINTS: you must know about the concept of parallel vector
SOLUTION: if vectors are parallel then,
$\begin{aligned} &\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} \\ & \end{aligned}$
$\text { Here, } a_{1}=3, b_{1}=-2, c_{1}=-4 \& a_{2}=18, b_{2}=-12, c_{2}=-m \\$
$\Rightarrow \frac{3}{18}=\frac{-2}{-12}=\frac{4}{-m}$
$\begin{aligned} &\Rightarrow \frac{1}{6}=\frac{1}{6}=\frac{4}{-m} \\ \end{aligned}$
$\Rightarrow \frac{1}{6}=\frac{4}{-m} \\$
$\Rightarrow m=-6 \times 4 \\$
$\Rightarrow m=-24$

Scalar or dot product exercise very short answer question 6

ANSWER: $0$
GIVEN: $|\vec{a}|=|\vec{b}| \&(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=?$
HINTS:Use the identity of $\left(a^{2}-b^{2}\right)$
SOLUTION:
$\begin{aligned} &|\vec{a}|=|\vec{b}| \& \text { ? } \\ & \end{aligned}$
$(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=|\vec{a}|^{2}-|\vec{b}|^{2} \qquad \qquad \qquad\left[\because(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})=\mathrm{a}^{2}-b^{2}\right] \\$
$(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=|\vec{b}|^{2}-|\vec{b}|^{2} \qquad \qquad\qquad[\because|\vec{a}|=|\vec{b}|] \\$
$=0$

Scalar or dot product exercise very short answer question 7

ANSWER: $\overrightarrow{|a|}= |\overrightarrow{b}|$
GIVEN: $(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=0$
HINTS:Use the identity of $\left(a^{2}-b^{2}\right)$
SOLUTION:
$\begin{aligned} &(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=0 \\ & \end{aligned}$
$\Rightarrow|\vec{a}|^{2}-|\vec{b}|^{2}=0 \qquad \qquad \qquad\left[\because(a+b)(a-b)=a^{2}-b^{2}\right] \\$
$\Rightarrow|\vec{a}|^{2}=|\vec{b}|^{2} \\$
$\Rightarrow|\vec{a}|=|\vec{b}|$

Scalar or dot product exercise very short answer question 8

ANSWER:$\vec{a} \& \vec{b}$ are parallel
GIVEN: $|\vec{a}+\vec{b}|=|\vec{a}|+|\vec{b}|$
HINTS:Use algebraic identities and formulas of cosθ
SOLUTION:
$\begin{aligned} &|\vec{a}+\vec{b}|=|\vec{a}|+|\vec{b}| \\ & \end{aligned}$
$(|\vec{a}+\vec{b}|)^{2}=(|\vec{a}|+|\vec{b}|)^{2}\qquad[\text { Squaring both sides }]$
$\begin{aligned} &\Rightarrow|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=|\vec{a}|^{2}+|\vec{b}|^{2}+2|\vec{a}||\vec{b}|\qquad\qquad\left[\because(\mathrm{a}+\mathrm{b})^{2}=\mathrm{a}^{2}+b^{2}+2 a b\right] \\ & \end{aligned}$
$\begin{aligned} & &\Rightarrow 2 \vec{a} \cdot \vec{b}=2|\vec{a}||\vec{b}| \\ & \end{aligned}$
$\Rightarrow \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}|$
Now we know that $\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$
$\begin{aligned} &\Rightarrow \cos \theta=\frac{|\vec{a}||\vec{b}|}{|\vec{a}||\vec{b}|} \qquad[\because \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}|] \\ & \end{aligned}$
$\Rightarrow \cos \theta=1 \quad \Rightarrow \theta=0 \qquad[\because \cos \theta=1]$
$\therefore \vec{a}$ is parallel to $\vec{b}$

Scalar or dot product exercise very short answer question 9

ANSWER:$\vec{a} \& \vec{b}$ are perpendicular
GIVEN: $|\vec{a}+\vec{b}|=|\vec{a}|-|\vec{b}|$
HINTS: Use the identity of $(a+b)^{2} \&(a-b)^{2}$
SOLUTION:
$\begin{aligned} &|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| \\ & \end{aligned}$
$(|\vec{a}+\vec{b}|)^{2}=(|\vec{a}-\vec{b}|)^{2}\qquad[\text { Squaring both sides }]$
$\begin{aligned} &\Rightarrow|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=|\vec{a}|^{2}+|\vec{b}|^{2}-2 \vec{a} \cdot \vec{b}\qquad\left[\because(\mathrm{a}+\mathrm{b})^{2}=\mathrm{a}^{2}+b^{2}-2 a b \&(\mathrm{a}-\mathrm{b})^{2}=\mathrm{a}^{2}+b^{2}-2 a b\right] \\ & \end{aligned}$
$\Rightarrow 2 \vec{a} \cdot \vec{b}=-2 \vec{a} \cdot \vec{b} \\$
$\Rightarrow \vec{a} \cdot \vec{b}=-\vec{a} \cdot \vec{b} \\$
$\Rightarrow \vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{b}=0 \Rightarrow 2 \vec{a} \cdot \vec{b}=0$
$\therefore \vec{a}$ is perpendicular to $\vec{b}$

