# NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra

Two chapters vector algebra & three-dimensional geometry has 17% weightage in 12th board maths final examination. The CBSE NCERT solutions for class 12 maths chapter 10 vector algebra will help you to score good marks in the exam. The concepts of vector algebra are very useful in both maths and physics. In chapter 10 vector algebra total of 54 questions in 4 exercises and 30 solved examples are given in the NCERT textbook. These NCERT questions are prepared and explained in a detailed manner to help students in their board exam as well as in competitive exams. Before discussing this chapter in detail, you need to understand what is a vector quantity and what is a scalar quantity.

Vector Quantity- Quantity which involves both the value magnitude and direction. Vector quantities like weight, velocity, acceleration, displacement, force, momentum, etc.

Scalar Quantity- Quantity which involves only one value (magnitude) which is a real number. Scalar quantities like distance,  length, time, mass, speed, area, temperature, work, money, volume, voltage, density, resistance, etc.

The focus of class 12 maths chapter 10 vector algebra is on vector quantities. To develop a grip on the topic, students must try to solve the miscellaneous exercise also. In the NCERT solutions for class 12 maths chapter 10 vector algebra, you will find solutions to miscellaneous exercises too.

## Topics of NCERT class 12 maths chapter 10 Vector Algebra

10.1 Introduction

10.2 Some Basic Concepts

10.3 Types of Vectors

10.5 Multiplication of a Vector by a Scalar

10.5.1 Components of a vector

10.5.2 Vector joining two points

10.5.3 Section formula

10.6 Product of Two Vectors

10.6.1 Scalar (or dot) product of two vectors

10.6.2 Projection of a vector on a line

10.6.3 Vector (or cross) product of two vectors

## Solutions of NCERT for class 12 maths chapter 10 vector algebra-Exercise: 10.1

Represent graphically a displacement of 40 km, $30 \degree$ east of north.

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

$\underset{OP}{\rightarrow}$ is displacement vector which $\left | \underset{OP}{\rightarrow} \right |$

= 40 km.

$\underset{OP}{\rightarrow}$ makes an angle of 30 degree east of north  as shown in figure.

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

10Kg

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

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

This is a scalar quantity as it has only magnitude.

This is a scalar quantity as it has only magnitude.

This is a scalar quantity as it has only magnitude.

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.

This is a scalar quantity as it has only magnitude.

(2)   distance

Distance is a scalar quantity as it has only magnitude.

(3)   force

Force is a vector quantity as it has both magnitude as well as direction.

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

(5)     work done

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

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

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

(3) Collinear but not equal

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

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

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 same magnitude are collinear.

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.

False, because two colinear vectors with the same magnitude can have opposite  direction

## CBSE NCERT solutions for class 12 maths chapter 10 vector algebra-Exercise: 10.2

(1)    $\vec a = \hat i + \hat j + \hat k$

Here

$\vec a = \hat i + \hat j + \hat k$

Magnitude of $\vec a$

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

(2)    $\vec b = 2 \hat i - 7 \hat j - 3 \hat k$

Here,

$\vec b = 2 \hat i - 7 \hat j - 3 \hat k$

Magnitude of $\vec b$

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

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

Here,

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

Magnitude of $\vec c$

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

Two different Vectors having the same magnitude are

$\vec a= 3\hat i+6\hat j+9\hat k$

$\vec b= 9\hat i+6\hat j+3\hat k$

The magnitude of both vector

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

Two different vectors having the same direction are:

$\vec a=\hat i+2\hat j+3\hat k$

$\vec b=2\hat i+4\hat j+6\hat k$

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

Hence when,

$x=2$ and

$y=3$

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

Now,

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

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

Given,

$\\ \vec a = \hat i - 2 \hat j + \hat k ,\\ \vec b = -2 \hat i + 4 \hat j + 5 \hat k \: \: and\: \: \: \\\vec c = \hat i - 6 \hat j - 7 \hat k$

Now, The sum of the vectors:

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

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

$\vec a +\vec b+\vec c =-4\hat j-\hat k$

Given

$\vec a = \hat i + \hat j + 2 \hat k$

Magnitude of $\vec a$

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

A unit vector in the direction of $\vec a$

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

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

A vector in direction of PQ

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

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

Magnitude of PQ

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

Now, unit vector in direction of PQ

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

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

Given

$\vec a = 2 \hat i - \hat j + 2 \hat k$

$\vec b = - \hat i + \hat j - \hat k$

Now,

$\vec a + \vec b=(2-1)\hat i+(-1+1)\hat j+ (2-1)\hat k$

$\vec a + \vec b=\hat i+\hat k$

Now a unit vector in the direction of $\vec a + \vec b$

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

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

Given a vector

$\vec a=5 \hat i - \hat j + 2 \hat k$

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

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

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

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

Let

$\vec a =2 \hat i -3 \hat j + 4 \hat k$

$\vec b=- 4 \hat i + 6 \hat j - 8 \hat k$

It can be seen that

$\vec b=- 4 \hat i + 6 \hat j - 8 \hat k=-2(2 \hat i -3 \hat j + 4 \hat k)=-2\vec a$

Hence here $\vec b=-2\vec a$

As we know

Whenever we have $\vec b=\lambda \vec a$, the vector $\vec a$ and $\vec b$ will be colinear.

Here $\lambda =-2$

Hence  vectors $2 \hat i -3 \hat j + 4 \hat k$ and $- 4 \hat i + 6 \hat j - 8 \hat k$  are collinear.

