RD Sharma Class 12 Exercise MCQ Vector or cross product Solutions Maths-Download PDF Online

Edited By Lovekush kumar saini | Updated on Jan 27, 2022 10:31 AM IST

The public exam preparations do not stop till the last day of school for the class 12 students. Every student in that category spends most of their time preparing for the board and JEE mains exam. It is equally important to choose the right solution book that is used for reference. Subjects like mathematics with the vector or cross-product concept are very much challenging for the 12th graders. Here is where the presence of the RD Sharma Class 12th Chapter 24 MCQ is inevitable.

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RD Sharma Class 12 Solutions Chapter 24MCQ Vector or Cross Product - Other Exercise
Vector or Cross Product Excercise: MCQ
RD Sharma Chapter-wise Solutions

RD Sharma Class 12 Solutions Chapter 24MCQ Vector or Cross Product - Other Exercise

Vector or Cross Product Excercise: MCQ

Vector or cross product exercise multiple choice questions question 1

Answer: $2\overrightarrow{a}^{2}$
Given: If $\overrightarrow{a}$ is any vector, then $(\vec{a} \times \hat{i})^{2}+(\vec{a} \times \hat{j})^{2}+(\vec{a} \times \hat{k})^{2}$
Hint: You must know about the vector products of orthonormal triad of unit vectors.
Explanation: Let $\overrightarrow{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}$
Now,
$\begin{aligned} &\vec{a} \times \hat{i}=\left(a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}\right) \times \hat{i} \\ &=a_{1}(\hat{i} \times \hat{i})+a_{2}(\hat{j} \times \hat{i})+a_{3}(\hat{k} \times \hat{i}) \quad[\therefore \hat{i} \times \hat{i}=0 ; \hat{j} \times \hat{i}=-\hat{k}, \hat{k} \times \hat{i}=\hat{j}] \end{aligned}$
$\begin{aligned} &=a_{1}(0)+a_{2}(-\hat{k})+a_{3}(\hat{j}) \\ &=0-a_{2} \hat{k}+a_{3} \hat{j} \\ &=a_{3} \hat{j}-a_{2} \hat{k} \end{aligned}$
On squaring,
$\begin{aligned} &\begin{aligned} &\left(a_{3} \hat{j}-a_{2} \hat{k}\right)^{2}=\left(a_{3} \hat{j}-a_{2} \hat{k}\right)\left(a_{3} \hat{j}-a_{2} \hat{k}\right) \\ &=\left(a_{3} \hat{j}\right)^{2}+\left(a_{2} \hat{k}\right)^{2}-2 a_{3} \hat{j} \cdot a_{2} \hat{k} \end{aligned} \quad\left[\begin{array}{l} \because \hat{j} \cdot \hat{j}=1 \\ \hat{k} \cdot \hat{k}=1 \\ \hat{j} \cdot \hat{k}=0 \end{array}\right]\\ &=a_{3}^{2}(1)+a_{2}^{2}(1)-2 a_{2} a_{3}(0)\\ &\left(a_{3} \hat{j}-a_{2} \hat{k}\right)^{2}=a_{2}^{2}+a_{3}^{2} \end{aligned}$
$\begin{gathered} =a_{3}^{2}(1)+a_{2}^{2}(1)-2 a_{2} a_{3}(0) \\ \left(a_{3} \hat{j}-a_{2} \hat{k}\right)^{2}=a_{2}^{2}+a_{3}^{2} \end{gathered}$ …………… (1)
Similarly,
$\begin{aligned} &\vec{a} \times \hat{j}=\left(a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}\right) \times \hat{j} \\ &=a_{1}(\hat{i} \times \hat{j})+a_{2}(\hat{j} \times \hat{j})+a_{3}(\hat{k} \times \hat{j}) \quad\left[\begin{array}{l} \because \hat{i} \times \hat{j}=\hat{k}, \\ \hat{j} \times \hat{j}=0, \\ \hat{k} \times \hat{j}=-\hat{i} \end{array}\right] \\ &=a_{1}(\hat{k})+a_{2}(0)+a_{3}(-\hat{i}) \\ &=a_{1} \hat{k}-a_{3} \hat{i} \end{aligned}$ ………………… (2)
Again
$\begin{aligned} &\vec{a} \times \hat{k}=\left(a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}\right) \times \hat{k} \\ &=a_{1}(\hat{i} \times \hat{k})+a_{2}(\hat{j} \times \hat{k})+a_{3}(\hat{k} \times \hat{k}) \quad\left[\begin{array}{l} \hat{i} \times \hat{k}=-\hat{j} \\ \hat{j} \times \hat{k}=\hat{i} \\ \hat{k} \times \hat{k}=0 \end{array}\right] \\ &=a_{1}(-\hat{j})+a_{2}(\hat{i})+a_{3}(0) \end{aligned}$
$\begin{aligned} &=a_{1}\hat{j}+a_{2}\hat{i} \end{aligned}$
On squaring
$\begin{aligned} &\left(a_{2} \hat{i}-a_{1} \hat{j}\right)^{2}=\left(a_{2} \hat{i}-a_{1} \hat{j}\right)\left(a_{2} \hat{i}-a_{1} \hat{j}\right)\\ &=\left(a_{2} \hat{i}\right)^{2}+\left(a_{1} \hat{j}\right)^{2}-2 a_{1} a_{2}(\hat{i} . \hat{j})\left[\begin{array}{l} \because \hat{i} . \hat{i}=1 \\ \hat{j} \cdot \hat{j}=1 \\ \hat{i} \cdot \hat{j}=0 \end{array}\right]\\ &=a_{2}^{2}(1)+a_{1}^{2}(1)-2 a_{1} a_{2}(0)\\ &=a_{2}^{2}+a_{1}^{2} \end{aligned}$ ……………… (3)
Now,
$\begin{aligned} (\vec{a} \times \hat{i})^{2}+&(\vec{a} \times \hat{j})^{2}+(\vec{a} \times \hat{k})^{2} \\ &=a_{2}^{2}+a_{3}^{2}+a_{1}^{2}+a_{3}^{2}+a_{2}^{2}+a_{1}^{2}[\because \operatorname{from}(1),(2),(3)] \\ &=2\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)=2 \vec{a}^{2} \end{aligned}$

