NCERT Solutions for Class 12 Maths Chapter 10 - Vector Algebra

Question: 1 Represent graphically a displacement of 40 km, ${30}^{\circ }$ east of north.

Answer:

Represent graphically a displacement of 40 km, ${30}^{\circ }$ east of north.

N, S, E, W are all 4 directions: north, south, east, west, respectively.

$\overrightarrow{OP}$ is displacement vector which $\left|\overrightarrow{OP}\right|$ = 40 km.

$\overrightarrow{OP}$ makes an angle of 30 degrees east of north as shown in the figure.

Question: 2 (1) Classify the following measures as scalars and vectors.

10Kg

Answer:

10kg is a scalar quantity as it has only magnitude.

Question: 2 (2) Classify the following measures as scalars and vectors. 2 meters north west

Answer:

This is a vector quantity as it has both magnitude and direction.

Question: 2 (3) Classify the following measures as scalars and vectors. ${40}^{\circ }$

Answer:

This is a scalar quantity as it has only magnitude.

Question: 2 (4) Classify the following measures as scalars and vectors. 40 watt

Answer:

This is a scalar quantity as it has only magnitude.

Question: 2 (5) Classify the following measures as scalars and vectors. ${10}^{-19}coulomb$

Answer:

This is a scalar quantity as it has only magnitude.

Question: 2 (6) Classify the following measures as scalars and vectors. $20m/s^2$

Answer:

This is a Vector quantity as it has magnitude as well as direction.by looking at the unit, we conclude that measure is acceleration which is a vector.

Question: 3 Classify the following as scalar and vector quantities.
(1) time period

Answer:

This is a scalar quantity as it has only magnitude.

Question: 3 Classify the following as scalar and vector quantities.

(2) distance

Answer:

Distance is a scalar quantity as it has only magnitude.

Question: 3 Classify the following as scalar and vector quantities.

(3) force

Answer:

Force is a vector quantity as it has magnitude and direction.

Question: 3 Classify the following as scalar and vector quantities.
(4) velocity

Answer:

Velocity is a vector quantity as it has both magnitude and direction.

Question: 3 Classify the following as scalar and vector quantities.

(5) work done

Answer:

Work done is a scalar quantity, as it is the product of two vectors.

Question: 4 In Fig 10.6 (a square), identify the following vectors.
(1) Coinitial

Answer:

Since vector $\overrightarrow{a}$ and vector $\overrightarrow{d}$ are starting from the same point, they are coinitial.

Question: 4 In Fig 10.6 (a square), identify the following vectors.
(2) Equal

Answer:

Since Vector $\overrightarrow{b}$ and Vector $\overrightarrow{d}$ both have the same magnitude and same direction, they are equal.

Question: 4 In Fig 10.6 (a square), identify the following vectors.

(3) Collinear but not equal

Answer:

Since vector $\overrightarrow{a}$ and vector $\overrightarrow{c}$ have the same magnitude but different direction, they are collinear and not equal.

Question: 5 Answer the following as true or false.
(1) $\overrightarrow{a}$ and $-\overrightarrow{a}$ are collinear.

Answer:

True, $\overrightarrow{a}$ and $-\overrightarrow{a}$ are collinear. they are both parallel to one line, hence they are colinear.

Question: 5 Answer the following as true or false.
(2) Two collinear vectors are always equal in magnitude.

Answer:

False, because colinear means they are parallel to the same line but their magnitude can be anything and hence, this is a false statement.

Question: 5 Answer the following as true or false.

(3) Two vectors having the same magnitude are collinear.

Answer:

False, because any two non-colinear vectors can have the same magnitude.

Question: 5 Answer the following as true or false.

(4) Two collinear vectors having the same magnitude are equal.

Answer:

False, because two collinear vectors with the same magnitude can have opposite directions.