Let

$\vec a=\hat i + 2 \hat j + 3 \hat k$

$\left |\vec a \right |=\sqrt{1^2+2^2+3^2}=\sqrt{14}$

Hence direction cosine of $\vec a$ are

$\left ( \frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}} ,\frac{3}{\sqrt{14}}\right )$

Given

point A=(1, 2, –3)

point B=(–1, –2, 1)

Vector joining A and B Directed from A to B

$\vec {AB}=(-1-1)\hat i +(-2-2)\hat j+(1-(-3))\hat k$

$\vec {AB}=-2\hat i +-4\hat j+4\hat k$

$\left | \vec {AB} \right |=\sqrt{(-2)^2+(-4)^2+4^2}=\sqrt{36}=6$

Hence Direction cosines of vector AB are

$\left ( \frac{-2}{6},\frac{-4}{6},\frac{4}{6} \right )=\left ( \frac{-1}{3},\frac{-2}{3},\frac{2}{3} \right )$

Let

$\vec a=\hat i + \hat j + \hat k$

$\left | \vec a \right |=\sqrt{1^2+1^2+1^2}=\sqrt{3}$

Hence direction cosines of this vectors is

$\left ( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right )$

Let $\alpha$$\beta$ and $\gamma$ be the angle made by x-axis, y-axis and z- axis respectively

Now as we know,

$cos\alpha=\frac{1}{\sqrt{3}}$,    $cos\beta=\frac{1}{\sqrt{3}}$     $and\:cos\gamma=\frac{1}{\sqrt{3}}$

Hence Given vector is equally inclined to axis OX,OY and OZ.

As we know

The position vector of the point R which divides the line segment PQ in ratio m:n internally:

$\vec r=\frac{m\vec b+n\vec a}{m+n}$

Here

position vector os P = $\vec a$ = $i + 2 j - k$

the position vector of Q  = $\vec b=- i + j + k$

m:n = 2:1

And Hence

$\vec r = \frac{2(-\hat i+\hat j +\hat k)+1(\hat i+2\hat j-\hat k)}{2+1}=\frac{-2\hat i+2\hat j +2\hat k+\hat i+2\hat j-\hat k}{3}$

$\vec r = \frac{-2\hat i+2\hat j +2\hat k+\hat i+2\hat j-\hat k}{3}=\frac{-\hat i+4\hat j+\hat k}{3}$

$\vec r = \frac{-\hat i}{3}+\frac{4\hat j}{3}+\frac{\hat k}{3}$

As we know

The position vector of the point R which divides the line segment PQ in ratio m:n externally:

$\vec r=\frac{m\vec b-n\vec a}{m-n}$

Here

position vector os P = $\vec a$ = $i + 2 j - k$

the position vector of Q  = $\vec b=- i + j + k$

m:n = 2:1

And Hence

$\vec r = \frac{2(-\hat i+\hat j +\hat k)-1(\hat i+2\hat j-\hat k)}{2-1}=\frac{-2\hat i+2\hat j +2\hat k-\hat i-2\hat j+\hat k}{1}$

$\vec r = -3\hat i +3\hat k$

Given

The position vector of point P = $2\hat i+3\hat j +4\hat k$

Position Vector of point Q = $4\hat i+\hat j -2\hat k$

The position vector of R which divides PQ in half is given by:

$\vec r =\frac{2\hat i+3\hat j +4\hat k+4\hat i+\hat j -2\hat k}{2}$

$\vec r =\frac{6\hat i+4\hat j +2\hat k}{2}=3\hat i+2\hat j +\hat k$

Given

the position vector of A, B, and C are

$\\\vec a = 3 \hat i - 4 \hat j - 4 \hat k ,\\\vec b = 2 \hat i - \hat j + \hat k \: \: and \\\vec c = \hat i - 3 \hat j - 5 \hat k$

Now,

$\vec {AB}=\vec b-\vec a=-\hat i+3\hat j+5\hat k$

$\vec {BC}=\vec c-\vec b=-\hat i-2\hat j-6\hat k$

$\vec {CA}=\vec a-\vec c=2\hat i-\hat j+\hat k$

$\left | \vec {AB} \right |=\sqrt{(-1)^2+3^2+5^2}=\sqrt{35}$

$\left | \vec {BC} \right |=\sqrt{(-1)^2+(-2)^2+(-6)^2}=\sqrt{41}$

$\left | \vec {CA} \right |=\sqrt{(2)^2+(-1)^2+(1)^2}=\sqrt{6}$

AS we can see

$\left | \vec {BC} \right |^2=\left | \vec {CA} \right |^2+\left | \vec {AB} \right |^2$

Hence ABC is a right angle triangle.

$A ) \overline{AB}+ \overline{BC}+ \overline{CA} = \vec 0 \\\\ B ) \overline{AB}+ \overline{BC}- \overline{AC} = \vec 0 \\\\ C ) \overline{AB}+ \overline{BC}- \overline{CA} = \vec 0 \\\\ D ) \overline{AB}- \overline{CB}+ \overline{CA} = \vec 0$

From triangles law of addition we have,

$\vec {AB}+\vec {BC}=\vec {AC}$

From here

$\vec {AB}+\vec {BC}-\vec {AC}=0$

also

$\vec {AB}+\vec {BC}+\vec {CA}=0$

Also

$\vec {AB}-\vec {CB}+\vec {CA}=0$

Hence options A,B and D are true SO,

Option C is False.

If two vectors are collinear then, they have same direction or are parallel or anti-parallel.
Therefore,
They can be expressed in the form $\vec{b}= \lambda \vec{a}$ where a and b are vectors and  $\lambda$ is some scalar quantity.

Therefore, (a) is true.
Now,
(b)  $\lambda$ is a scalar quantity so its value may be equal to $\pm 1$

Therefore,
(b) is also true.

C)   The vectors  and  are proportional,
Therefore, (c) is not true.

D)  The vectors  and  can have different magnitude as well as different directions.

Therefore, (d) is not true.

Therefore,   the correct options are (C) and (D).

## Question:1 Find the angle between two vectors $\vec a \: \:and \: \: \vec b$ with magnitudes $\sqrt 3 \: \:and \: \: 2$ , respectively having .$\vec a . \vec b = \sqrt 6$

Given

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

$\left | \vec b \right |=2$

$\vec a . \vec b = \sqrt 6$

As we know

$\vec a . \vec b = \left | \vec a \right |\left | \vec b \right |cos\theta$

where $\theta$ is the angle between two vectors

So,

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

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

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

Given two vectors

$\vec a=\hat i - 2 \hat j + 3 \hat k \: \:and \: \: \vec b=3 \hat i - 2 \hat j + \hat k$

Now As we know,

The angle between two vectors $\vec a$ and $\vec b$ is given by

$\theta=cos^{-1}\left ( \frac{\vec a.\vec b}{\left | \vec a \right |\left | \vec b \right |}\right )$

Hence the angle between $\vec a=\hat i - 2 \hat j + 3 \hat k \: \:and \: \: \vec b=3 \hat i - 2 \hat j + \hat k$

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

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

$\theta=cos^{-1}\frac{10}{14}$

$\theta=cos^{-1}\frac{5}{7}$

Let

$\vec a=\hat i - \hat j$

$\vec b=\hat i + \hat j$

Projection of vector $\vec a$ on $\vec b$

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

Hence, Projection of vector $\vec a$ on $\vec b$ is 0.