Vector or cross product exercises multiple choice questions question 2

Answer: $\overrightarrow{b}=\overrightarrow{c}$

Given: $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{a}.\overrightarrow{c}$ and $\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{a}\times \overrightarrow{c},\overrightarrow{a}\neq 0$

Hint: You must know about dot product and cross product.

Explanation:

$\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{a}.\overrightarrow{c}$ ………………. (1)

$\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{a}\times \overrightarrow{c}$ ………………. (2)

Squaring and adding (1) and (2)

$\begin{gathered} (\vec{a} \cdot \vec{b})^{2}+|\vec{a} \times \vec{b}|^{2}=(\vec{a} \cdot \vec{c})^{2}+|\vec{a} \times \vec{c}|^{2} \\ \end{gathered}$

$\Rightarrow|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta+|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta=|\vec{a}|^{2}|\vec{c}|^{2} \cos ^{2} \theta+|\vec{a}|^{2}|\vec{c}|^{2} \sin ^{2} \theta \quad\left[\begin{array}{l} \because \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \\ \vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \end{array}\right]$

$\begin{aligned} &\Rightarrow|\vec{a}|^{2}|\vec{b}|^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=|\vec{a}|^{2}|\vec{c}|^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right) \\ &\Rightarrow|\vec{a}|^{2}|\vec{b}|^{2}(1)=|\vec{a}|^{2}|\vec{c}|^{2}(1) \end{aligned} \quad\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$

$\begin{aligned} &\Rightarrow|\vec{a}|^{2}|\vec{b}|^{2}=|\vec{a}|^{2}|\vec{c}|^{2} \\ &\Rightarrow|\vec{b}|^{2}=|\vec{c}|^{2} \\ &\Rightarrow|\vec{b}|=|\vec{c}| \\ &\Rightarrow \vec{b}=\vec{c} \end{aligned}$

Vector or cross product exercises multiple choice questions question 3

Answer: $\frac{3}{2}\left ( \hat{i}+\hat{j} \right )$

Given: The vector $\overrightarrow{b}=3\hat{i}+4\hat{k}$ is to be written as the sum of a vector $\overrightarrow{a}$ parallel to $\overrightarrow{a}=\hat{i}+\hat{j}$ and $\overrightarrow{\beta }$

perpendicular to $\overrightarrow{a}$ . Then $\overrightarrow{a}=$

Hint: If $\overrightarrow{a }\parallel \overrightarrow{a }\Rightarrow \overrightarrow{a }$ and if $\overrightarrow{a }\perp \overrightarrow{a }\Rightarrow \overrightarrow{a }.\overrightarrow{a }=0$

Explanation:

Here $\overrightarrow{a}=\hat{i}+\hat{j}$ , $\overrightarrow{b}=3\hat{i}+4\hat{k}$

Also $\overrightarrow{b}=\overrightarrow{a}+\overrightarrow\beta$ ……………. (1) where $\overrightarrow{a}\parallel \overrightarrow{a}$ and $\overrightarrow{\beta }\perp \overrightarrow{a}$

