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

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

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

Lovekush kumar sainiUpdated on 27 Jan 2022, 05:03 PM IST

The practice sessions for the class 12 students go unending until their exams. Therefore, the students must make sure that they are using a proper solution book for their reference. The solutions for mathematical sums can be derived using various methods; it is in the hands of the teachers to make the students aware of the right method. When students struggle to solve the sums in the Scalar and Dot Product portion, they can use the RD Sharma Class 12th Chapter 23 MCQ book for clarification.

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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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