Let

$\vec a =\hat i + 3 \hat j + 7 \hat k$

$\vec b=7\hat i - \hat j + 8 \hat k$

The projection of $\vec a$ on  $\vec b$ is

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

Hence, projection of vector $\vec a$ on  $\vec b$ is

$\frac{60}{\sqrt{114}}$

Given

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

Now magnitude of $\vec a,\vec b \:and\: \vec c$

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

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

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

Hence, they all are unit vectors.

Now,

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

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

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

Hence all three are mutually perpendicular to each other.

Given in the question

$( \vec a + \vec b ). ( \vec a - \vec b )=8$

$\left | \vec a \right |^2-\left | \vec b \right |^2=8$

Since $|\vec a |\: \:= 8 \: \:|\vec b |$

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

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

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

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

So, answer of the question is

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

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

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

$=6\vec a.^2+11\vec a.\vec b-35\vec b^2$

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

Given two vectors $\vec a \: \: and \: \: \vec b$

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

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

Now Angle between $\vec a \: \: and \: \: \vec b$

$\theta=60^0$

Now As we know that

$\vec a.\vec b=\left | \vec a \right |\left | \vec b \right |cos\theta$

$\frac{1}{2}=\left | \vec a \right |\left | \vec a \right |cos60^0$

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

Hence, the magnitude of two vectors $\vec a \: \: and \: \: \vec b$

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

Given in the question that

$( \vec x -\vec a ) . ( \vec x + \vec a ) = 12$

And we need to find $\left | \vec x \right |$

$\left | \vec x \right |^2-\left | \vec a \right |^2 = 12$

$\left | \vec x \right |^2-1 = 12$

$\left | \vec x \right |^2 = 13$

$\left | \vec x \right | = \sqrt{13}$

So the value of $\left | \vec x \right |$ is $\dpi{100} \sqrt{13}$

Given in the question is

$\vec a = 2 \hat i + 2 \hat j + 3 \hat k , \vec b = - \hat i + 2 \hat j + \hat k \: \: and \: \: \vec c = 3 \hat i + \hat j$

and  $\vec a + \lambda \vec b$ is perpendicular to $\vec c$

and we need to find the value of $\lambda$,

so the value of $\vec a + \lambda \vec b$-

$\vec a + \lambda \vec b=2\hat i +2\hat j +3\hat k+\lambda (-\hat i+2\hat j+\hat k)$

$\vec a + \lambda \vec b=(2-\lambda)\hat i +(2+2\lambda)\hat j +(3+\lambda)\hat k$

As  $\vec a + \lambda \vec b$ is perpendicular to $\vec c$

$(\vec a + \lambda \vec b).\vec c=0$

$((2-\lambda)\hat i +(2+2\lambda)\hat j +(3+\lambda)\hat k)(3\hat i+\hat j)=0$

$3(2-\lambda)+2+2\lambda=0$

$6-3\lambda+2+2\lambda=0$

$\lambda=8$

the value of $\lambda=8$,

Given in the question that -

$\vec a \: \: \: and \: \: \vec b$ are two non-zero vectors

According to the question

$\left ( |\vec a | \vec b + |\vec b | \vec a\right )\left (|\vec a | \vec b - |\vec b | \vec a \right )$

$=|\vec a |^2 |\vec b|^2 - |\vec b |^2 |\vec a|^2+|\vec b||\vec a|\vec a.\vec b-|\vec a||\vec b|\vec b.\vec a=0$

Hence $|\vec a | \vec b + |\vec b | \vec a$  is perpendicular to   $|\vec a | \vec b - |\vec b | \vec a$.

Given in the question

$\\\vec a . \vec a = 0 \\|\vec a|^2=0$

$\\|\vec a|=0$

Therefore $\vec a$ is a zero vector. Hence any vector  $\vec b$ will satisfy $\vec a . \vec b = 0$

Given in the question

$\vec a , \vec b , \vec c$  are unit vectors    $\Rightarrow |\vec a|=|\vec b|=|\vec c|=1$

and $\vec a + \vec b + \vec c = \vec 0$

and we need to find the value of $\vec a . \vec b + \vec b. \vec c + \vec c . \vec a$

$(\vec a + \vec b + \vec c)^2 = \vec 0$

$\vec a^2 + \vec b^2 + \vec c ^2+2(\vec a . \vec b + \vec b. \vec c + \vec c . \vec a)=0$

$|\vec a|^2 + |\vec b|^2 + |\vec c |^2+2(\vec a . \vec b + \vec b. \vec c + \vec c . \vec a)=0$

$1+1+1+2(\vec a . \vec b + \vec b. \vec c + \vec c . \vec a)=0$

$\vec a . \vec b + \vec b. \vec c + \vec c . \vec a=\frac{-3}{2}$

Answer- the value of $\vec a . \vec b + \vec b. \vec c + \vec c . \vec a$ is $\dpi{80} \frac{-3}{2}$

Let

$\vec a=\hat i-2\hat j +3\hat k$

$\vec b=5\hat i+4\hat j +1\hat k$

we see that

$\vec a.\vec b=(\hat i-2\hat j +3\hat k)(5\hat i+4\hat j +1\hat k)=5-8+3=0$

we now observe that

$|\vec a|=\sqrt{1^2+(-2)^2+3^2}=\sqrt{14}$

$|\vec b|=\sqrt{5^2+4^2+1^2}=\sqrt{42}$

Hence here converse of the given statement is not true.