Vector Algebra Class 12 Chapter 10 Exercise 10.2
Page number: 354-355, Total Questions: 19

Question: 1 Compute the magnitude of the following vectors:

(1) $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$

Answer:

Here

$\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$

Magnitude of $\overrightarrow{a}$

$\overrightarrow{a}=\sqrt{1^2+1^2+1^2}=\sqrt{3}$

Question: 1 Compute the magnitude of the following vectors:

(2) $\overrightarrow{b}=2\hat{i}-7\hat{j}-3\hat{k}$

Answer:

Here,

$\overrightarrow{b}=2\hat{i}-7\hat{j}-3\hat{k}$

Magnitude of $\overrightarrow{b}$

$\left|\overrightarrow{b}\right|=\sqrt{2^2+(-7{)}^2+(-3{)}^2}=\sqrt{62}$

Question: 1 Compute the magnitude of the following vectors:

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

Answer:

Here,

$\overrightarrow{c}=\frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}-\frac{1}{\sqrt{3}}\hat{k}$

Magnitude of $\overrightarrow{c}$

$\left|\overrightarrow{c}\right|=\sqrt{{\left(\frac{1}{\sqrt{3}}\right)}^2+{\left(\frac{1}{\sqrt{3}}\right)}^2+{\left(\frac{1}{\sqrt{3}}\right)}^2}=1$

Question: 2 Write two different vectors having the same magnitude

Answer:

Two different Vectors having the same magnitude are

$\overrightarrow{a}=3\hat{i}+6\hat{j}+9\hat{k}$

$\overrightarrow{b}=9\hat{i}+6\hat{j}+3\hat{k}$

The magnitude of both vector

$\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\sqrt{9^2+6^2+3^2}=\sqrt{126}$

Question: 3 Write two different vectors having the same direction.

Answer:

Two different vectors having the same direction are:

$\overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k}$

$\overrightarrow{b}=2\hat{i}+4\hat{j}+6\hat{k}$

Question: 4 Find the values of x and y so that the vectors $2\hat{i}+3\hat{j}$ and $x\hat{i}+y\hat{j}$ are equal.

Answer:

$2\hat{i}+3\hat{j}$ will be equal to $x\hat{i}+y\hat{j}$ when their corresponding components are equal.

Hence when,

$x=2$ and $y=3$

Question: 5 Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (– 5, 7).

Answer:

Let point P = (2, 1) and Q = (– 5, 7).

Now,

$\overrightarrow{PQ}=(-5-2)\hat{i}+(7-1)\hat{j}=-7\hat{i}+6\hat{j}$

Hence, scalar components are (-7,6) and the vector is $-7\hat{i}+6\hat{j}$

Question: 6 Find the sum of the vectors $\overrightarrow{a}=\hat{i}-2\hat{j}+\hat{k},\overrightarrow{b}=-2\hat{i}+4\hat{j}+5\hat{k}and\overrightarrow{c}=\hat{i}-6\hat{j}-7\hat{k}$

Answer:

Given,

$$\begin{gathered} \overrightarrow{a}=\hat{i}-2\hat{j}+\hat{k}, \\ \overrightarrow{b}=-2\hat{i}+4\hat{j}+5\hat{k}and \\ \overrightarrow{c}=\hat{i}-6\hat{j}-7\hat{k} \end{gathered}$$

Now, The sum of the vectors:

$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\hat{i}-2\hat{j}+\hat{k}+-2\hat{i}+4\hat{j}+5\hat{k}+\hat{i}-6\hat{j}-7\hat{k}$

$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=(1-2+1)\hat{i}+(-2+4-6)\hat{j}+(1+5-7)\hat{k}$

$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=-4\hat{j}-\hat{k}$

Question: 7 Find the unit vector in the direction of the vector $\overrightarrow{a}=\hat{i}+\hat{j}+2\hat{k}$

Answer:

Given

$\overrightarrow{a}=\hat{i}+\hat{j}+2\hat{k}$

Magnitude of $\overrightarrow{a}$

$\left|\overrightarrow{a}\right|=\sqrt{1^2+1^2+2^2}=\sqrt{6}$

A unit vector in the direction of $\overrightarrow{a}$

$\overrightarrow{u}=\frac{\hat{i}}{\left|a\right|}+\frac{\hat{j}}{\left|a\right|}+\frac{2\hat{k}}{\left|a\right|}=\frac{\hat{i}}{\sqrt{6}}+\frac{\hat{j}}{\sqrt{6}}+\frac{2\hat{k}}{\sqrt{6}}$

Question: 8 Find the unit vector in the direction of vector $\overrightarrow{PQ}$ , where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively.

