RD Sharma Class 12 Exercise 23 MCQ Scalar and dot product Solutions Maths

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The RD Sharma Class 12th Chapter 23 MCQ reference book can be used to clarify any doubts regarding the Multiple Choice Questions of chapter 23.

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RD Sharma Class 12 Exercise 23 MCQ Scalar and dot product Solutions Maths - Download PDF Free Online

Updated on Jan 27, 2022 05:03 PM IST

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RD Sharma Class 12 Solutions Chapter23 MCQ Scaler and dot product - Other Exercise
Scalar or Dot Products Excercise: MCQ
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter23 MCQ Scaler and dot product - Other Exercise

Scalar or Dot Products Excercise: MCQ

Scaler and dot product exercise multiple choice question 1

Answer:
Option (c) $\cos \theta=\frac{-4}{5}$
Hint:
Apply Elimination method to find $\vec{a} \text { and } \vec{b}$, then find $|\vec{a}| \text { and }|\vec{b}|$
Given:
$\vec{a} \text { and } \vec{b}$satisfy the equation $2 \vec{a}+\vec{b}=\vec{p}$ and $\vec{a}+2 \vec{b}=\vec{q}$
Where, $\vec{p}=\hat{i}+\hat{j}, \vec{q}=\hat{i}-\hat{j}$ and $\theta$ is the angle between $\vec{a} \text { and } \vec{b}$
Solution:
$\begin{aligned} &2 \vec{a}+\vec{b}=\vec{p} \quad\quad\quad.....(i)\\ &\vec{a}+2 \vec{b}=\vec{q} \quad\quad\quad.....(ii)\times 2 \end{aligned}$
$\Rightarrow \quad \begin{aligned} &2 \vec{a}+\vec{b}=\vec{p} \\ &2 \vec{a}+4 \vec{b}=2 \vec{q} \end{aligned}$
____________

$\begin{aligned} &-3 \vec{b}=\vec{p}-2 \vec{q} \\ \Rightarrow \quad & \vec{b}=\frac{2 \vec{q}-\vec{p}}{3} \end{aligned}$
Put in equation (i)
$\begin{aligned} &2 \vec{a}+\frac{2 \vec{q}-\vec{p}}{3}=\vec{p} \Rightarrow 2 \vec{a}=\frac{3 \vec{p}-2 \vec{q}+\bar{p}}{3} \\\\ &\vec{a}=\frac{4 \vec{p}-2 \vec{q}}{3(2)} \Rightarrow \vec{a}=\frac{2 \vec{p}-\vec{q}}{3} \end{aligned}$
Now, $\vec{a}=\frac{2(\hat{i}+\hat{j})-(\hat{i}-\hat{j})}{3}=\frac{\hat{i}+3 \hat{j}}{3}$ $\left[\begin{array}{l} \because \vec{p}=\hat{i}+\hat{j} \\ \because \vec{q}=\hat{i}-\hat{j} \end{array}\right]$
$\begin{aligned} &\vec{b}=\frac{2(\hat{i}-\hat{j})-(\hat{i}+\hat{j})}{3}=\frac{\hat{i}-3 \hat{j}}{3} \\\\ &|\vec{a}|=\frac{\sqrt{1+9}}{3}=\frac{\sqrt{10}}{3} \\\\ &|\vec{b}|=\frac{\sqrt{1+9}}{3}=\frac{\sqrt{10}}{3} \end{aligned}$ $\left[\begin{array}{l} \because|\vec{a}|=\sqrt{(\hat{i})^{2}+(3 \hat{j})^{2}} \\\\ \because|\vec{b}|=\sqrt{(\hat{i})^{2}+(-3 \hat{j})^{2}} \end{array}\right]$
$\begin{aligned} &\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \\\\ &\frac{1}{9}\left[\hat{i}^{2}-(3 \hat{j})^{2}\right]=\frac{\sqrt{10}}{3} \times \frac{\sqrt{10}}{3} \cos \theta \\\\ &\frac{1}{9}[1-9]=\frac{10}{9} \cos \theta \end{aligned}$
$\begin{aligned} &\frac{-8}{9}=\frac{10}{9} \cos \theta \\\\ &\frac{-8}{10}=\cos \theta \\\\ &\cos \theta=\frac{-4}{5} \end{aligned}$
Hence, option (c) is correct