Given points,

A=(1, 2, 3),

B=(–1, 0, 0),

C=(0, 1, 2),

As need to find Angle between  $\overline{BA}\: \: and\: \: \overline{BC} ]$

$\vec {BA}=(1-(-1))\hat i+(2-0)\hat j+(3-0)\hat k=2\hat i+2\hat j+3\hat k$

$\vec {BC}=(0-(-1))\hat i+(1-0)\hat j+(2-0)\hat k=\hat i+\hat j+2\hat k$

Hence angle between them ;

$\theta=cos^{-1}(\frac{\vec {BA}.\vec {BC}}{\left | \vec {BA} \right |\left | \vec {BC} \right |})$

$\theta=cos^{-1}\frac{2+2+6}{\sqrt{17}\sqrt{6}}$

$\theta=cos^{-1}\frac{10}{\sqrt{102}}$

Answer- Angle between the vectors $\overline{BA}\: \: and\: \: \overline{BC}$ is  $\dpi{80} \theta=cos^{-1}\frac{10}{\sqrt{102}}$

Given in the question

A=(1, 2, 7), B=(2, 6, 3) and C(3, 10, –1)

To show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear

$\vec {AB}=(2-1)\hat i+(6-2)\hat j+(3-7)\hat k$

$\vec {AB}=\hat i+4\hat j-4\hat k$

$\vec {BC}=(3-2)\hat i+(10-6)\hat j+(-1-3)\hat k$

$\vec {BC}=\hat i+4\hat j-4\hat k$

$\vec {AC}=(3-1)\hat i+(10-2)\hat j+(-1-7)\hat k$

$\vec {AC}=2\hat i+8\hat j-8\hat k$

$|\vec {AB}|=\sqrt{1^2+4^2+(-4)^2}=\sqrt{33}$

$|\vec {BC}|=\sqrt{1^2+4^2+(-4)^2}=\sqrt{33}$

$|\vec {AC}|=\sqrt{2^2+8^2+(-8)^2}=2\sqrt{33}$

As we see that

$|\vec {AC}|=|\vec {AB}|+|\vec {BC}|$

Hence point A, B , and C are colinear.

Given the position vector of A, B , and C are

$2 \hat i - \hat j + \hat k , \hat i - 3 \hat j - 5 \hat k \: \: and \: \: 3 \hat i - 4 \hat j - 4 \hat k$

To show that the vectors $2 \hat i - \hat j + \hat k , \hat i - 3 \hat j - 5 \hat k \: \: and \: \: 3 \hat i - 4 \hat j - 4 \hat k$  form the vertices of a right angled triangle

$\vec {AB}=(1-2)\hat i + (-3-(-1))\hat j+(-5-1)\hat k=-1\hat i -2\hat j-6\hat k$

$\vec {BC}=(3-1)\hat i + (-4-(-3))\hat j+(-4-(-5))\hat k=-2\hat i -\hat j+\hat k$

$\vec {AC}=(3-2)\hat i + (-4-(-1))\hat j+(-4-(1))\hat k=\hat i -3\hat j-5\hat k$

$|\vec {AB}|=\sqrt{(-1)^2+(-2)^2+(-6)^2}=\sqrt{41}$

$|\vec {BC}|=\sqrt{(-2)^2+(-1)^2+(1)^2}=\sqrt{6}$

$|\vec {AC}|=\sqrt{(1)^2+(-3)^2+(-5)^2}=\sqrt{35}$

Here we see that

$|\vec {AC}|^2+|\vec {BC}|^2=|\vec {AB}|^2$

Hence A,B, and C are the vertices of a right angle triangle.

$\\A ) \lambda = 1 \\\\ B ) \lambda = -1 \\\\ C ) a = |\lambda | \\\\ D ) a = 1 / |\lambda |$

Given $\vec a$  is a nonzero vector of magnitude ‘a’ and  $\lambda$  a nonzero scalar

$\lambda \vec a$ is a unit vector when

$|\lambda \vec a|=1$

$|\lambda|| \vec a|=1$

$| \vec a|=\frac{1}{|\lambda|}$

Hence the correct option is D.

## Solutions of NCERT for class 12 maths chapter 10 vector algebra-Exercise: 10.4

Given in the question,

$\\ \vec a = \hat i - 7 \hat j + 7 \hat k \: \: and \: \: \\\vec b = 3 \hat i - 2 \hat j + 2 \hat k$

and we need to find $\dpi{100} |\vec a \times \vec b |$

Now,

$|\vec a \times \vec b | =\begin{vmatrix} \hat i &\hat j &\hat k \\ 1 &-7 &7 \\ 3& -2 &2 \end{vmatrix}$

$|\vec a \times \vec b | =\hat i(-14+14)-\hat j(2-21)+\hat k(-2+21)$

$|\vec a \times \vec b | =19\hat j+19\hat k$

So the value of $\dpi{100} |\vec a \times \vec b |$ is $19\hat j+19\hat k$

Given in the question

$\vec a = 3 \hat i + 2 \hat j + 2 \hat k \: \:and \: \: \vec b = \hat i + 2 \hat j - 2 \hat k$

$\vec a + \vec b =3\hat i +2\hat j+2\hat k+\hat i +2\hat j-2\hat k=4\hat i +4\hat j$

$\vec a - \vec b =3\hat i +2\hat j+2\hat k-\hat i -2\hat j+2\hat k=2\hat i +4\hat k$

Now , A vector which perpendicular to both $\vec a + \vec b \: \: and\: \: \vec a - \vec b$ is  $(\vec a + \vec b) \times (\vec a - \vec b)$

$(\vec a + \vec b) \times (\vec a - \vec b)=\begin{vmatrix} \hat i &\hat j &\hat k \\ 4&4 &0 \\ 2& 0& 4 \end{vmatrix}$

$(\vec a + \vec b) \times (\vec a - \vec b)= \hat i (16-0)-\hat j(16-0)+\hat k(0-8)$

$(\vec a + \vec b) \times (\vec a - \vec b)= 16\hat i -16\hat j-8\hat k$

And a unit vector in this direction :

$\vec u =\frac{16\hat i-16\hat j-8\hat k}{|16\hat i-16\hat j-8\hat k|}=\frac{16\hat i-16\hat j-8\hat k}{\sqrt{16^2+(-16)^2+(-8)^2}}$

$\vec u =\frac{16\hat i-16\hat j-8\hat k}{24}=\frac{2}{3}\hat i-\frac{2}{3}\hat j-\frac{1}{3}\hat k$

Hence Unit vector perpendicular to each of the vector $\vec a + \vec b \: \: and\: \: \vec a - \vec b$ is $\frac{2}{3}\hat i-\frac{2}{3}\hat j-\frac{1}{3}\hat k$.