Answer:

Given P = (1, 2, 3) and Q = (4, 5, 6)

A vector in the direction of PQ

$\overrightarrow{PQ}=(4-1)\hat{i}+(5-2)\hat{j}+(6-3)\hat{k}$

$\overrightarrow{PQ}=3\hat{i}+3\hat{j}+3\hat{k}$

Magnitude of PQ

$\left|\overrightarrow{PQ}\right|=\sqrt{3^2+3^2+3^2}=3\sqrt{3}$

Now, the unit vector in the direction of PQ

$\hat{u}=\frac{\overrightarrow{PQ}}{\left|\overrightarrow{PQ}\right|}=\frac{3\hat{i}+3\hat{j}+3\hat{k}}{3\sqrt{3}}$

$\hat{u}=\frac{\hat{i}}{\sqrt{3}}+\frac{\hat{j}}{\sqrt{3}}+\frac{\hat{k}}{\sqrt{3}}$

Question: 9 For given vectors, $\overrightarrow{a}=2\hat{i}-\hat{j}+2\hat{k}$ and $\overrightarrow{b}=-\hat{i}+\hat{j}-\hat{k}$ , find the unit vector in the direction of the vector $\overrightarrow{a}+\overrightarrow{b}$ .

Answer:

Given

$\overrightarrow{a}=2\hat{i}-\hat{j}+2\hat{k}$

$\overrightarrow{b}=-\hat{i}+\hat{j}-\hat{k}$

Now,

$\overrightarrow{a}+\overrightarrow{b}=(2-1)\hat{i}+(-1+1)\hat{j}+(2-1)\hat{k}$

$\overrightarrow{a}+\overrightarrow{b}=\hat{i}+\hat{k}$

Now a unit vector in the direction of $\overrightarrow{a}+\overrightarrow{b}$

$\overrightarrow{u}=\frac{\overrightarrow{a}+\overrightarrow{b}}{|\overrightarrow{a}+\overrightarrow{b}|}=\frac{\hat{i}+\hat{j}}{\sqrt{1^2+1^2}}$

$\overrightarrow{u}=\frac{\hat{i}}{\sqrt{2}}+\frac{\hat{j}}{\sqrt{2}}$

Question: 10 Find a vector in the direction of vector $5\hat{i}-\hat{j}+2\hat{k}$ which has magnitude 8 units.

Answer:

Given a vector

$\overrightarrow{a}=5\hat{i}-\hat{j}+2\hat{k}$

the unit vector in the direction of $5\hat{i}-\hat{j}+2\hat{k}$

$\overrightarrow{u}=\frac{5\hat{i}-\hat{j}+2\hat{k}}{\sqrt{5^2+(-1{)}^2+2^2}}=\frac{5\hat{i}}{\sqrt{30}}-\frac{\hat{j}}{\sqrt{30}}+\frac{2\hat{k}}{\sqrt{30}}$

A vector in direction of $5\hat{i}-\hat{j}+2\hat{k}$ and whose magnitude is 8 =

$8\overrightarrow{u}=\frac{40\hat{i}}{\sqrt{30}}-\frac{8\hat{j}}{\sqrt{30}}+\frac{16\hat{k}}{\sqrt{30}}$

Vector Algebra Class 12 Chapter 10 Exercise 10.3
Page number: 361-362, Total Questions: 18

Question: 1 Find the angle between two vectors $\overrightarrow{a}and\overrightarrow{b}$ with magnitudes $\sqrt{3}and2$ , respectively having . $\overrightarrow{a}.\overrightarrow{b}=\sqrt{6}$

Answer:

Given

$\left|\overrightarrow{a}\right|=\sqrt{3}$

$\left|\overrightarrow{b}\right|=2$

$\overrightarrow{a}.\overrightarrow{b}=\sqrt{6}$

As we know

$\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cos\theta$

where $\theta$ is the angle between two vectors

So,

$cos\theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{\sqrt{6}}{\sqrt{3}\times 2}=\frac{1}{\sqrt{2}}$

$\theta =\frac{\pi }{4}$

Hence, the angle between the vectors is $\frac{\pi }{4}$.