As $\vec{\alpha} \| \vec{a} \Rightarrow \vec{\alpha}=\lambda \vec{a} \Rightarrow \vec{\alpha}=\lambda(\hat{i}+\hat{j}) \Rightarrow \vec{\alpha}=\lambda \hat{i}+\lambda \hat{j} \quad[\because \vec{a} \| \vec{b} \Rightarrow \vec{a}=k \vec{b}]$

By (1)

$\begin{aligned} &\vec{\beta}=\vec{b}-\vec{\alpha} \\ &\Rightarrow \vec{\beta}=3 \hat{i}+4 \hat{k}-\lambda \hat{i}-\lambda \hat{j} \\ &\Rightarrow \vec{\beta}=(3-\lambda) \hat{i}-\lambda \hat{j}+4 \hat{k} \end{aligned}$

Now $\vec{\beta}.\vec{a}=0$ ………….. (2) $\because \vec{\beta}\perp \vec{a}$

$\begin{aligned} &\Rightarrow \vec{\beta} \cdot \vec{a}=(3-\lambda)(1)-\lambda(1)+4(0) \\ &=3-\lambda-\lambda \\ &=3-2 \lambda \end{aligned}$

$\begin{aligned} &\because \text { By (2) } \quad 3-2 \lambda=0 \Rightarrow 2 \lambda=3 \Rightarrow \lambda=\frac{3}{2} \\ &\Rightarrow \vec{\alpha}=\frac{3}{2} \vec{a}=\frac{3}{2}(\hat{i}+\hat{j}) \\ &\Rightarrow \vec{\alpha}=\frac{3}{2}(\hat{i}+\hat{j}) \end{aligned}$

Vector or cross product exercises multiple choice questions question 4

Answer: $\frac{(2 \hat{i}+\hat{j}+\hat{k})}{\sqrt{6}}$

Given: A unit vector is perpendicular to the plane passing through points

$P(\hat{i}-\hat{j}+2 \hat{k}), Q(2 \hat{i}-\hat{k}), R(2 \hat{j}+\hat{k})$

Hint: You must know how to find position vectors and magnitude of vectors

Explanation: Here

$\begin{gathered} \vec{p}=\hat{i}-\hat{j}+2 \hat{k} \\ \vec{q}=2 \hat{i}-\hat{k} \\ \vec{r}=2 \hat{j}+\hat{k} \\ \overrightarrow{P Q}=(P . V \text { of } Q)-(P . V \text { of } P) \end{gathered}$

$\begin{aligned} &=\vec{q}-\vec{p} \\ &=2 \hat{i}-\hat{k}-\hat{i}+\hat{j}-2 \hat{k} \\ &=\hat{i}+\hat{j}-3 \hat{k} \\ &\overline{Q R}=(\mathrm{P} . \mathrm{V} \text { of } \mathrm{R})-(\mathrm{P} \cdot \mathrm{V} \text { of } \mathrm{Q}) \\ &=2 \hat{j}+\hat{k}-2 \hat{j}+\hat{k} \\ &=-2 \hat{i}+2 \hat{j}+2 \hat{k} \\ &\frac{ }{P R}=(\mathrm{P} . \mathrm{V} \text { of } \mathrm{R})-(\mathrm{P} . \mathrm{V} \text { of } \mathrm{P}) \end{aligned}$

$\begin{aligned} &=2 \hat{j}+\hat{k}-\hat{i}+\hat{j}-2 \hat{k} \\ &=-\hat{i}+3 \hat{j}-\hat{k} \end{aligned}$

? Normal vector $(\vec{n})=\overrightarrow{P Q} \times \overrightarrow{P R}$

$=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -3 \\ -1 & 3 & -1 \end{array}\right| \\$

$=\hat{i}(-1+9)-\hat{j}(-1-3)+\hat{k}(3+1) \\$

$\vec{n}=8 \hat{i}+4 \hat{j}+4 \hat{k}$

$\vec{n}|=\sqrt{(8)^{2}+(4)^{2}+(4)^{2}}=\sqrt{64+16+16}=\sqrt{96}$

$\Rightarrow \hat{n}=\frac{\vec{n}}{|\vec{n}|}=\frac{1}{\sqrt{96}}(8 \hat{i}+4 \hat{j}+4 \hat{k})$

$=\frac{1}{4 \sqrt{6}} \times 4(2 \hat{i}+\hat{j}+\hat{k})$

$=\frac{(2 \hat{i}+\hat{j}+\hat{k})}{\sqrt{6}}$

Vector or cross product exercises multiple choice questions question 5

Answer: $\overrightarrow{a}.\overrightarrow{b}=0$
Given: $\overrightarrow{a},\overrightarrow{b}$ represents the diagonals of a rhombus.
Hint: You must know about the properties of rhombus
Explanation: By properties of rhombus
The diagonals of rhombus bisect each other
$\Rightarrow \overrightarrow{a}\perp \overrightarrow{b}$
And we know If the two vectors are perpendicular then their dot product is zero.
Here $\theta =90^{\circ}$
$\begin{aligned} &\because \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \\ &=|\vec{a}||\vec{b}| \cos 90^{\circ} \\ &=|\vec{a}| \vec{b} \mid \cdot 0 \\ &\vec{a} \vec{b}=0 \end{aligned}$