Scaler and dot product exercise multiple choice question 2

Answer:
Option (b) $\hat{i}$
Hint:
You must know about the concept of dot product
Given:
$\vec{a} \cdot \hat{i}=\vec{a} \cdot(\hat{i}+\hat{j})=\vec{a} \cdot(\hat{i}+\hat{j}+\hat{k})=1$
Solution:
$\vec{a} \cdot \hat{i}=\vec{a} \cdot(\hat{i}+\hat{j})=\vec{a} \cdot(\hat{i}+\hat{j}+\hat{k})=1$ ............(i)
$\begin{aligned} & \vec{a} \cdot \hat{i}=a_{1} \\\\ \Rightarrow & \vec{a} \cdot \hat{i}+\vec{a} \cdot \hat{j}=a_{1}+a_{2} \\\\ & \vec{a} \cdot \hat{i}+\vec{a} \cdot \hat{j}+\vec{a} \cdot \hat{k}=a_{1}+a_{2}+a_{3} \end{aligned}$
Putting in (i)
$a_{1}=a_{1}+a_{2}=a_{1}+a_{2}+a_{3}=1$ ...........(ii)
$Now,a_{1}=a_{1}+a_{2} \Rightarrow a_{2}=0\; \; and\\\\ a_{1}+a_{2}=a_{1}+a_{2}+a_{3} \Rightarrow a_{3}=0$
Put in (ii)
$\begin{aligned} &a_{1}=a_{1}=a_{1}=1 \quad\left[\because a_{2}=0, a_{3}=0\right] \\\\ &\Rightarrow a_{1}=\hat{i} \end{aligned}$
Hence, option (b) is correct

Scaler and dot product exercise multiple choice question 3

Answer:
Option (d) $\frac{\pi }{3}$
Hint:
You must know the formula of $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$
Given:
$\vec{a}+\vec{b}+\vec{c}=0,|\vec{a}|=3,|\vec{b}|=5,|\vec{c}|=7$
Solution:
$\begin{aligned} &\vec{a}+\vec{b}+\vec{c}=0 \\ &\vec{a}+\vec{b}=-\vec{c} \end{aligned}$
Squaring on both sides,
$\begin{aligned} &(\vec{a}+\vec{b})^{2}=(-\vec{c})^{2} \\\\ &(\vec{a})^{2}+2 \vec{a} \cdot \vec{b}+(\vec{b})^{2}=\vec{c}^{2} \quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right] \end{aligned}$
Now, $|\vec{a}|=\vec{a}=3,|\vec{b}|=\vec{b}=5,|\vec{c}|=\vec{c}=7$
$(3)^{2}+2(3) \cdot(5) \cos \theta+(5)^{2}=(7)^{2} \quad[\because \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta]$
$\begin{aligned} &9+30 \cos \theta+25=49 \\\\ &30 \cos \theta=49-34 \\\\ &\cos \theta=\frac{15}{30} \quad\left[\because \cos \frac{\pi}{3}=\frac{1}{2}\right] \end{aligned}$
$\begin{aligned} &\cos \theta=\frac{1}{2} \\\\ &\theta=\frac{\pi}{3} \end{aligned}$
Hence, option (d) is correct

Scaler and dot product exercise multiple choice question 4

Answer:
Option (c) write what is given in option c from book pdf
Hint:
If $\vec{a} \text { and } \vec{b}$ are unit vectors then $|\vec{a}|=1 \text { and }|\vec{b}|=1$
Given:
$\vec{a} \text { and } \vec{b}$ are two unit vectors and $\alpha$ is an angle between them
Solution:
According to the given question,
$|\vec{a}|+|\vec{b}|=1$
Squaring both side, we get
$|\vec{a}|^{2}+2 \vec{a} \cdot \vec{b}+|\vec{b}|^{2}=1 \quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$
As both are unit vectors
$1+2|\vec{a}| \cdot|\vec{b}| \cos \alpha+1=1 \quad[\because \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta]$
$\begin{aligned} &2 \cos \alpha=1-2 \\\\ &\cos \alpha=\frac{-1}{2} \quad\left[\because \cos \left(\frac{-\pi}{3}\right)=\cos \left(\pi-\frac{\pi}{3}\right)\right] \end{aligned}$
$\alpha=\frac{2 \pi}{3}$

Scaler and dot product exercise multiple choice question 5

Answer: Option (b) unit vector
Hint: The formula of Magnitude of a vector$=\sqrt{a^{2}+b^{2}+c^{2}}$
Given:
$\operatorname{Vector}(\cos \alpha \cos \beta) \hat{\imath}+(\cos \alpha \sin \beta) \hat{\jmath}+(\sin \alpha) \hat{k}$
Solution: $(\cos \alpha \cos \beta) \hat{\imath}+(\cos \alpha \sin \beta) \hat{\jmath}+(\sin \alpha) \hat{k}$
Magnitude of a vector $=\sqrt{a^{2}+b^{2}+c^{2}}$
Magnitude of a vector$=\sqrt{\left(\cos ^{2} \alpha \cos ^{2} \beta\right)+\left(\cos ^{2} \alpha \sin ^{2} \beta\right)+\left(\sin ^{2} \alpha\right)}$
$=\sqrt{\cos ^{2} \alpha\left(\cos ^{2} \beta+\sin ^{2} \beta\right)+\sin ^{2} \alpha} \quad\left[\because \cos ^{2} x+\sin ^{2} x=1\right]$
$\begin{aligned} &=\sqrt{\cos ^{2} \alpha+\sin ^{2} \alpha} \\ &=\sqrt{1} \\ &=1 \end{aligned}$
Hence, it is unit vector