Question: 2 Find the angle between the vectors $\hat{i}-2\hat{j}+3\hat{k}and3\hat{i}-2\hat{j}+\hat{k}$

Answer:

Given two vectors

$\overrightarrow{a}=\hat{i}-2\hat{j}+3\hat{k}and\overrightarrow{b}=3\hat{i}-2\hat{j}+\hat{k}$

Now, As we know,

The angle between two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by

$\theta =co{s}^{-1}\left(\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}\right)$

Hence the angle between $\overrightarrow{a}=\hat{i}-2\hat{j}+3\hat{k}and\overrightarrow{b}=3\hat{i}-2\hat{j}+\hat{k}$

⇒ $\theta =co{s}^{-1}\left(\frac{(\hat{i}-2\hat{j}+3\hat{k}).(3\hat{i}-2\hat{j}+\hat{k})}{|\hat{i}-2\hat{j}+3\hat{k}||3\hat{i}-2\hat{j}+\hat{k}|}\right)$

⇒ $\theta =co{s}^{-1}\left(\frac{3+4+3}{\sqrt{1^2+(-2{)}^2+3^3}\sqrt{3^2+(-2{)}^2+1^2}}\right)$

⇒ $\theta =co{s}^{-1}\frac{10}{14}$

⇒ $\theta =co{s}^{-1}\frac{5}{7}$

Question: 3 Find the projection of the vector $\hat{i}-\hat{j}$ on the vector $\hat{i}+\hat{j}$

Answer:

Let

$\overrightarrow{a}=\hat{i}-\hat{j}$

$\overrightarrow{b}=\hat{i}+\hat{j}$

Projection of vector $\overrightarrow{a}$ on $\overrightarrow{b}$

$\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{b}\right|}=\frac{(\hat{i}-\hat{j})(\hat{i}+\hat{j})}{|\hat{i}+\hat{j}|}=\frac{1-1}{\sqrt{2}}=0$

Hence, the Projection of vector $\overrightarrow{a}$ on $\overrightarrow{b}$ is 0.

Question: 4 Find the projection of the vector $\hat{i}+3\hat{j}+7\hat{k}$ on the vector $7\hat{i}-\hat{j}+8\hat{k}$

Answer:

Let

$\overrightarrow{a}=\hat{i}+3\hat{j}+7\hat{k}$

$\overrightarrow{b}=7\hat{i}-\hat{j}+8\hat{k}$

The projection of $\overrightarrow{a}$ on $\overrightarrow{b}$ is

$\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{b}\right|}=\frac{(\hat{i}+3\hat{j}+7\hat{k})(7\hat{i}-\hat{j}+8\hat{k})}{|7\hat{i}-\hat{j}+8\hat{k}|}=\frac{7-3+56}{\sqrt{7^2+(-1{)}^2+8^2}}=\frac{60}{\sqrt{114}}$

Hence, projection of vector $\overrightarrow{a}$ on $\overrightarrow{b}$ is $\frac{60}{\sqrt{114}}$.

Question: 5 Show that each of the given three vectors is a unit vector: $\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k}),\frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k}),\frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k})$ Also, show that they are mutually perpendicular to each other.

Answer:

Given

$$\begin{gathered} \overrightarrow{a}=\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k}), \\ \overrightarrow{b}=\frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k}), \\ \overrightarrow{c}=\frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k}) \end{gathered}$$

Now magnitude of $\overrightarrow{a},\overrightarrow{b}and\overrightarrow{c}$

$\left|\overrightarrow{a}\right|=\frac{1}{7}\sqrt{2^2+3^2+6^2}=\frac{\sqrt{49}}{7}=1$

$\left|\overrightarrow{b}\right|=\frac{1}{7}\sqrt{3^2+(-6{)}^2+2^2}=\frac{\sqrt{49}}{7}=1$

$\left|\overrightarrow{c}\right|=\frac{1}{7}\sqrt{6^2+2^2+(-3{)}^2}=\frac{\sqrt{49}}{7}=1$

Hence, they are all unit vectors.