Vector or cross product exercises multiple choice questions question 6

Answer: 300
Given: Vector $\overrightarrow{a}$ & $\overrightarrow{b}$ are inclined at angle $\theta =120^{\circ}$ . If $\left | \overrightarrow{a} \right |=1 , \left | \overrightarrow{b} \right |=2$ then $\left | \left ( \overrightarrow{a}+3\overrightarrow{b} \right )\times \left ( 3\overrightarrow{a}-\overrightarrow{b} \right ) \right |$ is equal to
Hint: You must know about cross product.

Explanation:
$\begin{aligned} |\vec{a}|=1,|\vec{b}|=2, \theta=120^{\circ} \\ \end{aligned}$
$\qquad|(\vec{a}+3 \vec{b}) \times(3 \vec{a}-\vec{b})|^{2}=|3(\vec{a} \times \vec{a})-(\vec{a} \times \vec{b})+9(\vec{b} \times \vec{a})-3(\vec{b} \times \vec{b})|^{2} \\$
$=|3(0)+(\vec{b} \times \vec{a})+9(\vec{b} \times \vec{a})-3(0)|^{2} \quad[\because(\vec{a} \times \vec{a})=0 \text { and }(\vec{b} \times \vec{b})=0] \\$
$=|10(\vec{b} \times \vec{a})|^{2} \\$
$=|\vec{b} \times \vec{a}|=|\vec{b}| \vec{a} \mid \sin \theta$
Substitute,
$\begin{aligned} |\vec{a}|=1,|\vec{b}|=2, \theta=120^{\circ} \\ \end{aligned}$
${l} =|\vec{b}||\vec{a}| \sin 120^{\circ}$ $\left[\begin{array}{l} \because \sin 120^{\circ}=\sin \left(180^{\circ}-60^{\circ}\right) \\ =\sin 60^{\circ}=\frac{\sqrt{3}}{2} \end{array}\right]$
$\\ =(2)(1)\left(\frac{\sqrt{3}}{2}\right) \\ =\sqrt{3}$
$\qquad|(\vec{a}+3 \vec{b}) \times(3 \vec{a}-\vec{b})|^{2}=\left | 10\left ( \sqrt{3} \right ) \right |^{2}=300$

Vector or cross product exercises multiple choice questions question 7

Answer: $\hat{i}$
Given: $\vec{a}=\hat{i}+\hat{j}-\hat{k} ; \vec{b}=-\hat{i}+2 \hat{j}+2 \hat{k} ; \vec{c}=-\hat{i}+2 \hat{j}-\hat{k}$
Hint: You must know about normal to the two vectors
Explanation: Let
$\begin{gathered} \vec{d}=\vec{a}+\vec{b} \\ \vec{e}=\vec{b}-\vec{c} \\ \vec{d}=\vec{a}+\vec{b}=(\hat{i}+\hat{j}-\hat{k})+(-\hat{i}+2 \hat{j}+2 \hat{k}) \\ =0 \hat{i}+3 \hat{j}+\hat{k}=3 \hat{j}+\hat{k} \\ \vec{e}=\vec{b}-\vec{c}=-\hat{i}+2 \hat{j}+2 \hat{k}-(-\hat{i}+2 \hat{j}-\hat{k}) \\ =0 \hat{i}+4 \hat{j}+3 \hat{k}=4 \hat{j}+3 \hat{k} \end{gathered}$
A unit vector normal to the unit vector is $\frac{\overrightarrow{d}\times \overrightarrow{e}}{\left |\overrightarrow{d}\times \overrightarrow{e} \right |}$
Now,
$\begin{aligned} &\vec{d} \times \vec{e}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 0 & 3 & 1 \\ 0 & 4 & 3 \end{array}\right|=\hat{i}(9-4)=5 \hat{i} \\ &|\vec{d} \times \vec{e}|=\sqrt{(5)^{2}+(0)^{2}+(0)^{2}}=\sqrt{25}=5 \end{aligned}$

Unit vector = $\frac{\overrightarrow{d}\times \overrightarrow{e}}{\left |\overrightarrow{d}\times \overrightarrow{e} \right |}=\frac{5\hat{i}}{5}=\hat{i}$