Scaler and dot product exercise multiple choice question 6

Answer:
Option (c) $-\frac{5}{\sqrt{162}}$
Hint: You must know the formula of $\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$
Given:
$\vec{P}=\hat{i}+3 \hat{j}-7 \hat{k} \text { and } \vec{Q}=5 \hat{i}-2 \hat{j}+4 \hat{k}$
Solution:
Let $O$ be the origin
$\begin{aligned} &\overrightarrow{P Q}=\overrightarrow{O Q}-\overrightarrow{O P} \\\\ &\overrightarrow{P Q}=5 \hat{i}-2 \hat{j}+4 \hat{k}-(\hat{i}+3 \hat{j}-7 \hat{k}) \\\\ &\overrightarrow{P Q}=4 \hat{i}-5 \hat{j}+11 \hat{k} \end{aligned}$
We know that unit vector along $y-\text { axis is } \hat{j}$
Therefore, angle between $\overrightarrow{P Q} \text { and } y-\text { axis }$
$\cos \theta=\frac{(\hat{4}-5 \hat{j}+11 \hat{k}) \hat{j}}{\sqrt{4^{2}+(-5)^{2}+11^{2}}}$
$\begin{aligned} &\cos \theta=\frac{0-5+0}{\sqrt{16+25+121}} \quad[\because \hat{i} \cdot \hat{j}=0, \hat{k} \cdot \hat{j}=0, \hat{j} \cdot \hat{j}=1] \\\\ &\cos \theta=\frac{-5}{\sqrt{162}} \end{aligned}$
Hence, option (c) is correct

Scaler and dot product exercise multiple choice question 7

Answer: Option (a) $\sqrt{3}$
Hint: Use the formula of $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$
Given:
$\vec{a} \text { and } \vec{b}$ are unit vector
Solution:
$\vec{a} \text { and } \vec{b}$ are unit vector
$\therefore|\vec{a}|=1 \text { and }|\vec{b}|=1$
We know that,
$\begin{aligned} &-1<\cos \theta<1 \\\\ &-1<\vec{a} \cdot \vec{b}<1 \end{aligned} \quad[\because \cos \theta=\vec{a} \cdot \vec{b}]$
So $\vec{a} \cdot \vec{b}$ cannot be greater than one.
Hence, option (a) is correct

Scaler and dot product exercise multiple choice question 8

Answer:
Option (b) an ellipse
Hint:
You must know the value of $\vec{a} \cdot \vec{b}=0$ , when $\vec{a} \text { and } \vec{b}$are perpendicular
Given:
$\hat{i}-2 x \hat{j}+3 y \hat{k} \text { and } \hat{i}+2 x \hat{j}-3 y \hat{k}$ are perpendicular, locus $(x, y)=?$
Solution:
$a=\hat{i}-2 x \hat{j}+3 y \hat{k} \text { and } b=\hat{i}+2 x \hat{j}-3 y \hat{k}$
If vectors are perpendicular, then $\vec{a} \cdot \vec{b}=0$
$\begin{aligned} &(\hat{i}-2 x \hat{j}+3 y \hat{k}) \cdot(\hat{i}+2 x \hat{j}-3 y \hat{k})=0 \\\\ &1-4 x^{2}-9 y^{2}=0 \\\\ &4 x^{2}+9 y^{2}=1 \end{aligned}$
This is the equation of ellipse
Hence, option (b) is correct

Scaler and dot product exercise multiple choice question 9

Answer:
Option (b) $=\frac{\vec{a} \times(\vec{b} \times \vec{a})}{|\vec{a}|^{2}}$
Hint:
You must know the formula of $\vec{b}$ perpendicular to $\vec{a}$
Given:
Vector component of $\vec{b}$ perpendicular to $\vec{a}=?$
Solution:
Vector $\vec{b}$ is perpendicular to $\vec{a}$
$=\frac{\vec{a} \times(\vec{b} \times \vec{a})}{|\vec{a}|^{2}}$
Hence, option (b) is correct