Now,

$\overrightarrow{a}.\overrightarrow{b}=\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k})\frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})=\frac{1}{49}(6-18+12)=0$

$\overrightarrow{b}.\overrightarrow{c}=\frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})\frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k})=\frac{1}{49}(18-12-6)=0$

$\overrightarrow{c}.\overrightarrow{a}=\frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k})\frac{1}{7}(2\hat{i}+3\hat{j}-6\hat{k})=\frac{1}{49}(12+6-18)=0$

Hence, all three are mutually perpendicular to each other.

Question: 6 Find $|\overrightarrow{a}|and|\overrightarrow{b}|$ , if $(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b})=8and|\overrightarrow{a}|=8|\overrightarrow{b}|$ .

Answer:

Given in the question

$(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b})=8$

${\left|\overrightarrow{a}\right|}^2-{\left|\overrightarrow{b}\right|}^2=8$

Since $|\overrightarrow{a}|=8|\overrightarrow{b}|$

⇒ ${\left|\overrightarrow{8b}\right|}^2-{\left|\overrightarrow{b}\right|}^2=8$

⇒ ${\left|\overrightarrow{63b}\right|}^2=8$

⇒ ${\left|\overrightarrow{b}\right|}^2=\frac{8}{63}$

⇒ $\left|\overrightarrow{b}\right|=\sqrt{\frac{8}{63}}$

So, the answer is

$\left|\overrightarrow{a}\right|=8\left|\overrightarrow{b}\right|=8\sqrt{\frac{8}{63}}$

Question: 7 Evaluate the product $(3\overrightarrow{a}-5\overrightarrow{b}).(2\overrightarrow{a}+7\overrightarrow{b})$ .

Answer:

To evaluate the product $(3\overrightarrow{a}-5\overrightarrow{b}).(2\overrightarrow{a}+7\overrightarrow{b})$

$(3\overrightarrow{a}-5\overrightarrow{b}).(2\overrightarrow{a}+7\overrightarrow{b})=6\overrightarrow{a}.\overrightarrow{a}+21\overrightarrow{a}.\overrightarrow{b}-10\overrightarrow{b}.\overrightarrow{a}-35\overrightarrow{b}.\overrightarrow{b}$

$=6\overrightarrow{a}{.}^2+11\overrightarrow{a}.\overrightarrow{b}-35{\overrightarrow{b}}^2$

$=6{\left|\overrightarrow{a}\right|}^2+11\overrightarrow{a}.\overrightarrow{b}-35{\left|\overrightarrow{b}\right|}^2$

Question: 8 Find the magnitude of two vectors $\overrightarrow{a}and\overrightarrow{b}$ , having the same magnitude and such that the angle between them is ${60}^{\circ }$ and their scalar product is 1/2

Answer:

Given two vectors $\overrightarrow{a}and\overrightarrow{b}$

$\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|$

$\overrightarrow{a}.\overrightarrow{b}=\frac{1}{2}$

Now Angle between $\overrightarrow{a}and\overrightarrow{b}$

$\theta ={60}^0$

Now, As we know that

$\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cos\theta$

$\frac{1}{2}=\left|\overrightarrow{a}\right|\left|\overrightarrow{a}\right|cos{60}^0$

${\left|a\right|}^2=1$

Hence, the magnitude of two vectors $\overrightarrow{a}and\overrightarrow{b}$

$\left|a\right|=\left|b\right|=1$

Question: 9 Find $|\overrightarrow{x}|$ , if for a unit vector $\overrightarrow{a},(\overrightarrow{x}-\overrightarrow{a}).(\overrightarrow{x}+\overrightarrow{a})=12$

Answer:

Given in the question that

$(\overrightarrow{x}-\overrightarrow{a}).(\overrightarrow{x}+\overrightarrow{a})=12$

And we need to find $\left|\overrightarrow{x}\right|$

${\left|\overrightarrow{x}\right|}^2-{\left|\overrightarrow{a}\right|}^2=12$

⇒ ${\left|\overrightarrow{x}\right|}^2-1=12$

⇒ ${\left|\overrightarrow{x}\right|}^2=13$

⇒ $\left|\overrightarrow{x}\right|=\sqrt{13}$

So the value of $\left|\overrightarrow{x}\right|$ is $\sqrt{13}$.