Vector or cross product exercises multiple choice questions question 8

Answer $\frac{1}{\sqrt{3}}\left ( \hat{i}-\hat{j}+\hat{k} \right )$
Given: A unit vector perpendicular to both $\hat{i}+\hat{j}$ and $\hat{j}+\hat{k}$
Hint: You must know about cross product
Explanation: Let $\overrightarrow{a}=\hat{i}+\hat{j}$ ; $\overrightarrow{b}=\hat{j}+\hat{k}$
A unit vector perpendicular to both $\frac{\overrightarrow{a}\times \overrightarrow{b}}{\left | \overrightarrow{a}\times \overrightarrow{b} \right |}$
Now,
$\begin{aligned} &\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right| \\ &=\hat{i}(1-0)-\hat{j}(1-0)+\hat{k}(1-0) \\ &= & \hat{i}-\hat{j}+\hat{k} \\ &|\vec{a} \times \vec{b}|=\sqrt{(1)^{2}+(-1)^{2}+(1)^{2}}=\sqrt{3} \\ &\therefore \text { Unit vector }=\frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}} \\ & \end{aligned}$

$=\frac{1}{\sqrt{3}}(\hat{i}-\hat{j}+\hat{k})$

Vector or cross product exercises multiple choice questions question 9

Answer: $10\hat{i}+3\hat{j}+11\hat{k}$
Given: $\overrightarrow{a}=2\hat{i}-3\hat{j}-\hat{k}$ ; $\overrightarrow{b}=\hat{i}+4\hat{j}-2\hat{k}$
Hint: You must know about cross product
Explanation: $\overrightarrow{a}=2\hat{i}-3\hat{j}-\hat{k}$ ; $\overrightarrow{b}=\hat{i}+4\hat{j}-2\hat{k}$
Then,
$\begin{aligned} &\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & -1 \\ 1 & 4 & -2 \end{array}\right| \\ &=\hat{i}(6+4)-\hat{j}(-4+1)+\hat{k}(8+3) \\ &=10 \hat{i}+3 \hat{j}+11 \hat{k} \end{aligned}$

Vector or cross product exercises multiple choice questions question 10

Answer: $\hat{i}.\hat{i}=1$
Given: $\hat{i},\hat{j},\hat{k}$ are unit vectors
Hint: Check all the options
Explanation:

$\hat{i}.\hat{j}=1$ which is wrong $\because \hat{i}.\hat{j}=0$
$\hat{i}.\hat{i}=1$ which is true
$\hat{i}\times \hat{j}=1$ which is wrong $\because \hat{i}\times \hat{j}=\hat{k}$
$\hat{i}\times \left ( \hat{j}\times \hat{k} \right )=1$ which is wrong

$\because \hat{i}\times \left ( \hat{-i} \right )=0$

Vector or cross product exercises multiple choice questions question 11

Answer: $\frac{2}{\sqrt{7}}$
Given: $\theta$ is the angle between the vectors $2\hat{i}-2\hat{j}+4\hat{k}$ & $3\hat{i}+\hat{j}+2\hat{k}$ , then $\sin \theta$
Hint: Using $\sin \theta =\frac{\overrightarrow{a}\times \overrightarrow{b}}{\left | \overrightarrow{a}\times \overrightarrow{b} \right |}$
Solution:
$\vec{a}=2 \hat{i}-2 \hat{j}+4 \hat{k} ; \vec{b}=3 \hat{i}+\hat{j}+2 \hat{k} \\ \Rightarrow|\vec{a}|=\sqrt{2^{2}+(-2)^{2}+(4)^{2}}=\sqrt{4+4+16}=\sqrt{24} \\ \Rightarrow|\vec{b}|=\sqrt{(3)^{2}+(1)^{2}+(2)^{2}}=\sqrt{9+1+4}=\sqrt{14} \$
$\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 4 \\ 3 & 1 & 2 \end{array}\right| \\$
$=\hat{i}(-4-4)-\hat{j}(4-12)+\hat{k}(2+6) \\$
$=-8 \hat{i}+8 \hat{j}+8 \hat{k} \\$
$|\vec{a} \times \vec{b}|=\sqrt{(-8)^{2}+(8)^{2}+(8)^{2}}=\sqrt{64+64+64}=\sqrt{192} \\ \qquad$
$\because \sin \theta=\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}=\frac{\sqrt{192}}{\sqrt{24} \sqrt{14}}=\frac{\sqrt{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3}}{\sqrt{2 \times 2 \times 2 \times 3} \sqrt{2 \times 7}}=\frac{\sqrt{4}}{\sqrt{7}}=\frac{2}{\sqrt{7}}$
$\qquad \sin \theta=\frac{2}{\sqrt{7}}$