Scalar or dot product exercise very short answer question 10

ANSWER: $4$
GIVEN: $|\vec{a}|=|\vec{b}|, \theta=60^{\circ} \quad \& \quad \vec{a} \cdot \vec{b}=8$
HINTS: use the formula of cosθ
SOLUTION:Let,
$$$ |\vec{a}|=|\vec{b}|=x$
We know that
$\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$
$\begin{aligned} &\Rightarrow \cos 60^{\circ}=\frac{8}{xx x}\qquad\left[\because \theta=60^{\circ}, \vec{a} \cdot \vec{b}=8, \overline{\mid a}|=| \vec{b} \mid=x\right] \\ & \end{aligned}$
$\Rightarrow \frac{1}{2}=\frac{8}{x^{2}} \qquad\left[\because \cos 60^{\circ}=\frac{1}{2}\right] \\$
$\Rightarrow x^{2}=16 \\$
$\Rightarrow x=4 \\$
$\therefore|\vec{a}|=\vec{|b|}=4$

Scalar or dot product exercise very short answer question 11

ANSWER: $\vec{b}$ is any non zero vector
GIVEN: $\vec{a} \cdot \vec{a}=0 \text { \& } \vec{a} \cdot \vec{b}=0$
HINTS: You must know about the concept of non zero vector
SOLUTION:Let,
$\begin{aligned} &\vec{a} \cdot \vec{a}=0 \\ &\Rightarrow \vec{(a)}^{2}=0 \\ &\Rightarrow \vec{a}=0 \\ &\vec{a} \cdot \vec{b}=0 \end{aligned}$
$\vec{b}$ is a non zero vector

Scalar or dot product exercise very short answer question 12

ANSWER: $3$
GIVEN: $|\vec{b}|=1 \quad \&(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=8$
HINTS:Use the identity of $\left(a^{2}-b^{2}\right)$
SOLUTION:
$\begin{aligned} &(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=8 \\ & \end{aligned}$
$\Rightarrow|\vec{a}|^{2}-|\vec{b}|^{2}=8\left[\because(a+b)(a-b)=a^{2}-b^{2}\right] \\$
$\Rightarrow|\vec{a}|^{2}-1=8 \quad\left[\because|\vec{b}|^{2}=1\right] \\$
$\Rightarrow|\vec{a}|^{2}=9 \\$
$\Rightarrow|\vec{a}|=3$

Scalar or dot product exercise very short answer question 13

ANSWER: $\sqrt{3}$
GIVEN: $|\vec{a}|=|\vec{b}|=1 \quad \&|\vec{a}+\vec{b}|=1$
HINTS: Use the identity of $(a+b)^{2} \&(a-b)^{2}$
SOLUTION:
$\begin{aligned} &|\vec{a}+\vec{b}|=1 \\ & \end{aligned}$
$|\vec{a}+\vec{b}|^{2}=1^{2}\qquad[\text { Squaring both sides }] \\$
$\Rightarrow|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=1\qquad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right] \\$
$\Rightarrow 1+1+2 \vec{a} \cdot \vec{b}=1 \qquad[\because|\vec{a}|=|\vec{b}|=1] \\$
$\Rightarrow 2 \vec{a} \cdot \vec{b}=-1 \Rightarrow \vec{a} \cdot \vec{b}=-1$
Now
$\begin{aligned} &|\vec{a}-\vec{b}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}-2 \vec{a} \cdot \vec{b}\qquad\left[\because(\mathrm{a}-\mathrm{b})^{2}=\mathrm{a}^{2}+b^{2}-2 a b\right] \\ & \end{aligned}$
$\Rightarrow|\vec{a}-\vec{b}|^{2}=1+1-2 \frac{1}{2}=1 \qquad\left[\because|\vec{a}|=|\vec{b}|=-\frac{1}{2}\right] \\$
$\Rightarrow|\vec{a}-\vec{b}|=\sqrt{3}$

Scalar or dot product exercise very short answer question 14

ANSWER: $5$
GIVEN: $|\vec{a}|=2|\vec{b}|=5 \quad \& \vec{a} \cdot \vec{b}=2$
HINTS:Use the identity of $\left ( a-b \right )^{2}$
SOLUTION:
$\begin{aligned} &|\vec{a}-\vec{b}|^{2}=|\vec{a}|^{2}-|\vec{b}|^{2}-2 \vec{a} \cdot \vec{b} \qquad\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right] \\ & \end{aligned}$
$\Rightarrow|\vec{a}-\vec{b}|^{2}=(2)^{2}+(5)^{2}-2 \times 2 \qquad\left[\because|\vec{a}|^{2}=2 \&|\vec{b}|^{2}=5 \& \vec{a} \cdot \vec{b}=2\right]$
$\begin{aligned} &\Rightarrow|\vec{a}-\vec{b}|^{2}=4+25-4 \\ & \end{aligned}$
$\Rightarrow|\vec{a}-\vec{b}|^{2}=25 \\$
$\Rightarrow|\vec{a}-\vec{b}|=5$

Scalar or dot product exercise very short answer question 15

ANSWER:- $\frac{1}{\sqrt{2}}$
GIVEN: $\vec{a}=\hat{i}-\hat{j} \quad \& \quad \vec{b}=-\hat{j}+\hat{k}$
HINTS: use the formula of projection of $\vec{a} \text { and } \vec{b}$
SOLUTION:
We know of Projection of $\vec{a} \text { on } \vec{b}=\frac{{\vec{a} \cdot \vec{b}}}{{|\vec{b}|}}$
$\begin{aligned} &=\frac{(\hat{i}-\hat{j}) \cdot(-\hat{j}+\hat{k})}{\hat{i}-\hat{j}+\hat{k}} \\ & \end{aligned}$
$=\frac{1}{\sqrt{1+1}} \\$
$=\frac{1}{\sqrt{2}}$