Given in the question,

angle between $\vec a$ and $\hat i$ :

$\alpha =\frac{\pi}{3}$

angle between $\vec a$ and $\hat j$

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

angle with $\vec a$ and $\hat k$:

$\gamma =\theta$

Now, As we know,

$cos^2\alpha+cos^2\beta+cos^2\gamma=1$

$cos^2\frac{\pi}{3}+cos^2\frac{\pi }{4}+cos^2\theta=1$

$\left ( \frac{1}{2} \right )^2+\left ( \frac{1}{\sqrt{2}} \right )^2+cos^2\theta=1$

$cos^2\theta=\frac{1}{4}$

$cos\theta=\frac{1}{2}$

$\theta=\frac{\pi}{3}$

Now components of $\vec a$ are:

$\left ( cos\frac{\pi}{3},cos\frac{\pi}{2},cos\frac{\pi}{3} \right )=\left ( \frac{1}{2},\frac{1}{\sqrt{2}},\frac{1}{2} \right )$

To show that $( \vec a - \vec b ) \times (\vec a + \vec b ) = 2 ( \vec a \times \vec b )$

LHS=

$\\( \vec a - \vec b ) \times (\vec a + \vec b )=( \vec a - \vec b ) \times (\vec a)+( \vec a - \vec b ) \times (\vec b)$

$( \vec a - \vec b ) \times (\vec a + \vec b )= \vec a \times \vec a-\vec b \times\vec a+\vec a \times \vec b-\vec b \times \vec b$

As product of a vector with itself is always Zero,

$( \vec a - \vec b ) \times (\vec a + \vec b )= 0-\vec b \times\vec a+\vec a \times \vec b-\0$

As cross product of a and b is equal to negative of cross product of b and a.

$( \vec a - \vec b ) \times (\vec a + \vec b )= \vec a \times\vec b+\vec a \times \vec b$

$( \vec a - \vec b ) \times (\vec a + \vec b )= 2(\vec a \times\vec b)$= RHS

LHS is equal to RHS, Hence Proved.

Given in the question

$( 2 \hat i + 6 \hat j + 27 \hat k ) \times ( \hat i + \lambda j + \mu \hat k ) = \vec 0$

and we need to find values of $\lambda$ and $\mu$

$\begin{vmatrix} \hat i &\hat j & \hat k\\ 2& 6&27 \\ 1& \lambda &\mu \end{vmatrix}=0$

$\hat i (6\mu-27\lambda)-\hat j(2\mu-27)+\hat k(2\lambda-6)=0$

From Here we get,

$6\mu-27\lambda=0$

$2\mu-27=0$

$2\lambda -6=0$

From here, the value of  $\lambda$ and $\mu$ is

$\lambda = 3 , \: and \: \mu=\frac{27}{2}$

Given in the question

$\vec a . \vec b = 0$ and $\vec a \times \vec b = 0$

When $\vec a . \vec b = 0$, either $|\vec a| =0,\:or\: |\vec b|=0,\vec a\: and \:\vec b$ are perpendicular to each other

When $\vec a \times \vec b = 0$ either $|\vec a| =0,\:or\: |\vec b|=0,\vec a\: and \:\vec b$are parallel to each other

Since two vectors can never be both parallel and perpendicular at same time,we conclude that

$|\vec a| =0\:or\: |\vec b|=0$

Given in the question

$\\\vec a=\vec a_1 \hat i + \vec a_2 \hat j + \vec a_3 \hat k , \\\vec b=\vec b_ 1 \hat i + \vec b_ 2 \hat j + \vec b_3 \hat k , \\\vec c=\vec c_ 1 \hat i + \vec c_ 2 \hat j + \vec c_ 3 \hat k$

We need to show that $\vec a \times ( \vec b + \vec c ) = \vec a \times \vec b + \vec a \times \vec c$

Now,

$\vec a \times ( \vec b + \vec c ) =(\vec a_1 \hat i + \vec a_2 \hat j + \vec a_3 \hat k)\times(\vec b_ 1 \hat i + \vec b_ 2 \hat j + \vec b_3 \hat k +\vec c_ 1 \hat i + \vec c_ 2 \hat j + \vec c_ 3 \hat k)$

$=(\vec a_1 \hat i + \vec a_2 \hat j + \vec a_3 \hat k)\times((\vec b_ 1+\vec c_1) \hat i + (\vec b_ 2+\vec c_2) \hat j +( \vec b_3 +\vec c_3)\hat k)$

$=\begin{vmatrix} \hat i &\hat j &\hat k \\ a_1&a_2 &a_3 \\ (b_1+c_1)&(b_2+c_2) &(b_3+c_3) \end{vmatrix}$

$\\=\hat i(a_2(b_3+c_3)-a_3(b_2+c_2))-\hat j(a_1(b_3+c_3)- a_3(b_1+c_1))+\hat k (a_1(b_2+c_2)-a_2(a_1(b_2+c_3)))$

$\\=\hat i(a_2(b_3+c_3)-a_3(b_2+c_2))-\hat j(a_1(b_3+c_3)-a_3(b_1+c_1))+\hat k (a_1(b_2+c_2)-a_2(a_1(b_2+c_3)))$

Now

$\vec a \times \vec b + \vec a \times \vec c=\begin{vmatrix} \hat i &\hat j & \hat k\\ a_1&a_2 &a_3 \\ b_1&b_2 &b_3 \end{vmatrix}+\begin{vmatrix} \hat i &\hat j & \hat k\\ a_1&a_2 &a_3 \\ c_1&c_2 &c_3 \end{vmatrix}$

$\vec a \times \vec b + \vec a \times \vec c=\hat i(a_2b_3-a_3b_2)-\hat j (a_1b_3-a_3b_1)+\hat k(a_1b_2-b_1a_2)+\hat i(a_2c_3-a_3c_2)-\hat j (a_1c_3-a_3c_1)+\hat k(a_1c_2-c_1a_2)$

$\\=\hat i(a_2(b_3+c_3)-a_3(b_2+c_2))-\hat j(a_1(b_3+c_3)-a_3(b_1+c_1))+\hat k (a_1(b_2+c_2)-a_2(a_1(b_2+c_3)))$

Hence they are equal.