Scaler and dot product exercise multiple choice question 10

Answer:
Option (c) $\sqrt{593}$
Hint:
Use the formula $|\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}$
Given:
$|\vec{a}|=2 \sqrt{2},|\vec{b}|=3$ and angle between $\vec{a} \text { and } \vec{b} \text { is } \frac{\pi}{4}$
$\overrightarrow{A D}=\overrightarrow{B C}=\vec{a}-3 \vec{b} \text { and } \overrightarrow{A B}=\overrightarrow{D C}=5 \vec{a}+2 \vec{b}$
Solution:
Here $|\vec{a}|=2 \sqrt{2},|\vec{b}|=3$
First we have to find both diagonals
Length of the two diagonal will be
$\begin{aligned} &d_{1}=|(5 \vec{a}+2 \vec{b})+(\vec{a}-3 \vec{b})| \\\\ &d_{2}=|(5 \vec{a}+2 \vec{b})-(\vec{a}-3 \vec{b})| \\\\ &\Rightarrow d_{1}=|6 \vec{a}-\vec{b}| \text { and } d_{2}=|4 \vec{a}+5 \vec{b}| \end{aligned}$
$\begin{aligned} &d_{1}=\sqrt{36 a^{2}+b^{2}-2 \times 6 \times|\vec{a}| \cdot|\vec{b}| \cdot \cos \frac{\pi}{4}} \\\\ &=\sqrt{36 \times 8+9-12 \times 2 \sqrt{2} \times 3 \times \frac{1}{\sqrt{2}}} \end{aligned}$
$\begin{aligned} &=\sqrt{238+9-72} \\\\ &=\sqrt{225}=15 \end{aligned}$
And again
$\begin{aligned} &d_{2}=\sqrt{|4 \vec{a}|^{2}+|5 \vec{b}|^{2}+2 \times|4 \vec{a}| \times|5 \vec{b}| \cos \frac{\pi}{4}} \\\\ &=\sqrt{16 \times 8+25 \times 9+40 \times 2 \sqrt{2} \times 3 \times \frac{1}{\sqrt{2}}} \end{aligned}$
$\begin{aligned} &=\sqrt{128+225+240} \\\\ &=\sqrt{593}>15 \\\\ &\therefore d_{2}>d_{1} \end{aligned}$
Hence length of the longest diagonal $=\sqrt{593}$

Scaler and dot product exercise multiple choice question 11

Answer:
Option (d) $a=\frac{1}{|\lambda|}$
Hint:
Use the concept of unit vector, $\lambda \vec{a}=1$
Given:
$\vec{a}$ is a non-zero vector, $|\vec{a}|=a$ and $\lambda$ is a non-zero scalar and $\lambda \vec{a}$ is a unit vector
Solution:
$\lambda \vec{a}$ is a unit vector
$\begin{aligned} &\lambda \vec{a}=1 \\\\ &|\lambda||\vec{a}|=1 \\\\ &|\vec{a}|=\frac{1}{|\lambda|}[\because|\vec{a}|=a] \\\\ &a=\frac{1}{|\lambda|} \end{aligned}$
Hence, option (d) is correct

Scaler and dot product exercise multiple choice question 12

Answer:
Option (b) $0 \leq \theta \leq \frac{\pi}{2}$
Hint:
You must know the formula of $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$ and range of $\theta \text { in } \cos \theta$
Given:
$\theta$ is the angle between $\vec{a} \text { and } \vec{b}, \vec{a} \cdot \vec{b} \geq 0$
Solution:
$\begin{aligned} &\vec{a} \cdot \vec{b} \geq 0 \\\\ &|\vec{a}||\vec{b}| \cos \theta \geq 0 \\\\ &\cos \theta \geq 0 \quad[\because \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta] \end{aligned}$
$0 \leq \theta \leq \frac{\pi}{2}$
Hence, option (b) is correct

Scaler and dot product exercise multiple choice question 13

Answer:
Option (b) $0<x<\frac{1}{2}$
Hint:
Use the formula of finding the angle between two vectors $\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$
Given:
$\vec{a}=2 x^{2} \hat{i}+4 x \hat{j}+\hat{k} \text { and } \vec{b}=7 \hat{i}-2 \hat{j}+x \hat{k}$ angle between them is obtuse and angle between $\vec{b}$ and $z-axis$ is acute and less than $\frac{\pi }{6}$
Solution:
$\vec{a}=2 x^{2} \hat{i}+4 x \hat{j}+\hat{k}, \vec{b}=7 \hat{i}-2 \hat{j}+x \hat{k}$
Let $\cos A$ be the angle between $\vec{a} \text { and } \vec{b}$
Since, $A$ is obtuse angle
$\begin{aligned} &\cos A<0 \\\\ &\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}<0 \mid \quad[\because \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta] \end{aligned}$
$\frac{\left(2 x^{2} \hat{i}+4 x \hat{j}+\hat{k}\right)(7 \hat{i}-2 \hat{j}+x \hat{k})}{\sqrt{\left(\left(2 x^{2}\right)^{2}+(4 x)^{2}+1^{2}\right)\left((7)^{2}+(-2)^{2}+x^{2}\right)}}<0$
$\begin{aligned} &\frac{14 x^{2}-8 x+x}{\sqrt{\left(4 x^{4}+16 x^{2}+1\right)\left(49+4+x^{2}\right)}}<0 \\\\ &14 x^{2}-7 x<0 \\\\ &7 x(2 x-1)<0 \\\\ &x<0 \; \; \; \; x<\frac{1}{2} \end{aligned}$
Hence, option (b) is correct