Scalar or dot product exercise very short answer question 16

ANSWER: $2$
GIVEN: $\frac{|\vec{a}+\vec{b}|^{2}+|\vec{a}-\vec{b}|^{2}}{|\vec{a}|^{2}+|\vec{b}|^{2}+}=?$
HINTS:Use algebraic identities
SOLUTION:
$\frac{|\vec{a}+\vec{b}|^{2}+|\vec{a}-\vec{b}|^{2}}{|\vec{a}|^{2}+|\vec{b}|^{2}+}=$ $\frac{|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}+|\vec{a}|^{2}+\left.|\vec{b}|\right|^{2}-2 \vec{a} \cdot \vec{b}}{{|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}}}$ $\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b \quad \&(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$
$\Rightarrow \frac{|\vec{a}+\vec{b}|^{2}+|\vec{a}-\vec{b}|^{2}}{|\vec{a}|^{2}+|\vec{b}|^{2}}=\frac{2|\vec{a}|^{2}+2|\vec{b}|^{2}}{|\vec{a}|^{2}+|\vec{b}|^{2}}$
$\Rightarrow \frac{|\vec{a}+\vec{b}|^{2}+|\vec{a}-\vec{b}|^{2}}{|\vec{a}|^{2}+|\vec{b}|^{2}}=2$

Scalar or dot product exercise very short answer question 17

ANSWER:$3,-4,12$
GIVEN: $\vec{r}=3 \hat{i}-4 \hat{j}+12 \hat{k}$
HINTS: Find the projection on each coordinate axis one by one.
SOLUTION:We know that,
Component along x-axis= $\hat{i}$
Component along y-axis= $\hat{j}$
Component along z-axis= $\hat{k}$
Now projection of : $\vec{r}$ on x axis = $\frac{\vec{r\cdot \hat{i}}}{\left | \hat{i} \right |}$
$\begin{aligned} &=\frac{(3 \hat{i}-4 \hat{j}+12 \hat{k}) \hat{.}}{\sqrt{1}} \\ & \end{aligned}$
$=\frac{3-0+0}{1}=3\qquad[\because \hat{i} \times \hat{i}=1, \hat{i} \times \hat{j}=0, \hat{i} \times \hat{k}=0]$
projection of $\vec{r}$ on y axis $\frac{\vec{r\cdot \hat{j}}}{\left | \hat{j} \right |}$
$\begin{aligned} &=\frac{(\hat{i}-4 \hat{j}+12 \hat{k}) \hat{. j}}{\sqrt{1}} \\ & \end{aligned}$
$=\frac{0-4+0}{1}=-4\qquad[\because \hat{i} \times \hat{i}=0, \hat{j} \times \hat{j}=1, \hat{j} \times \hat{k}=0]$
projection of $\vec{r}$ on y axis $\frac{\vec{r\cdot \hat{k}}}{\left | \hat{k} \right |}$
$\begin{aligned} &=\frac{(3 \hat{i}-4 \hat{j}+12 \hat{k}) \cdot \hat{k}}{\sqrt{1}} \\ & \end{aligned}$
$=\frac{0-0+12}{1}=12\qquad[\because \hat{i} \times \hat{k}=0, \hat{i} \times \hat{k}=0, \hat{k} \times \hat{k}=1]$
Hence ,required answer is $3,-4,12$

Scalar or dot product exercise very short answer question 18

ANSWER: $\frac{(\vec{a} \cdot \vec{b})}{|\vec{a}|^{2}} \times \vec{a}$
GIVEN: Component of $\vec{b}$ along $\vec{a}$
HINTS:you must know the concept of $\vec{b}$ along $\vec{a}$
SOLUTION:The component of $\vec{b}$ along $\vec{a}$
$= \frac{(\vec{a} \cdot \vec{b})}{|\vec{a}|^{2}} \times \vec{a}$

Scalar or dot product exercise very short answer question 19

ANSWER:-$\vec{a}$
GIVEN: $\vec{a}$ is any vector and $(\vec{a} \hat{i}) \hat{i}+(\vec{a} \cdot \hat{j}) \hat{j}+(\vec{a} \cdot \hat{k}) \hat{k}$
HINTS: you must know about the general form of any vector
SOLUTION:we know the general form of any vector.
$\therefore \vec{a}=\hat{x i}+y \hat{j}+z \hat{k}\cdot \cdot \cdot \cdot (i)$
Now
$\begin{aligned} &(\vec{a} \hat{i}) \hat{i}+(\vec{a} \cdot \hat{j}) \hat{j}+(\vec{a} \cdot \hat{k}) \hat{k}=[x \hat{i}+y \hat{j}+z \hat{k}]\cdot \hat{i}+[x \hat{i}+y \hat{j}+z \hat{k}]\cdot \hat{j}+[\hat{x} \hat{i}+y \hat{j}+z \hat{k}] \cdot \hat{k} \\ & \end{aligned}$
$(\vec{a} \hat{i}) \hat{i}+(\vec{a} \cdot \hat{j}) \hat{j}+(\vec{a} \cdot \hat{k}) \hat{k}=(x+0+0) \cdot \hat{i}+(0+y+0) \cdot \hat{j}+(0+0+z) \cdot \hat{k}$
$\begin{aligned} &(\vec{a} \cdot \hat{i}) \hat{i}+(\vec{a} \cdot \hat{j}) \hat{j}+(\vec{a} \cdot \hat{k}) \hat{k}=x \hat{i}+y \hat{j}+z \hat{k} \\ & \end{aligned}$
$(\vec{a}\cdot \hat{i}) \hat{i}+(\vec{a} \cdot \hat{j}) \hat{j}+(\vec{a} \cdot \hat{k}) \hat{k}=\vec{a} \quad[\text { use }(\mathrm{i})]$