No, the converse of the statement is not true, as there can be two non zero vectors, the cross product of whose are zero. they are colinear vectors.

Consider an example

$\vec a=\hat i +\hat j + \hat k$

$\vec b =2\hat i +2\hat j + 2\hat k$

Here $|\vec a| =\sqrt{1^2+1^2+1^2}=\sqrt{3}$

$|\vec b| =\sqrt{2^2+2^2+2^2}=2\sqrt{3}$

$\vec a \times \vec b=\begin{vmatrix} \hat i &\hat j &\hat k \\ 1&1 &1 \\ 2&2 &2 \end{vmatrix}=\hat i(2-2)-\hat j(2-2)+\hat k(2-2)=0$

Hence converse of the given statement is not true.

Given in the question

vertices A=(1, 1, 2), B=(2, 3, 5) and C=(1, 5, 5). and we need to find the area of the triangle

$AB=(2-1)\hat i+(3-1)\hat j+(5-2)\hat k=\hat i+2\hat j+3\hat k$

$BC=(1-2)\hat i+(5-3)\hat j+(5-5)\hat k=-\hat i+2\hat j$

Now as we know

Area of triangle

$A=\frac{1}{2}|\vec {AB}\times\vec {BC}|=\frac{1}{2}|(\hat i+2\hat j +3\hat k)\times(-\hat i+2\hat j)|$

$\\A=\frac{1}{2}\begin{vmatrix} \hat i &\hat j &\hat k \\ 1 &2 &3 \\ -1 &2 &0 \end{vmatrix}=\frac{1}{2}|\hat i(0-6)-\hat j(0-(-3))+\hat k(2-(-2))| \\A=\frac{1}{2}|-6\hat i-3\hat j+4\hat k|$

$A=\frac{1}{2}*\sqrt{(-6)^2+(-3)^2+(4)^2}=\frac{\sqrt{61}}{2}$

The area of the triangle is $\dpi{100} \frac{\sqrt{61}}{2}$ square units

Given in the question

$\vec a = \hat i - \hat j + 3 \hat k$

$\vec b = 2\hat i -7 \hat j + \hat k$

Area of parallelogram with adjescent side $\vec a$ and $\vec b$,

$A=|\vec a\times\vec b|=|(\vec i-\vec j+3\vec k)\times (2\hat i-7\hat j+\hat k)|$

$A=\begin{vmatrix} \hat i& \hat j & \hat k\\ 1&-1 &3 \\ 2&-7 &1 \end{vmatrix}=|\hat i(-1+21)-\hat j (1-6)+\hat k (-7+2)|$

$A=|\hat i(20)-\hat j (-5)+\hat k (-5)|=\sqrt{20^2+5^2+(-5)^2}$

$A=\sqrt{450}=15\sqrt{2}$

The area of the parallelogram whose adjacent sides are determined by the vectors   $\vec a = \hat i - \hat j + 3 \hat k$ and $\vec b = 2\hat i -7 \hat j + \hat k$  is $A=\sqrt{450}=15\sqrt{2}$

$\\A ) \pi /6 \\\\ B ) \pi / 4 \\\\ C ) \pi / 3 \\\\ D ) \pi /2$

Given in the question,

$|\vec a| = 3 \: \: and\: \: |\vec b | = \frac{\sqrt 2 }{3}$

As given $\vec a \times \vec b$  is a unit vector, which means,

$|\vec a \times \vec b|=1$

$|\vec a| | \vec b|sin\theta=1$

$3*\frac{\sqrt{2}}{3}sin\theta=1$

$sin\theta=\frac{1}{\sqrt{2}}$

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

Hence the angle between two vectors is $\frac{\pi}{4}$. Correct option is B.

$- \hat i + \frac{1}{2} \hat j + 4 \hat k , \hat i + \frac{1}{2} \hat j + 4 \hat k , \hat i - \frac{1}{2}\hat j + 4 \hat k \: \: and \: \: - \hat i - \frac{1}{2} \hat j + 4 \hat k$

(A)1/2

(B) 1

(C) 2

(D) 4

Given 4 vertices of rectangle are

$\\\vec a=- \hat i + \frac{1}{2} \hat j + 4 \hat k , \\\vec b=\hat i + \frac{1}{2} \hat j + 4 \hat k , \\\vec c= \hat i - \frac{1}{2}\hat j + 4 \hat k \: \: and \: \: \\\vec d= - \hat i - \frac{1}{2} \hat j + 4 \hat k$

$\vec {AB}=\vec b-\vec a=(1+1)\hat i+(\frac{1}{2}-\frac{1}{2})\hat j+(4-4)\hat k=2\hat i$

$\vec {BC}=\vec c-\vec b=(1-1)\hat i+(-\frac{1}{2}-\frac{1}{2})\hat j+(4-4)\hat k=-\hat j$

Now,

Area of the Rectangle

$A=|\vec {AB}\times\vec {BC}|=|2\hat i \times (-\hat j)|=2$

Hence option C is correct.

## CBSE NCERT solutions for class 12 maths chapter 10 vector algebra-Miscellaneous Exercise

As we know

a unit vector in XY-Plane making an angle $\theta$ with x-axis :

$\vec r=cos\theta \hat i+sin\theta \hat j$

Hence for $\theta = 30^0$

$\vec r=cos(30^0) \hat i+sin(30^0) \hat j$

$\vec r=\frac{\sqrt{3}}{2} \hat i+\frac{1}{2} \hat j$

Answer- the unit vector in XY-plane, making an angle of $30 \degree$ with the positive direction of x-axis is

$\vec r=\frac{\sqrt{3}}{2} \hat i+\frac{1}{2} \hat j$

Given in the question

$P(x_1, y_1, z_1) \: \: and \: \: Q(x_2, y_2, z_2).$

And we need to finrd the scalar components and magnitude of the vector joining the points P and Q

$\vec {PQ}=(x_2-x_1)\hat i +(y_2-y_1)\hat j+(z_2-z_1)\hat k$

Magnitiude of vector PQ

$|\vec {PQ}|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

Scalar components are

$(x_2-x_1),(y_2-y_1),(z_2-z_1)$

As the girl walks 4km towards west

Position vector =  $-4\hat i$

Now as she moves 3km in direction 30 degree east of north.