Vector or cross product exercises multiple choice questions 12

Answer: 20
Given: $\left | \overrightarrow{a}\times \overrightarrow{b} \right |=4\left | \overrightarrow{a}\overrightarrow{b} \right |=2$
Hint: Using
$\left | \overrightarrow{a}\times \overrightarrow{b} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\sin \theta ;$
$\left | \overrightarrow{a}. \overrightarrow{b} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\cos \theta$
Explanation:
$\left | \overrightarrow{a}\times \overrightarrow{b} \right |=4$
$\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\sin \theta=4$ ……………………. (1)
Also $\left | \overrightarrow{a}.\overrightarrow{b} \right |=2$
$\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\cos \theta=2$ ……………………. (2)
Squaring and adding (1) and (2)
$\begin{aligned} &|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta+|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta=(4)^{2}+(2)^{2} \\ &\end{aligned}$
$|\vec{a}|^{2}|\vec{b}|^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=16+4$ $\quad\left[\sin ^{2} \theta+\cos ^{2} \theta=1\right] \\$
$|\vec{a}|^{2}|\vec{b}|^{2}(1)=20 \\$
$|\vec{a}|^{2}|\vec{b}|^{2}=20$

Vector or cross product exercises multiple choice questions question 13

Answer: $\left | \overrightarrow{a} \right |^{2}\left | \overrightarrow{b} \right |^{2}-\left ( \overrightarrow{a}.\overrightarrow{b} \right )^{2}$
Given: $\left ( \overrightarrow{a}\times \overrightarrow{b} \right )^{2}$
Hint: Using $\left | \overrightarrow{a}\times \overrightarrow{b} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\sin \theta$
Explanation: We know,
$\left | \overrightarrow{a}\times \overrightarrow{b} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\sin \theta$
Squaring on both sides
$\begin{aligned} &(\vec{a} \times \vec{b})^{2}=(|\vec{a}||\vec{b}| \sin \theta)^{2}=|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta \\ &\Rightarrow(\vec{a} \times \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2}\left(1-\cos ^{2} \theta\right) \\ &\Rightarrow(\vec{a} \times \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta \end{aligned}$ ………………….. (*)
We know $\left | \overrightarrow{a}. \overrightarrow{b} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\cos \theta$
On squaring both sides
$\left ( \overrightarrow{a}.\overrightarrow{b} \right )^{2}=\left | \overrightarrow{a} \right |^{2}\left | \overrightarrow{b} \right |^{2}\cos \theta$
Put in (*)
$\begin{aligned} &\Rightarrow(\vec{a} \times \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta \\ &(\vec{a} \times \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2} \end{aligned}$

Vector or cross product exercises multiple choice questions question 14

Answer: 1

Given: $\begin{aligned} &\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j}) \\ & \end{aligned}$

Hint: You must know about the vector product of orthonormal triad of unit vectors

Explanation:

$\begin{aligned} &\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j}) \\ & \\ & \end{aligned}$

$=\hat{i} \cdot(\hat{i})+\hat{j} \cdot(-\hat{j})+\hat{k} \cdot(\hat{k})$ ${\left[\begin{array}{l} \because \hat{j} \times \hat{k}=\hat{i} \\ \hat{i} \times \hat{k}=-\hat{j} \\ \hat{i} \times \hat{j}=\hat{k} \end{array}\right]}$

$=|\hat{i}|^{2}=|\hat{j}|^{2}+|\hat{k}|^{2}$ $\left[\begin{array}{l} \because|\hat{i}|^{2}=1 \\ |\hat{j}|^{2}=1 \\ |\hat{k}|^{2}=1 \end{array}\right]$

$=1-1+1=1$

Vector or cross product exercises multiple choice questions question 15

Answer: $\frac{\pi }{4}$
Given: $\theta$ is the angle between any two vectors $\overrightarrow{a}$ & $\overrightarrow{b}$ , $\left | \overrightarrow{a}.\overrightarrow{b} \right |=\left | \overrightarrow{a}\times \overrightarrow{b} \right |$
Hint: Using $\left | \overrightarrow{a}\times \overrightarrow{b} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\sin \theta$ and $\left | \overrightarrow{a}. \overrightarrow{b} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\cos \theta$
Explanation: We are given
$\left | \overrightarrow{a}\times \overrightarrow{b} \right |=\left | \overrightarrow{a}.\overrightarrow{b} \right |$ $\left | \overrightarrow{a}\times \overrightarrow{b} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\sin \theta$ and $\left | \overrightarrow{a}. \overrightarrow{b} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\cos \theta$
$\Rightarrow \left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\sin \theta=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\cos \theta$
$\Rightarrow \sin \theta =\cos \theta$
Divide by $\cos \theta$
$\frac{\sin \theta }{\cos \theta }=\frac{\cos \theta }{\cos \theta }$
$\tan \theta =1\Rightarrow \theta =\frac{\pi }{4}$