Scalar or dot product exercise very short answer question 20

ANSWER:- $\theta = \frac{\pi}{3}$
GIVEN:$\theta=\left(0, \frac{\pi}{3}\right) \text { and } \vec{a}=(\sin \theta \hat{i}+(\cos \theta \hat{j}) \text { and } \vec{b}=\hat{i}-\sqrt{3} \hat{j}+2 \hat{k}$ are perpendicular
HINTS: use the formula of perpendicular vectors
SOLUTION:if vectors are perpendicular,then
$\begin{aligned} &\Rightarrow[(\sin \theta \hat{i}+(\cos \theta) \hat{j}] \cdot[\hat{i}-\sqrt{3} \hat{j}+2 \hat{k}]=0 \\ & \end{aligned}$
$\Rightarrow \sin \theta-\sqrt{3} \cos \theta=0$
Divide the equation by $\cos \theta$
$\begin{aligned} &\Rightarrow \frac{\sin \theta}{\cos \theta}-\frac{\sqrt{3} \cos \theta}{\cos \theta}=\frac{0}{\cos \theta} \\ & \end{aligned}$
$\Rightarrow \tan \theta-\sqrt{3}=0 \Rightarrow \tan \theta=\sqrt{3} \\$
$\Rightarrow \tan \theta-\tan \frac{\pi}{3} \Rightarrow \theta=\frac{\pi}{3}$

Scalar or dot product exercise very short answer question 21

Answer :- $1$
Hint :- You must know the formula of projection of $\vec{a} \& \vec{b}$
Given :- $\hat{\imath}+\hat{\jmath}+\hat{k} \& \hat{\jmath}$
Solution :- projection $(\hat{\imath}+\hat{\jmath}+\hat{k})$ along $\hat{j}$
$=\frac{(\hat{\imath}+\hat{\jmath}+\hat{k})(\hat{j})}{|\hat{j}|}$
$\Rightarrow$ Projection of $(\hat{\imath}+\hat{\jmath}+\hat{k})$ along $\hat{j}$ $=\frac{0+1+0}{\sqrt{1}}=1$

Scalar or dot product exercise very short answer question 22

Answer :- $\hat{i}$
Hint :- Use the general form of any vector
Given :- $\vec{a} \cdot \hat{\imath}=\vec{a}(\hat{\imath}+\hat{\jmath})=\vec{a}(\hat{l}+\hat{\jmath}+\hat{k})=1$
Solution :-$\vec{a}=x \hat{i}+y \hat{\jmath}+z \hat{k}$ (?general form)
$\begin{aligned} &\Rightarrow(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot \hat{l}=(x \hat{\imath}+y \hat{\jmath}+z \hat{k})(\hat{\imath}+\hat{\jmath})=(x \hat{\imath}+y \hat{\jmath}+z \hat{k})(\hat{\imath}+\hat{\jmath}+\hat{k})=1 \\ & \end{aligned}$
$\Rightarrow x+0+0=x+y+0=x+y+z=1 \\$
$\Rightarrow x=1, y+x=1 \& x+y+z=1$
Now,
$\begin{aligned} &\Rightarrow y+1=0(\because x=1) \\ & \end{aligned}$
$\Rightarrow y=1$
and $x+y+z=1$
$\begin{aligned} &\Rightarrow 1+0+z=1 \quad(\because x=1 \& y=0) \\ & \end{aligned}$
$\Rightarrow z=0$

Scalar or dot product exercise very short answer question 23

Answer :-$\theta = \frac{\pi}{2}$
Hint :- You must know how to find the angle between two vectors
Given :- $|\vec{a}|=|\vec{b}|=1$
Solution :- Let the angle between
$\begin{aligned} &\Rightarrow \cos \theta=\frac{(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})}{|\vec{a}+\vec{b}| \cdot|\vec{a}-\vec{b}|} \\ & \end{aligned}$
$\Rightarrow \cos \theta=\frac{|\vec{a}|^{2}-|\vec{b}|^{2}}{|\vec{a}+\vec{b}| \cdot|\vec{a}-\vec{b}|}\qquad\qquad\left[\because(\mathrm{a}+\mathrm{b}) \cdot(\mathrm{a}-\mathrm{b})=a^{2}-b^{2}\right]$
$\Rightarrow \cos \theta=\frac{1-1}{|\vec{a}+\vec{b}| \cdot|\vec{a}-\vec{b}|}\qquad[\because|\vec{a}|=|\vec{b}|=1]$
$\begin{aligned} \Rightarrow &\cos \theta=\frac{0}{|\vec{a}+\vec{b}| \cdot|\vec{a}-\vec{b}|} \\ & \end{aligned}$
$\Rightarrow \cos \theta=0 \\$
$\Rightarrow \cos \theta=\cos \frac{\pi}{2} \\$
$\Rightarrow \theta=\frac{\pi}{2}$