Scaler and dot product exercise multiple choice question 14

Answer:
Option (c) $\sqrt{3} a$
Hint:
Use the formula of$f(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a \cdot b+2 b . c+2 c . a$
Given:
$\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular and $|\vec{a}|=|\vec{b}|=|\vec{c}|=a$
Solution:
$|\vec{a}|=|\vec{b}|=|\vec{c}|=a \text { and } \vec{a}, \vec{b}, \vec{c}$are mutually perpendicular.
$\therefore \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}=0$
Now,
$\begin{aligned} &(\vec{a}+\vec{b}+\vec{c})^{2}=(\vec{a})^{2}+(\vec{b})^{2}+(\vec{c})^{2}+2 \vec{a} \cdot \vec{b}+2 \vec{b} \cdot \vec{c}+2 \vec{c} \cdot \vec{a} \\\\ &(\vec{a}+\vec{b}+\vec{c})^{2}=a^{2}+a^{2}+a^{2} \end{aligned}$
$\begin{aligned} &(\vec{a}+\vec{b}+\vec{c})^{2}=3 a^{2} \\\\ &|\vec{a}+\vec{b}+\vec{c}|=\sqrt{3} a \end{aligned}$
Hence, option (c) is correct

Scaler and dot product exercise multiple choice question 15

Answer:
Option (c) 14
Hint:
If $\vec{a} \text { and } \vec{b}$ are perpendicular, then $\vec{a} \cdot \vec{b}=0$
Given:
$3 \hat{i}+\lambda \hat{j}+\hat{k} \text { and } 2 \hat{i}-\hat{j}+8 \hat{k}$ are perpendicular.
Solution:
Let $\vec{a}=3 \hat{i}+\lambda \hat{j}+\hat{k} \text { and } \vec{b}=2 \hat{i}-\hat{j}+8 \hat{k}$
If $\vec{a} \text { and } \vec{b}$ are perpendicular, then $\vec{a} \cdot \vec{b}=0$
$\begin{aligned} &(3 \hat{i}+\lambda \hat{j}+\hat{k}) \cdot(2 \hat{i}-\hat{j}+8 \hat{k})=0 \\\\ &6-\lambda+8=0 \\\\ &-\lambda+14=0 \\\\ &-\lambda=-14 \\\\ &\lambda=14 \end{aligned}$
Hence option (c) is correct.

Scaler and dot product exercise multiple choice question 16

Answer:
Option (a) 1
Hint:
The projection of $\vec{p} \text { and } \vec{q} \text { is } \frac{\bar{p} \cdot \vec{q}}{|\vec{q}|}$
Given:
Projection of the vector $\hat{i}+\hat{j}+\hat{k}$ along the vector $\vec{j}$
Solution:
$\text { Let } \vec{p}=\hat{i}+\hat{j}+\hat{k} \text { and } \vec{q}=\hat{j}$
The projection of $\vec{p} \text { and } \vec{q} \text { is } \frac{\bar{p} \cdot \vec{q}}{|\vec{q}|}$
$\begin{aligned} &=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot \hat{j}}{|\hat{j}|} \\\\ &=\frac{0+1+0}{1} \\\\ &=1 \end{aligned}$
Hence, option (a) is correct.

Scaler and dot product exercise multiple choice question 17

Answer:
Option (b) $a=4, b=4, c=5$
Hint:
Use the formula of $\vec{a} \text { and } \vec{b}$ , when $\vec{a} \text { and } \vec{b}$ are perpendicular $\vec{a} \cdot \vec{b}=0$
Given:
$2 \hat{i}+3 \hat{j}-4 \hat{k} \text { and } \hat{a} \hat{i}+b \hat{j}+c \hat{k}$ are perpendicular
Solution:
Given that,
$2 \hat{i}+3 \hat{j}-4 \hat{k} \text { and } \hat{a} \hat{i}+b \hat{j}+c \hat{k}$ are perpendicular
$\begin{aligned} \therefore &(2 \hat{i}+3 \hat{j}-4 \hat{k}) \cdot(\hat{a} \hat{i}+b \hat{j}+c \hat{k})=0 \\\\ & 2 a+3 b-4 c=0 \end{aligned}$
For option (a),
$\begin{aligned} &a=2, b=3, c=-4 \\\\ &2(2)+3(3)-4(-4) \\\\ &=4+9+16=29 \\\\ &29 \neq 0 \end{aligned}$
Hence, option (a) is not correct
For option (b),
$\begin{aligned} &a=4, b=4, c=5 \\\\ &2(4)+3(4)-4(5) \\\\ &=8+12-20 \\\\ &0=0 \end{aligned}$
Hence, option (b) is correct.

Scaler and dot product exercise multiple choice question 18

Answer:
Option (c) 0
Hint:
Use the identity $\left(a^{2}-b^{2}\right)=(a-b)(a+b)$
Given:
$|\vec{a}|=|\vec{b}|,(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=?$
Solution:
$\begin{aligned} &(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=|\vec{a}|^{2}-|\vec{b}|^{2} \quad\left[\because(a+b)(a-b)=a^{2}-b^{2}\right] \\\\ &(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=0 &{[\because|\vec{a}|=|\vec{b}|]} \end{aligned}$
Hence, option (c) is correct.