$-4\hat i+3sin30^0\hat i+3cos30^0\hat j$

$-4\hat i+\frac{3}{2}\hat i+3\frac{\sqrt{3}}{2}\hat j$

$\frac{-5}{2}\hat i+3\frac{\sqrt{3}}{2}\hat j$

hence final position vector is;

$\frac{-5}{2}\hat i+3\frac{\sqrt{3}}{2}\hat j$

No, if $\vec a = \vec b + \vec c$ then we can not conclude that  $|\vec a| =| \vec b |+| \vec c |$ .

the condition    $\vec a = \vec b + \vec c$ satisfies in the triangle.

also, in a triangle,  $|\vec a| <| \vec b |+| \vec c |$

Since, the condition $|\vec a| =| \vec b |+| \vec c |$ is contradicting with the triangle inequality, if$\vec a = \vec b + \vec c$  then we can not conclude that  $|\vec a| =| \vec b |+| \vec c |$

Given in the question,

a unit vector, $\vec u=x ( \hat i+ \hat j + \hat k )$

We need to find the value of x

$|\vec u|=1$

$|x ( \hat i+ \hat j + \hat k )|=1$

$x\sqrt{1^2+1^2+1^2}=1$

$x\sqrt{3}=1$

$x=\frac{1}{\sqrt{3}}$

The value of x   is $\dpi{80} \frac{1}{\sqrt{3}}$

Given two vectors

$\vec a = 2 \hat i + 3 \hat j - \hat k \: \: and \: \: \vec b = \hat i - 2 \hat j + \hat k$

Resultant of $\vec a$ and $\vec b$:

$\vec R = \vec a +\vec b$$=2 \hat i + 3 \hat j - \hat k + \hat i - 2 \hat j + \hat k=3\hat i + \hat j$

Now, a unit vector in the direction of $\vec R$

$\vec u =\frac{3\hat i+\hat j}{\sqrt{3^2+1^2}}=\frac{3}{\sqrt{10}}\hat i+\frac{1}{\sqrt{10}}\hat j$

Now, a unit vector of magnitude in direction of$\vec R$

$\vec v=5\vec u =5*\frac{3}{\sqrt{10}}\hat i+5*\frac{1}{\sqrt{10}}\hat j=\frac{15}{\sqrt{10}}\hat i+\frac{5}{\sqrt{10}}\hat j$

Hence the required vector is $\frac{15}{\sqrt{10}}\hat i+\frac{5}{\sqrt{10}}\hat j$

Given in the question,

$\vec a = \hat i + \hat j + \hat k , \vec b = 2 \hat i - \hat j + 3 \hat k \: \: and\: \: \vec c = \hat i - 2 \hat j + \hat k$

Now,

let vector $\vec V=2\vec a - \vec b + 3 \vec c$

$\vec V=2(\hat i +\hat j +\hat k) - (2\hat i-\hat j+3\hat k)+ 3 (\hat i-2\hat j+\hat k)$

$\vec V=3\hat i-3\hat j+2\hat k$

Now, a unit vector in direction of $\vec V$

$\vec u =\frac{3\hat i-3\hat j+2\hat k}{\sqrt{3^2+(-3)^2+2^2}}=\frac{3}{\sqrt{22}}\hat i-\frac{3}{\sqrt{22}} \hat j+\frac{2}{\sqrt{22}}\hat k$

Now,

A unit vector parallel to $\vec V$

$\vec u =\frac{3}{\sqrt{22}}\hat i-\frac{3}{\sqrt{22}} \hat j+\frac{2}{\sqrt{22}}\hat k$

OR

$-\vec u =-\frac{3}{\sqrt{22}}\hat i+\frac{3}{\sqrt{22}} \hat j-\frac{2}{\sqrt{22}}\hat k$

Given in the question,

points A(1, – 2, – 8), B(5, 0, –2) and C(11, 3, 7)

$\vec {AB }=(5-1)\hat i+(0-(-2))\hat j+(-2-(-8))\hat k=4\hat i+2\hat j+6\hat k$

$\vec {BC }=(11-5)\hat i+(3-0)\hat j+(7-(-2))\hat k=6\hat i+3\hat j+9\hat k$

$\vec {CA }=(11-1)\hat i+(3-(-2))\hat j+(7-(-8))\hat k=10\hat i+5\hat j+15\hat k$

now let's calculate the magnitude of the vectors

$|\vec {AB }|=\sqrt{4^2+2^2+6^2}=\sqrt{56}=2\sqrt{14}$

$|\vec {BC }|=\sqrt{6^2+3^2+9^2}=\sqrt{126}=3\sqrt{14}$

$|\vec {CA }|=\sqrt{10^2+5^2+15^2}=\sqrt{350}=5\sqrt{14}$

As we see that AB = BC + AC, we conclude that three points are colinear.

we can also see from here,

Point B divides AC in the ratio  2 : 3.

Given, two vectors $\vec P=( 2 \vec a + \vec b ) \: \:and \: \:\vec Q= ( \vec a - 3 \vec b )$

the point  R which divides line segment PQ in ratio 1:2 is given by

$=\frac{2(2\vec a +\vec b)-(\vec a-3\vec b)}{2-1}=4\vec a +2\vec b -\vec a+3\vec b=3\vec a+5\vec b$

Hence position vector of R is $3\vec a+5\vec b$.