Scalar or dot product exercise very short answer question 24

Answer :- $\sqrt{2}$
Hint :- Use the identity $\left ( a+b\right )^{2}$
Given :- $|\vec{a}|=|\vec{b}|=1 \& \vec{a} \cdot \vec{b}=0$
Solution :- $|\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b} \qquad\qquad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 \mathrm{ab}\right]$
$\begin{aligned} &\Rightarrow|\vec{a}+\vec{b}|^{2}=(1)^{2}+(1)^{2}+2 * 0 \qquad[\because|\vec{a}|=|\vec{b}|=1 \& \vec{a} \cdot \vec{b}=0] \\ & \end{aligned}$
$\Rightarrow|\vec{a}+\vec{b}|^{2}=2 \\$
$\Rightarrow|\vec{a}+\vec{b}|=\sqrt{2}$

Scalar or dot product exercise very short answer question 25

Answer :- $\sqrt{3}$
Hint :- Use the identity $(a+b+c)^{2}$
Given :- $|\vec{a}|=|\vec{b}|=|\vec{c}|=1, \vec{a} \cdot \vec{b}=0, \vec{b} \cdot \vec{c}=0 \text { and } \vec{c} \cdot \vec{a}=0$
Solution :- $|\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$
$\left[\because(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a)\right]$
$\begin{aligned} &\Rightarrow|\vec{a}+\vec{b}+\vec{c}|^{2}\! =\! 1+1+1+2 *(0+0+0)\qquad[\because|\vec{a}|=|\vec{b}|=|\vec{c}|\! =\! 1, \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}= \\ &0] \end{aligned}$
$\begin{aligned} &\Rightarrow|\vec{a}+\vec{b}+\vec{c}|^{2}=3+0 \\ & \end{aligned}$
$\Rightarrow|\vec{a}+\vec{b}+\vec{c}|=\sqrt{3}$

Scalar or dot product exercise very short answer question 26

Answer :- $\cos ^{-1}\left(\frac{-1}{3}\right)$
Hint :- You must know the formula of $\vec{a}\cdot \vec{b}$
Given :- $\vec{a}=\hat{\imath}-\hat{\jmath}+\hat{k}\: \&\: \vec{b}=\hat{\imath}+\hat{\jmath}-\hat{k}$
Solution :- we know that $\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$
$\begin{aligned} &\Rightarrow \cos \theta=\frac{(\hat{\imath}-\hat{j}+\hat{k})(\hat{i}+\hat{j}-\hat{k})}{\sqrt{1+1+1} \sqrt{1+1+1}} \\ & \end{aligned}$
$\Rightarrow \cos \theta=\frac{1-1-1}{\sqrt{3} * \sqrt{3}} \\$
$\Rightarrow \cos \theta=\frac{-1}{3} \\$
$\Rightarrow \theta=\cos ^{-1}\left(\frac{-1}{3}\right)$

Scalar or dot product exercise very short answer question 27

Answer :- $\lambda= \frac{5}{2}$

Hint :- You must know the formula for perpendicular vectors

Given :- $\vec{a}=2 \hat{\imath}+\lambda \hat{\jmath}+\hat{k} \& \vec{b}=\hat{\imath}-2 \hat{\jmath}+3 \hat{k}$

Solution :- If vectors are perpendicular than

$\begin{aligned} &\Rightarrow(2 \hat{\imath}+\lambda \hat{\jmath}+\hat{k}) \cdot(\hat{\imath}-2 \hat{\jmath}+3 \hat{k})=0 \\ & \end{aligned}$

$\Rightarrow 2-2 \lambda+3=0 \\$

$\Rightarrow-2 \lambda+5=0 \\$

$\Rightarrow-2 \lambda=-5 \\$

$\Rightarrow \lambda=\frac{5}{2}$

Scalar or dot product exercise very short answer question 28

Answer :-$\frac{8}{7}$
Hint :- Use the formula of projection of $\vec{a} \& \vec{b}$
Given :- $\vec{a} \cdot \vec{b}=8 \& \vec{b}=2 \hat{\imath}+6 \hat{\jmath}+3 \hat{k}$
Solution :- Projection of $\vec{a} \text { on } \vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\qquad[\because \vec{a} \cdot \vec{b}=8]$
$\Rightarrow$ Projection of $\vec{a} \text { on } \vec{b}=\frac{8}{\sqrt{(2)^{2}+(6)^{2}+(3)^{2}}}=\frac{8}{\sqrt{4+36+9}}=\frac{8}{\sqrt{49}}$
$\Rightarrow$ Projection of $\vec{a} \text { on } \vec{b}=\frac{8}{7}$

Scalar or dot product exercise very short answer question 29

Answer :-$23$
Hint :- Apply the formula of parallel vectors
Given :-$\vec{a}=3 \hat{\imath}+2 \hat{\jmath}+9 \hat{k} \& \vec{b}=\hat{\imath}+\mathrm{p} \hat{\jmath}+3 \hat{k}$ are parallel vectors
Solution :- If vectors are parallel then $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Here, $a_{1}=3, b_{1}=2, c_{1}=9, a_{2}=1, b_{2}=p, c_{2}=3$
$\begin{aligned} &\Rightarrow \frac{3}{1}=\frac{2}{p}=\frac{9}{3} \\ \end{aligned}$
$\Rightarrow \quad 3=\frac{2}{p}=3 \\$
$\Rightarrow \frac{2}{p}=3 \\$
$\Rightarrow 2=3 p \\$
$\Rightarrow p=\frac{2}{3}$