Scaler and dot product exercise multiple choice question 19

Answer:
Option (a) $2 \sin \frac{\theta}{2}$
Hint:
Use the identity $(a-b)^{2}=a^{2}-2 a b+b^{2}$
Given:
$\vec{a} \text { and } \vec{b}$ are unit vectors inclined at an angle $\theta$
Solution:
$\begin{aligned} &|\vec{a}|=|\vec{b}|=1 \text { [unit vectors }] \\\\ &\vec{a} \cdot \vec{b}=|\vec{a}| \vec{b} \mid \cos \theta \\\\ &\vec{a} \cdot \vec{b}=\cos \theta \end{aligned}$
Now,
$\begin{aligned} &|\vec{a}-\vec{b}|^{2}=(\vec{a})^{2}+(\vec{b})^{2}-2 \vec{a} \cdot \vec{b} \\\\ &|\vec{a}-\vec{b}|^{2}=1-2 \cos \theta+1 \\\\ &|\vec{a}-\vec{b}|^{2}=2-2 \cos \theta \quad\left[\because \cos \theta=1-2 \sin ^{2} \frac{\theta}{2}\right] \end{aligned}$
$\begin{aligned} &|\vec{a}-\vec{b}|^{2}=2(1-\cos \theta) \\\ &|\vec{a}-\vec{b}|^{2}=2\left(2 \sin ^{2} \frac{\theta}{2}\right) \\\ &|\vec{a}-\vec{b}|^{2}=4 \sin ^{2} \frac{\theta}{2} \\\ &|\vec{a}-\vec{b}|=2 \sin \frac{\theta}{2} \end{aligned}$
Hence, option (a) is correct.

Scaler and dot product exercise multiple choice question 20

Answer:
Option (c) 4
Hint:
Magnitude of a vector $=\sqrt{a^{2}+b^{2}+c^{2}}$
Given:
$\vec{a} \text { and } \vec{b}$ are unit vectors and the greatest value of $\sqrt{3}|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|=?$
Solution:
$\sqrt{3}|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|=\sqrt{3} \sqrt{|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}}+\sqrt{|\vec{a}|^{2}+|\vec{b}|^{2}-2 \vec{a} \cdot \vec{b}}$

$\sqrt{3}|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|=\sqrt{3} \sqrt{1+1+2 \cos \theta}+\sqrt{1+1-2 \cos \theta} \quad[\because|\vec{a}|=|\vec{b}|$$=1]$

$=\sqrt{3} \sqrt{2+2 \cos \theta}+\sqrt{2-2 \cos \theta}$

$\begin{aligned} &=\sqrt{3} \sqrt{2(1+\cos \theta)}+\sqrt{2(1-\cos \theta)} \\\\ &=\sqrt{3} \sqrt{2\left(2 \cos ^{2} \frac{\theta}{2}\right)}+\sqrt{2\left(2 \sin ^{2} \frac{\theta}{2}\right)} \end{aligned}$
$\begin{array}{ll} =\sqrt{3}\left(2 \cos \frac{\theta}{2}\right)+2 \sin \frac{\theta}{2} & {\left[\because \cos \theta=1-\sin ^{2} \frac{\theta}{2}\right]} \\\\ =2 \times 2\left(\frac{\sqrt{3}}{2} \cos \frac{\theta}{2}+\frac{1}{2} \sin \frac{\theta}{2}\right) & {[\because \sin x \cos y+\cos x \sin y=\sin (x+y)]} \end{array}$
$\begin{aligned} &=4\left(\sin \frac{\pi}{3} \cos \frac{\theta}{2}+\cos \frac{\pi}{3} \sin \frac{\theta}{2}\right) \\\\ &=4\left[\sin \left(\frac{\pi}{3}+\frac{\theta}{2}\right)\right] \end{aligned}$
Now,
$\begin{aligned} &-1 \leq \sin \left(\frac{\pi}{3}+\frac{\theta}{2}\right) \leq 1 \\\\ &-4 \leq 4\left[\sin \left(\frac{\pi}{3}+\frac{\theta}{2}\right)\right] \leq 4 \end{aligned}$
Hence, maximum value of $\sqrt{3}|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}| \text { is } 4$
So, option (C) is correct.

Scaler and dot product exercise multiple choice question 21

Answer:
Option (c) $R-[-4, 7]$
Hint:
You must know the formula for finding angle between two vectors $\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$
Given:
$x \hat{i}+3 \hat{j}-7 \hat{k} \text { and } \hat{x} \hat{i}-x \hat{j}+4 \hat{k}$, angle between them is acute.
Solution:
$\begin{aligned} &\frac{(\hat{x} \hat{i}+3 \hat{j}-7 \hat{k})(\hat{x}-x \hat{j}+4 \hat{k})}{\sqrt{\left(x^{2}+9+49\right)\left(2 x^{2}+16\right)}}=\cos \theta \\\\ &\frac{x^{2}-3 x-28}{\sqrt{\left(x^{2}+58\right)\left(2 x^{2}+16\right)}}=\cos \theta \end{aligned}$ $\left[\because \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\cos \theta\right]$
Angle between the vectors is acute
$\begin{aligned} &\theta<\frac{\pi}{2} \Rightarrow \cos \theta>0 \\\\ &\frac{x^{2}-3 x-28}{\sqrt{\left(x^{2}+56\right)\left(2 x^{2}+16\right)}}>0 \end{aligned}$
$\begin{aligned} &x^{2}-3 x-28>0 \\ &x^{2}-7 x+4 x-28>0 \\ &x(x-7)+4(x-7)>0 \\ &(x-7)(x+4)>0 \\ &x \in R-[-4,7] \end{aligned}$
Hence, option (c) is correct.