Vector or cross product exercise multiple choice questions question 16

Answer: 16

Given: $\left | \overrightarrow{a} \right |=10$ & $\left | \overrightarrow{b} \right |=2$ and $\overrightarrow{a}.\overrightarrow{b} =12$

Hint: Using the result $\left | \overrightarrow{a}\times \overrightarrow{b} \right |^{2}=\left | \overrightarrow{a} \right |^{2}\left |\overrightarrow{b} \right |^{2}-\left | \overrightarrow{a}.\overrightarrow{b} \right |^{2}$

Explanation:

Here $\left | \overrightarrow{a} \right |=10$ & $\left | \overrightarrow{b} \right |=2$ and $\overrightarrow{a}.\overrightarrow{b} =12$

We know,

$\begin{aligned} |\vec{a} \times \vec{b}|^{2} &=|\vec{a}|^{2}|\vec{b}|^{2}-|\vec{a} \cdot \vec{b}|^{2} \\ &=(10)^{2}(2)^{2}-(12)^{2} \\ &=100 \times 4-144 \\ &=400-144 \\ &=256 \\ |\vec{a} \times \vec{b}|^{2} &=256 \\ |\vec{a} \times \vec{b}| &=\sqrt{256}=16 \\ |\vec{a} \times \vec{b}| &=16 \end{aligned}$

Vector or cross product exercise multiple choice questions question 17

Answer: two

Given: $\vec{a}=2 \hat{i}+\hat{j}+2 \hat{k} ; \vec{b}=\hat{j}+\hat{k}$

Hint: You must know about unit vector perpendicular to the vectors

Explanation: $\vec{a}=2 \hat{i}+\hat{j}+2 \hat{k} ; \vec{b}=\hat{j}+\hat{k}$

Unit vector perpendicular to both $\vec{a}$ & $\vec{b}$ can

be drawn in two direction.

i.e., upward or downward

i.e., $\vec{a}\times \vec{b}$ and $-\left ( \vec{a}\times \vec{b} \right )$

$\because$ No of vectors is two

Vector or cross product exercise multiple choice questions question 18

Answer: $12\sqrt{3}$
Given: $\left | \overrightarrow{a} \right |=8$ , $\left | \overrightarrow{b} \right |=3$ & $\left | \overrightarrow{a}\times \overrightarrow{b} \right |=12$
Hint: Use result $\quad|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-|\vec{a} \cdot \vec{b}|^{2} \\$
Explanation: Here $\left | \overrightarrow{a} \right |=8$ , $\left | \overrightarrow{b} \right |=3$ & $\left | \overrightarrow{a}\times \overrightarrow{b} \right |=12$
We know,
$\begin{aligned} \quad|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-|\vec{a} \cdot \vec{b}|^{2} \end{aligned}$
$\begin{aligned} \\ \Rightarrow(12)^{2}=(8)^{2}(3)^{2}-(\vec{a} \cdot \vec{b})^{2}\end{aligned}$
$\begin{aligned} \\ \Rightarrow 144=64 \times 9-(\vec{a} \cdot \vec{b})^{2} \\ \end{aligned}$
$\begin{aligned} \Rightarrow(\vec{a} \cdot \vec{b})^{2}=576-144=432 \\ \\ \end{aligned}$
$\begin{aligned}\Rightarrow(\vec{a} \cdot \vec{b})^{2}=432 \\ \\ \end{aligned}$
$\Rightarrow(\vec{a} \cdot \vec{b})=\sqrt{432}=12 \sqrt{3}$

Vector or cross product exercise multiple choice questions question 19

Answer: $\frac{1}{2}\sqrt{229}$
Given: The vectors from origin O to the points A and B are $\overrightarrow{a}=2\hat{i}-3\hat{j}+2\hat{k};\overrightarrow{b}=2\hat{i}+3\hat{j}+\hat{k}$ respectively.
Hint: Area of
Explanation: Here $\overrightarrow{a}=2\hat{i}-3\hat{j}+2\hat{k};\overrightarrow{b}=2\hat{i}+3\hat{j}+\hat{k}$
$\overrightarrow{O A}=( P. V of A- P.V of O)$
$=(2 \hat{i}-3 \hat{j}+2 \hat{k})-(0 \hat{i}+0 \hat{j}+0 \hat{k})$
$=2 \hat{i}-3 \hat{j}+2 \hat{k}$
$\overrightarrow{A B}=(P . V \text { of } B-P . V \text { of } A) \\$
$=(2 \hat{i}+3 \hat{j}+\hat{k})-(2 \hat{i}-3 \hat{j}+2 \hat{k}) \\$
$=0 \hat{i}+6 \hat{j}-\hat{k}$
$Area of \triangle O A B=\frac{1}{2}|\overrightarrow{O A} \times \overrightarrow{A B}|$
$\begin{aligned} &\text { Now, }(\overrightarrow{O A} \times \overrightarrow{A B})=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -2 \\ 0 & 6 & -1 \end{array}\right| \\ \end{aligned}$
$= \hat{i}(3-12)-\hat{j}(-2-0)+\hat{k}(12+0) \\$
$=-9 \hat{i}+2 \hat{j}+12 \hat{k} \\$
$\Rightarrow|\overrightarrow{O A} \times \overrightarrow{A B}|=\sqrt{(-9)^{2}+(2)^{2}+(12)^{2}}=\sqrt{81+4+144}=\sqrt{229}$
$Area of \triangle O A B=\frac{1}{2}\sqrt{229}$

