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

Edited By Lovekush kumar saini | Updated on Jan 27, 2022 10:30 AM IST

The Class 12 students of the CBSE board schools opt for the RD Sharma solution books for their reference. This set of books are the most preferred whenever the students face challenges in solving mathematical problems. For example, the sums in the Vector or Cross Products consume more time while rechecking. Here is where the RD Sharma Class 12th VSA books lend a helping hand to the students.

JEE Main 2025: Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

JEE Main 2025: Maths Formulas | Study Materials

JEE Main 2025: Syllabus | Preparation Guide | High Scoring Topics

This Story also Contains

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

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

Vector or Cross Product Excercise: VSA

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, Vector or Cross Product, is easy but has complexities while checking the answers. The Very Short Answer (VSA) must be answered using shortcut methods or direct formula application. The RD Sharma Class 12 Chapter 24 VSA book has the right methods to solve these sums to save time. There are 31 VSA questions in this chapter; the concepts presented here are the vector product of two vectors, the value of vectors by solving, finding the angle between two vectors, unit vector, dot product, cross product and finding the area of the parallelogram with the given vectors.

Most of the students cause a blunder by elaborating these sums and spending their time on the VSA section during examinations. Only a good solution book will provide the methods that let students complete it in less time. The RD Sharma books are based on the NCERT pattern making the CBSE students adapt to it easily. The RD Sharma Class 12th VSA book consists of numerous VSA questions for practice. Once a student gains practice with this preparatory book, he can face the exam easily.

To understand the concept clearly, the students can refer to the solution book available for exercise 24.1. An abundance of practice questions on Very Short Answers for this chapter is also given in the Class 12 RD Sharma Chapter 24 VSA Solution book. Therefore, the students not only get access to the answer key, but they can also use the additional questions for the exam preparation. All the expert-provided answers will let them grasp the vector concepts quickly.

And as a bonus benefit, the RD Sharma Class 12 Solutions Vector or Cross Product VSA reference material is now available at the Career360 website for free. RD Sharma Solutions Everyone can access this best set of solution books and download it to their device for referring later when they encounter doubts. Many CBSE School students have benefitted by learning the Vector or Cross Product topic from the RD Sharma Class 12th VSA solution book.

The staff who have spent their time preparing the RD Sharma Class 12 Solutions Chapter 24 VSA book are experts in the domain. This reference material provides guidance for the students to learn and achieve more marks in the public exam.

RD Sharma Chapter wise Solutions

Frequently Asked Questions (FAQs)

1. What is the most prescribed book to learn the methods used to solve VSA questions in chapter 24?

The Very Short Answers or the VSA part need to be solved using shortcut methods to save time. And for this, the most recommended book is the RD Sharma Class 12th VSA reference material.

2. How many questions are given in the VSA Section in the class 12 mathematics, chapter 24 textbook?

There are thirty-one questions given in the Very Short Answers Section in chapter 24. You can refer to the RD Sharma Class 12th VSA for the right answer keys.

3. Can the CBSE Schools students use the RD Sharma solution books?

As the RD Sharma solution materials are based on the NCERT pattern, most CBSE schools themselves recommend this set of solution books to their students.

4. How many exercises do the class 12 mathematics chapter 24 consist of?

The class 12 mathematics chapter 24 has only one exercise, ex 24.1.

5. Where can I find the additional questions for the chapter 24 VSA section?

The additional practice questions for this chapter are available in the RD Sharma Class 12 Chapter 24 VSA book.