Scalar or dot product exercise very short answer question 30

Answer :-$3$

Hint :- Apply the formula of perpendicular vectors

Given :-$\vec{a}=2 \hat{\imath}+\lambda \hat{\jmath}+3 \hat{k} \& \vec{b}=3 \hat{\imath}+2 \hat{\jmath}-4 \hat{k}$are perpendicular

Solution :- If vectors are perpendicular then $\vec{a} \cdot \vec{b}=0$
$\begin{aligned} &\Rightarrow(2 \hat{\imath}+\lambda \hat{\jmath}+3 \hat{k})(3 \hat{\imath}+2 \hat{\jmath}-4 \hat{k})=0 \\ & \end{aligned}$
$\Rightarrow 6+2 \lambda-12=0 \\$
$\Rightarrow 2 \lambda-6=0 \\$
$\Rightarrow 2 \lambda=6 \\$
$\Rightarrow \lambda=\frac{6}{2} \\$
$\Rightarrow \lambda=3$

Scalar or dot product exercise very short answer question 31

Answer :-$\frac{3}{2}$
Hint :- Apply the formula of projection of $\vec{b}\: and \: \vec{a}$
Given :- $|\vec{a}|=2 \&|\vec{b}|=3 \& \vec{a} \cdot \vec{b}=3$
Solution :- projection of $\vec{b} \text { on } \vec{a}=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}$
$\Rightarrow$ projection of $\vec{b} \text { on } \vec{a}=\frac{3}{2}[\because \vec{a} \cdot \vec{b}=3 \&|\vec{a}|=2]$

Scalar or dot product exercise very short answer question 32

Answer :-$\frac{\pi}{4}$
Hint :- Apply the formula of cosθ
Given :-$|\vec{a}|=\sqrt{3} \&|\vec{b}|=2 \& \vec{a} \cdot \vec{b}=\sqrt{6}$
Solution :- we know that $\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$
$\begin{aligned} &\Rightarrow \cos \theta=\frac{\sqrt{6}}{\sqrt{3} * 2}\qquad[\because|\vec{a}|=\sqrt{3} \&|\vec{b}|=2 \& \vec{a} \cdot \vec{b}=\sqrt{6} \text { given }] \\ & \end{aligned}$
$\Rightarrow \cos \theta=\frac{\sqrt{2}}{2}$
$\begin{aligned} &\Rightarrow \cos \theta=\frac{1}{\sqrt{2}} \\ & \end{aligned}$
$\Rightarrow \theta=\frac{\pi}{4} \qquad\left[\because \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right]$

Scalar or dot product exercise very short answer question 33

Answer: $5$
Given: $(\hat{\imath}+3 \hat{\jmath}+7 \hat{k}) \&(2 \hat{\imath}-3 \hat{\jmath}+6 \hat{k})$
Hint: Use the formula of projection of a on $\vec{a} \text { on } \vec{b}$
Solution: projection of the vector $=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
$\begin{aligned} &=\frac{(\hat{i}+3 \hat{\jmath}+7 \hat{\mathrm{k}})(2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})}{\sqrt{2^{2}+6^{2}+3^{2}}} \\ & \end{aligned}$
$=\frac{2-9+42}{\sqrt{4+9+36}}=\frac{44-9}{\sqrt{49}}=\frac{35}{7}$
$\Rightarrow$ projection of the vector $= 5$

Scalar or dot product exercise very short answer question 34

Answer: $\lambda= 5$
Given:$\vec{a}=\lambda \hat{i}+\hat{\jmath}+4 \hat{k} \& \vec{b}=2 \hat{i}+6 \hat{\jmath}+3 \hat{k}$ and projection of a on $\vec{a} \text { on } \vec{b}=4 \text { units. }$
Hint: you must know about the concept of projection of a on $\vec{a} \text { on } \vec{b}$
Solution: projection of a on $\vec{a} \text { on } \vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
$\begin{aligned} &\Rightarrow 4=\frac{(\lambda \hat{i}+\hat{j}+4 \hat{k}) \cdot(2 \hat{i}+6 \hat{j}+3 \hat{k})}{\sqrt{2^{2}+6^{2}+3^{2}}} \\ & \end{aligned}$
$\Rightarrow 4=\frac{2 \lambda+6+12}{\sqrt{4+36+9}}$
$\begin{aligned} &\Rightarrow \quad 4=\frac{2 \lambda+18}{\sqrt{49}} \\ & \end{aligned}$
$\Rightarrow 4=\frac{2 \lambda+18}{\sqrt{7}}$
$\begin{aligned} &\Rightarrow 28=2 \lambda+18 \\ & \end{aligned}$
$\Rightarrow 28-18=2 \lambda \\$
$\Rightarrow 2 \lambda=10 \\$
$\Rightarrow \lambda=5$

Scalar or dot product exercise very short answer question 35

Answer:$\lambda= \frac{5}{2}$
Given:$\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+\lambda \hat{\jmath}+\hat{\mathrm{k}} \& \overrightarrow{\mathrm{b}}=\hat{\imath}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ and $\vec{a} \& \vec{b}$ are perpendicular
Hint: Use the formula of perpendicular vectors.
Solution: if vectors are perpendicular, then,
$\begin{aligned} &\Rightarrow(2 \hat{\imath}+\lambda \hat{\jmath}+\hat{k}) \cdot(\hat{\imath}-2 \hat{\jmath}+3 \hat{k})=0 \\ & \end{aligned}$
$\Rightarrow 2-2 \lambda+3=0 \\$
$\Rightarrow-2 \lambda+5=0 \\$
$\Rightarrow-2 \lambda=-5 \\$
$\Rightarrow 2 \lambda=5 \\$
$\Rightarrow \lambda=\frac{5}{2}$