Vector or cross product exercise multiple choice questions question 20

Answer: $\hat{i}.\hat{k}=0$

Given: $\hat{i},\hat{j},\hat{k}$ are unit vectors along three mutually perpendicular directions

Hint: Verify all the options

Explanation:

$\hat{i}.\hat{j}=1$ which is false

We know if $\hat{i},\hat{j},\hat{k}$ are unit vector then $\hat{i}.\hat{j}=0$

$\hat{i}\times \hat{j}=1$ which is false $\because$ of $\hat{i}\times \hat{j}=\hat{k}$
$\hat{i}.\hat{k}=0$ which is true
$\hat{i}\times \hat{k}=0$ which is false $\because$ of $\hat{i}\times \hat{k}=-\hat{j}$

Vector or cross product exercise multiple choice questions question 21

Answer: $3\sqrt{5}$
Given: $\begin{aligned} &\overrightarrow{O A}=\hat{i}+2 \hat{j}+3 \hat{k} ; \overrightarrow{O B}=-3 \hat{i}-2 \hat{j}+\hat{k} \\ \end{gathered}$
Hint:Area of $OAB=\frac{1}{2}\left | \overrightarrow{AO}\times \overrightarrow{AB} \right |$
Explanation:
$\begin{aligned} &\overrightarrow{O A}=\hat{i}+2 \hat{j}+3 \hat{k} ; \overrightarrow{O B}=-3 \hat{i}-2 \hat{j}+\hat{k} \\ \end{gathered}$
$\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A} \\$
$=-3 \hat{i}-2 \hat{j}+\hat{k}-(\hat{i}+2 \hat{j}+3 \hat{k}) \\$
$=-4 \hat{i}-4 \hat{j}-2 \hat{k}$
$Area of \triangle O A B=\frac{1}{2}|\overrightarrow{O A} \times \overrightarrow{A B}|$
$\overrightarrow{O A} \times \overrightarrow{A B}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -4 & -4 & -2 \end{array}\right| \\$
$=\hat{i}(-4+12)-\hat{j}(-2+12)+\hat{k}(-4+8) \\$
$=8 \hat{i}-10 \hat{j}+4 \hat{k} \\$
$|\overrightarrow{O A} \times \overrightarrow{A B}|=\sqrt{(8)^{2}+(-10)^{2}+(4)^{2}}=\sqrt{64+100+16}=\sqrt{180}$
$\text { Area of } \triangle \mathrm{OAB}=\frac{1}{2} \sqrt{180}=\frac{2 \sqrt{45}}{2}=\sqrt{45}=\sqrt{9 \times 5}=3 \sqrt{5} \text { units }$

Chapter 24, of class 12 mathematics, Vector or Cross Product, has only one exercise, ex 24.1. And there are only 21 questions in the MCQ part given in the textbook. This might seem easy when compared to the other chapters, but the real challenge is when the students try to solve those questions. The concepts included in the MCQ part are the cross-product of two vectors, non-zero parallel vectors, diagonals of a rhombus, area of a triangle, etc. The RD Sharma Class 12 Chapter 24 MCQ reference guide helps the students and leads them in the right direction.

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RD Sharma Chapter-wise Solutions

Frequently Asked Questions (FAQs)

1. Which is the best reference book to clarify the doubts regarding the Chapter 24 MCQ section?

The RD Sharma Class 12th Chapter 24 MCQ book is the best solution material for the students to clarify their doubts regarding the multiple-choice questions of the respective chapters.

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The RD Sharma solution books provide the solved sums for every question asked in the textbook. Therefore, all the answers can be found in these books.

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4. How many multiple-choice questions are asked in the Vector or Cross Product chapter in class 12 mathematics?

There are 21 questions asked in the Vector or Cross Product chapter in mathematics. You can use the RD Sharma Class 12th Chapter 24 MCQ solution guide to refer to the correct answers.

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