Scaler and dot product exercise Multiple choice question 22

Answer:
Option (d) $\frac{2 \pi}{3}<\theta<\pi$
Hint:
You must know the formula of magnitude of vector $=\sqrt{a^{2}+b^{2}+c^{2}}$
Given:
$\vec{a} \text { and } \vec{b}$ are unit vectors inclined at an angle $\theta$ such that $|\vec{a}+\vec{b}|<1$
Solution:
$\begin{aligned} &|\vec{a}+\vec{b}|<1 \\\\ &|\vec{a}+\vec{b}|^{2}<1 \end{aligned}$
$\begin{aligned} &|\vec{a}|^{2}+|\vec{b}|^{2}+2|\vec{a}| \cdot|\vec{b}| \cos \theta<1 \\\\ &1+1+2 \cos \theta<1 \\\\ &2(1+\cos \theta)<1 \end{aligned}$
$\begin{aligned} &2\left(2 \cos ^{2} \frac{\theta}{2}\right)<1 \\\\ &\cos ^{2} \frac{\theta}{2}<\frac{1}{4} \end{aligned}$ $\begin{aligned} &{[\because|\vec{a}|=|\vec{b}|=1]} \\\\ &{\left[\because \cos \theta=2 \cos ^{2} \frac{\theta}{2}-1\right]} \end{aligned}$
$\left|\cos \frac{\theta}{2}\right|<\frac{1}{2}$
We know $\theta$ is always lies between $\left [ -\pi ,\pi \right ]$
$\frac{2 \pi}{3}<\theta<\pi$
Hence, option (d) is correct.

Scaler and dot product exercise multiple choice question 23

Answer:
Option (d)$-1$
Hint:
Use the identity $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a \cdot b+2 b \cdot c+2 c \cdot a$
Given:
$|\vec{a}|=|\vec{b}|=|\vec{c}|=1 \text { and }|\vec{a}+\vec{b}+\vec{c}|=1$
$\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{c}$ makes angle $\alpha$ and $\beta$ with $\vec{a}$ and $\vec{b}$ respectively
Solution:
$\begin{aligned} &|\vec{a}+\vec{b}+\vec{c}|=1 \\\\ &(\vec{a}+\vec{b}+\vec{c})^{2}=(\vec{a})^{2}+(\vec{b})^{2}+(\vec{c})^{2}+2 \vec{a} \cdot \vec{b}+2 \vec{b} \cdot \vec{c}+2 \vec{c} \cdot \vec{a} \end{aligned}$
$1=1+1+1+0+2 \cos \alpha+2 \cos \beta$ $\begin{aligned} &{[\because|\vec{a}|=|\vec{b}|=|\vec{c}|=1]} \\ &{[\because \vec{a} \cdot \vec{b}=0]} \end{aligned}$
$\begin{aligned} &2 \cos \alpha+2 \cos \beta=-2 \\ &\cos \alpha+\cos \beta=-1 \end{aligned}$
Hence, option (d) is correct.

Scaler and dot product exercise multiple choice question 24

Answer:
Option (b) $=\frac{(\vec{a} \cdot \vec{b}) \vec{b}}{|\vec{b}|^{2}}$
Hint:
You must know the concept of orthogonal projection
Given:
Orthogonal projection of $\vec{a}$ on $\vec{b}$
Solution:
Orthogonal projection of $\vec{a}$ on $\vec{b}$ $=\frac{(\vec{a} \cdot \vec{b}) \vec{b}}{|\vec{b}|^{2}}$
Hence, option (b) is correct.

Scaler and dot product exercise multiple choice question 25

Answer:
Option (d) $\frac{\pi}{3}$
Vector a and b are perpendicular if their dot product is zero.
Given:
$\theta$ is an acute angle
$\sin \hat{\theta} \hat{i}+\cos \theta \hat{j} \text { and } \hat{i}-\sqrt{3} \hat{j}$ are perpendicular
$(\sin \theta \hat{i}+\cos \theta \hat{j}) \cdot(\hat{i}-\sqrt{3} \hat{j})=0$
Both are perpendicular to each other
$\begin{aligned} &\sin \theta-\sqrt{3} \cos \theta=0 \\\\ &\sin \theta=\sqrt{3} \cos \theta \end{aligned}$
$\begin{aligned} &\frac{\sin \theta}{\cos \theta}=\sqrt{3} \\\\ &\tan \theta=\sqrt{3} \end{aligned}$ $\begin{aligned} &{\left[\because \frac{\sin \theta}{\cos \theta}=\tan \theta\right]} \\ &{\left[\because \tan \frac{\pi}{3}=\sqrt{3}\right]} \end{aligned}$
$\theta=\frac{\pi}{3}$