Scalar or dot product exercise very short answer question 36

Answer:$\frac{8}{7}$
Given: $7 \hat{\imath}+\hat{\jmath}-4 \hat{\mathrm{k}} \& 2 \hat{\mathrm{i}}+6 \hat{\jmath}+3 \hat{\mathrm{k}}$
Hint: Apply the formula of projector of the vectors.
Solution: projection of the vector $=\frac{\vec{a} \cdot \vec{b}}{\left|b^{-3}\right|}$

$\begin{aligned} &\frac{(7 \hat{\imath}+\hat{\jmath}-4 \hat{\mathrm{k}}) \cdot(2 \hat{\mathrm{i}}+6 \hat{\jmath}+3 \hat{\mathrm{k}})}{\sqrt{2^{2}+6^{2}+3^{2}}} \\ & \end{aligned}$
$=\frac{14+6-12}{\sqrt{4+36+9}}$
$=\frac{20-12}{\sqrt{49}}$
projection of the vector $= \frac{8}{7}$

Scalar or dot product exercise very short answer question 37

Answer: $\lambda = \frac{5}{2}$
Given:$\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+\lambda \hat{\mathrm{j}}+\hat{\mathrm{k}} \& \overrightarrow{\mathrm{b}}=\hat{\mathrm{\imath}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ and $\vec{a} \& \vec{b}$ are perpendicular
Hint: Use the formula of perpendicular vectors.
Solution: if vectors are perpendicular, then,
$\begin{aligned} &\Rightarrow(2 \hat{\imath}+\lambda \hat{\jmath}+\hat{\mathrm{k}}) \cdot(\hat{\imath}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=0 \\ & \end{aligned}$
$\Rightarrow 2-2 \lambda+3=0 \\$
$\Rightarrow-2 \lambda+5=0 \\$
$\Rightarrow-2 \lambda=-5 \\$
$\Rightarrow 2 \lambda=5 \\$
$\Rightarrow \lambda=\frac{5}{2}$

Scalar or dot product exercise very short answer question 38

Answer:$2$
Given: $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-2 \hat{\mathrm{\jmath}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{\jmath}}-2 \hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{c}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}$
Hint: Find $\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}$ and then apply the formula of projection.
Solution: $\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\hat{\mathrm{\imath}}+2 \hat{\mathrm{\jmath}}-2 \hat{\mathrm{k}}+2 \hat{\mathbf{i}}-\hat{\mathrm{\jmath}}+4 \hat{\mathrm{k}}$
$\Rightarrow \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=3 \hat{\mathrm{i}}+\hat{\jmath}+2 \hat{\mathrm{k}}$
Projection of $\vec{b}+\vec{c} \text { on } \vec{a}=(\vec{b}+\vec{c}) \times \frac{\vec{a}}{|\vec{a}|}$
$=\frac{(3 \hat{i}+\hat{j}+2 \hat{k})(2 \hat{i}-2 \hat{j}+\hat{k})}{|2 \hat{\imath}-2 \hat{\jmath}+\widehat{k}|}$
$\begin{aligned} &=\frac{6-2+2}{\sqrt{2^{2}+(-2)^{2}+1^{2}}} \\ & \end{aligned}$
$=\frac{6}{\sqrt{4+4+1}}=\frac{6}{\sqrt{9}}=\frac{6}{3}$
$\Rightarrow$ Projectio of $(\vec{b}+\vec{c}) \text { on } \vec{a}=2$

Scalar or dot product exercise very short answer question 39

Answer:$12$
Given: $|\vec{a}|=5, \vec{a} \cdot \vec{b}=0 \text { and }|\vec{a}+\vec{b}|=13$
Hint: Apply the identity $(a+b)^{2}$
Solution: $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}=|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}+2 \cdot \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}$
$\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$
$\Rightarrow(13)^{2}=(5)^{2}+(|\vec{b}|)^{2}+2 \times 0$
$\left[\because|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}=13,|\overrightarrow{\mathrm{a}}|^{2}=5 \& \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0\right]$
$\begin{aligned} &\Rightarrow 169-25+|\vec{b}|^{2} \\ & \end{aligned}$
$\Rightarrow 169-25=|\vec{b}|^{2} \\$
$\Rightarrow|\vec{b}|^{2}=144 \\$
$\Rightarrow|\vec{b}|^{2}=12$

Scalar or dot product exercise very short answer question 40

Answer:$\theta = \frac{\pi}{6}$
Given: $|\vec{a} \times \vec{b}|=1,|\vec{a}|=3: \text { and }|\vec{b}|=\frac{2}{3}$
Hint: Apply the identity $\left|\vec{a} \times \vec{b}^{2}\right|$
Solution: We know that $|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}}| \sin \theta .|\hat{\mathrm{n}}|$
$\Rightarrow 1=\left|3 x \frac{2}{3} x \sin \theta \cdot 1\right|\qquad\left[\because|\vec{a}|=3,|\overrightarrow{\mid b}|=\frac{2}{3} \&|\vec{a} \times \vec{b}|=1\right]$
$\begin{aligned} &\Rightarrow 1=2 \sin \theta \\ & \end{aligned}$
$\Rightarrow \sin \theta=\frac{1}{2} \\$
$\Rightarrow \theta=\frac{\pi}{6} \qquad\left[\because \sin \frac{\pi}{6}=\frac{1}{2}\right]$

Scalar or dot product exercise very short answer question 41

Answer: $\frac{2\pi}{3}$
Given: $|\vec{a}|=|\vec{b}|=1 \text { and }|\vec{a}+\vec{b}|=1$
Hint: Apply the identity $(a+b)^{2}$
Solution: $|\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+2 \cdot \vec{a} \cdot \vec{b}$
$\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$
$\Rightarrow(1)^{2}=(1)^{2}+(1)^{2}+2 \cdot \vec{a} \cdot \vec{b}$
$[\because|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=1,|\overrightarrow{\mathrm{a}}|=1 \&|\overrightarrow{\mathrm{b}}|=1]$
$\begin{aligned} &\Rightarrow 1=2+2 \vec{a} \cdot \vec{b} \\ & \end{aligned}$
$\Rightarrow-1=2 \vec{a} \cdot \vec{b} \\$
$\Rightarrow \vec{a} \cdot \vec{b}=\frac{-1}{2}$
Now, we know $\operatorname{Cos} \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$
$\begin{aligned} &\Rightarrow|\vec{a}||\vec{b}| \operatorname{Cos} \theta=\frac{-1}{2}\qquad\left[\because \vec{a} \cdot \vec{b}=\frac{-1}{2}\right] \\ & \end{aligned}$
$\Rightarrow \operatorname{Cos} \theta=\frac{-1}{2}\qquad[\because|\vec{a}|=|\vec{b}|=1] \\$
$\Rightarrow \theta=\frac{2 \pi}{3} \qquad\left[\because \operatorname{Cos} \frac{2 \pi}{3}=\frac{-1}{2}\right]$

Scalar or dot product exercise very short answer question 42

Answer: $\frac{\pi}{6}$
Given: $|\vec{a}|=|\vec{b}|=1 \text { and }(\sqrt{3} \vec{a}-\vec{b})=1$
Hint: Apply the identity $(a-b)^{2}$
Solution: $(\sqrt{3} \vec{a}-\vec{b})=1$
$\begin{aligned} &\Rightarrow(\sqrt{3} \vec{a}-\vec{b})^{2}=1^{2} \quad(\text { squaring both sides }) \\ & \end{aligned}$
$\Rightarrow 3|\vec{a}|^{2}+|\vec{b}|^{2}-2(\sqrt{3} \vec{a}-\vec{b})=1$
$\begin{aligned} &{\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]} \\ & \end{aligned}$
$\Rightarrow(3 \times 1)^{2}+(1)^{2}-2 \sqrt{3} \vec{a} \cdot \vec{b}=1 \quad[\because|\vec{a}|=|\vec{b}|=1]$
$\begin{aligned} &\Rightarrow \quad 3+1-2 \sqrt{3} \vec{a} \cdot \vec{b}=1 \\ & \end{aligned}$
$\Rightarrow-2 \sqrt{3} \vec{a} \cdot \vec{b}=-3 \\$
$\Rightarrow \quad \vec{a} \cdot \vec{b}=\frac{+3}{+2 \sqrt{3}} \\$
$\Rightarrow \vec{a} \cdot \vec{b}=\frac{\sqrt{3}}{2}$
$\begin{array}{ll} \Rightarrow|\vec{a}||\vec{b}| \operatorname{Cos} \theta=\frac{\sqrt{3}}{2} &\qquad {\left[\because \operatorname{Cos} \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\right]} \\ \end{array}$
$\Rightarrow \operatorname{Cos} \theta=\operatorname{Cos} \frac{\pi}{6} \qquad\left[\because|\vec{a}|=|\vec{b}|=1 \& \operatorname{Cos} \frac{\pi}{6}=\frac{\sqrt{3}}{2}\right] \\$
$\Rightarrow \theta=\frac{\pi}{6}$

Scalar or dot product exercise very short answer question 43

Answer:$3$
Given: $|\vec{a}|=|\vec{b}|, \theta=60^{\circ} \quad \vec{a} \cdot \vec{b}=\frac{9}{2}$
Hint: Apply the formula of $\operatorname{Cos} \theta$
Solution: We know $\operatorname{Cos} \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$
$\begin{array}{ll} \Rightarrow \operatorname{Cos} \theta=60^{\circ}=\frac{\frac{9}{2}}{|\vec{a}||\vec{b}|} &\qquad {\left[\because \theta=60^{\circ}, \vec{a} \cdot \vec{b}=\frac{9}{2} \text { and }|\vec{a}|=|\vec{b}|\right]} \\ \end{array}$
$\Rightarrow \frac{1}{2}=\frac{9}{2|\vec{a}|^{2}} \qquad {\left[\because \operatorname{Cos} \theta=\frac{1}{2}\right]}$
$\begin{aligned} &\Rightarrow \frac{2|\vec{a}|^{2}}{2}=9 \\ & \end{aligned}$
$\Rightarrow|\vec{a}|^{2}=9 \mid \\$
$\Rightarrow|\vec{a}|=3=|\vec{b}| \qquad[\because|\vec{a}|=|\vec{b}|]$

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