Scaler and dot product exercise multiple choice question 26

Answer:
Option (b) $\frac{\pi }{3}$
Hint:
You must know the formula to find angle between two vectors $\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$
Given:
$|\vec{a}|=\sqrt{3},|\vec{b}|=4 \text { and } \vec{a} \cdot \vec{b}=2 \sqrt{3}$
Solution:
$\begin{aligned} &\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}=\cos \theta \\\\ &\frac{2 \sqrt{3}}{\sqrt{3} \cdot 4}=\cos \theta \end{aligned}$
$\cos \theta=\frac{1}{2} \quad\left[\because \cos \frac{\pi}{3}=\frac{1}{2}\right]$
$\begin{aligned} &\cos \theta=\cos \frac{\pi}{3} \\\\ &\theta=\frac{\pi}{3} \end{aligned}$
Hence, option (b) is correct.

Scaler and dot product exercise multiple choice question 27

Answer:
Option (c) $-\frac{3}{2}$
Hint:
You must know the formula of $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a \cdot b+2 b . c+2 c . a$
Given:
$|\vec{a}|=|\vec{b}|=|\vec{c}|=1 \text { and } \vec{a}+\vec{b}+\vec{c}=0$
Solution:
$\vec{a}+\vec{b}+\vec{c}=0$
We know that,
$(\vec{a}+\vec{b}+\vec{c})^{2}=(\vec{a})^{2}+(\vec{b})^{2}+(\vec{c})^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$
$0=1+1+1+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$ $\begin{aligned} &{[\because|\vec{a}|=|\vec{b}|=|\vec{c}|=1]} \\ &{[\because \vec{a}+\vec{b}+\vec{c}=0]} \end{aligned}$
$\frac{-3}{2}=\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$
Hence, option (c) is correct

Scaler and dot product exercise multiple choice question 28

Answer:
Option (c) $-19$
Hint:
Apply the identity$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a \cdot b+2 b . c+2 c . a$
Given:
$|\vec{a}|=2,|\vec{b}|=3,|\vec{c}|=5 \text { and } \vec{a}+\vec{b}+\vec{c}=0$
Solution:
$\vec{a}+\vec{b}+\vec{c}=0$
We know that,
$|\vec{a}+\vec{b}+\vec{c}|^{2}=(\vec{a})^{2}+(\vec{b})^{2}+(\vec{c})^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$
$\left[\because(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a)\right]$
$0=(2)^{2}+(3)^{2}+(5)^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \quad\left[\begin{array}{l} \because \vec{a}+\vec{b}+\vec{c}=0 \\ \because|\vec{a}|=2,|\vec{b}|=3,|\vec{c}|=5 \end{array}\right]$
$\begin{aligned} &0=4+9+25+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \\\\ &\frac{-38}{2}=\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a} \\\\ &\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=-19 \end{aligned}$

Scaler and dot product exercise multiple choice question 29

Answer:
Option (a) $\frac{\pi }{6}$
Hint:
You must know the identity $(a-b)^{2}=a^{2}+b^{2}-2 a b$
Given:
$|\vec{a}|=|\vec{b}|=1 \text { and } \sqrt{3} \vec{a}-\vec{b}$ to be a unit vector
Solution:
$\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}=\cos \theta$ $[\because|\vec{a}|=|\vec{b}|=1]$
$\vec{a} \cdot \vec{b}=\cos \theta$ According to the given question,
$\begin{aligned} &|\sqrt{3} \vec{a}-\vec{b}|=1 \Rightarrow|\sqrt{3} \vec{a}-\vec{b}|^{2}=1 \\\\ &3|\vec{a}|^{2}+|\vec{b}|^{2}-2 \sqrt{3}(\vec{a} \cdot \vec{b})=1 \\\\ &3+1-2 \sqrt{3} \cos \theta=1 \end{aligned}$
$\begin{aligned} &-2 \sqrt{3} \cos \theta=-3 \\\\ &2 \sqrt{3} \cos \theta=3 \\\\ &\cos \theta=\frac{3}{2 \sqrt{3}} \end{aligned}$ $\left[\because \frac{\sqrt{3}}{2}=\cos \frac{\pi}{6}\right]$
$\begin{aligned} &\cos \theta=\frac{\sqrt{3}}{2} \\\\ &\theta=\frac{\pi}{6} \end{aligned}$

Scaler and dot product exercise multiple choice question 30

Answer:
Option (c) $\frac{-2}{3}$
Hint:
Use the formula of projection $\vec{a} \text { on } \vec{b}$
Given:
Projection of $\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k} \text { on } \vec{b}=2 \hat{i}+\lambda \hat{k}$ is zero
Solution:
$\begin{aligned} &\vec{a} \cdot \vec{b}=0 \\\\ &(\hat{i}-2 \hat{j}+3 \hat{k}) \cdot(2 \hat{i}+\lambda \hat{k})=0 \\\\ &2-0+3 \lambda=0 \end{aligned}$
$\begin{aligned} &3 \lambda=-2 \\\\ &\lambda=\frac{-2}{3} \end{aligned}$
Hence, option (c) is correct

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