Now, Position vector of the midpoint of RQ

$=\frac{( 3\vec a + 5\vec b + \vec a - 3 \vec b )}{2}=2\vec a+\vec b$

which is the position vector of Point P . Hence, P is the mid-point of RQ

Given, two adjacent sides of the parallelogram

$2 \hat i - 4 \hat j + 5 \hat k \: \:and \: \: \hat i - 2 \hat j - 3 \hat k$

The diagonal will be the resultant of these two vectors. so

resultant R:

$\vec R=2 \hat i - 4 \hat j + 5 \hat k \: +\: \hat i - 2 \hat j - 3 \hat k=3\hat i-6\hat j+2\hat k$

Now unit vector in direction of R

$\vec u=\frac{3\hat i-6\hat j+2\hat k}{\sqrt{3^2+(-6)^2+2^2}}=\frac{3\hat i-6\hat j+2\hat k}{\sqrt{49}}=\frac{3\hat i-6\hat j+2\hat k}{7}$

Hence unit vector along the diagonal of the parallelogram

$\vec u={\frac{3}{7}\hat i-\frac{6}{7}\hat j+\frac{2}{7}\hat k}$

Now,

Area of parallelogram

$A=(2 \hat i - 4 \hat j + 5 \hat k )\: \times \: \: (\hat i - 2 \hat j - 3 \hat k)$

$A=\begin{vmatrix} \hat i &\hat j &\hat k \\ 2& -4 &5 \\ 1&-2 &-3 \end{vmatrix}=|\hat i(12+10)-\hat j(-6-5)+\hat k(-4+4)|=|22\hat i +11\hat j|$

$A=\sqrt{22^2+11^2}=11\sqrt{5}$

Hence the area of the parallelogram is $11\sqrt{5}$.

Let a vector $\vec a$ is equally inclined to axis OX, OY  and OZ.

let direction cosines of this vector be

$cos\alpha,cos\alpha \:and \:cos\alpha$

Now

$cos^2\alpha+cos^2\alpha +cos^2\alpha=1$

$cos^2\alpha=\frac{1}{3}$

$cos\alpha=\frac{1}{\sqrt{3}}$

Hence direction cosines are:

$\left ( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right )$

Given,

$\vec a = \hat i + 4 \hat j + 2 \hat k , \vec b = 3 \hat i - 2 \hat j + 7 \hat k \: \:and \: \: \vec c = 2 \hat i - \hat j + 4 \hat k$

Let $\vec d=d_1\hat i+d_2\hat j +d_3\hat k$

now, since it is given that d is perpendicular to $\vec a$ and $\vec b$, we got the condition,

$\vec b.\vec d=0$   and   $\vec a.\vec d=0$

$(\hat i+4\hat j +2\hat k)\cdot(d_1\hat i +d_2\hat j+d_3\hat k)=0$   And  $(3\hat i-2\hat j +7\hat k)\cdot(d_1\hat i +d_2\hat j+d_3\hat k)=0$

$d_1+4d_2+2d_3=0$     And $3d_1-2d_2+7d_3=0$

here we got 2 equation and 3 variable. one more equation will come from the condition:

$\vec c . \vec d = 15$

$(2\hat i-\hat j +4\hat k)\cdot(d_1\hat i +d_2\hat j+d_3\hat k)=15$

$2d_1-d_2+4d_3=15$

so now we have three equation and three variable,

$d_1+4d_2+2d_3=0$

$3d_1-2d_2+7d_3=0$

$2d_1-d_2+4d_3=15$

On solving this three equation we get,

$d_1=\frac{160}{3},d_2=-\frac{5}{3}\:and\:d_3=-\frac{70}{3}$,

Hence Required vector :

$\vec d=\frac{160}{3}\hat i-\frac{5}{3}\hat j-\frac{70}{3}\hat k$

Let, the sum of vectors$2\hat i + 4 \hat j -5 \hat k$  and $\lambda \hat i + 2 \hat j +3 \hat k$ be

$\vec a=(\lambda +2)\hat i + 6 \hat j -2 \hat k$

unit vector along $\vec a$

$\vec u=\frac{(\lambda +2)\hat i + 6 \hat j -2 \hat k}{\sqrt{(\lambda+2)^2+6^2+(-2)^2}}=\frac{(\lambda +2)\hat i + 6 \hat j -2 \hat k}{\sqrt{\lambda^2+4\lambda+44}}$

Now, the scalar product of this with $\hat i + \hat j + \hat k$

$\vec u.(\hat i+\hat j +\hat k)=\frac{(\lambda +2)\hat i + 6 \hat j -2 \hat k}{\sqrt{\lambda^2+4\lambda+44}}.(\hat i+\hat j +\hat k)$

$\vec u.(\hat i+\hat j +\hat k)=\frac{(\lambda +2) + 6 -2 }{\sqrt{\lambda^2+4\lambda+44}}=1$

$\frac{(\lambda +2) + 6 -2 }{\sqrt{\lambda^2+4\lambda+44}}=1$

$\frac{(\lambda +6) }{\sqrt{\lambda^2+4\lambda+44}}=1$

${(\lambda +6) }}={\sqrt{\lambda^2+4\lambda+44}$

squaring both the side,

${(\lambda^2 +12\lambda + 36) }}={\lambda^2+4\lambda+44}$

$\lambda =1$

Given

$|\vec a|=|\vec b|=|\vec c|$ and

$\vec a.\vec b=\vec b.\vec c=\vec c.\vec a=0$

Now, let vector $\vec a+\vec b +\vec c$is inclined to $\vec a , \vec b \: \: and \: \: \vec c$ at $\theta_1,\theta_2\:and\:\theta_3$ respectively.

Now,

$cos\theta_1=\frac{(\vec a+\vec b+\vec c).\vec a}{|\vec a+\vec b+\vec c||\vec a|}=\frac{\vec a.\vec a +\vec a.\vec b +\vec c.\vec a}{|\vec a+\vec b+\vec c||\vec a|}=\frac{\vec a.\vec a}{|\vec a+\vec b+\vec c||\vec a|}=\frac{|\vec a|}{|\vec a+\vec b+\vec c|}$

$cos\theta_2=\frac{(\vec a+\vec b+\vec c).\vec b}{|\vec a+\vec b+\vec c||\vec b|}=\frac{\vec a.\vec b +\vec b.\vec b +\vec c.\vec b}{|\vec a+\vec b+\vec c||\vec b|}=\frac{\vec b.\vec b}{|\vec a+\vec b+\vec c||\vec b|}=\frac{|\vec b|}{|\vec a+\vec b+\vec c|}$

$cos\theta_3=\frac{(\vec a+\vec b+\vec c).\vec c}{|\vec a+\vec b+\vec c||\vec c|}=\frac{\vec a.\vec c +\vec b.\vec c +\vec c.\vec c}{|\vec a+\vec b+\vec c||\vec c|}=\frac{\vec c.\vec c}{|\vec a+\vec b+\vec c||\vec c|}=\frac{|\vec c|}{|\vec a+\vec b+\vec c|}$

Now, Since,$|\vec a|=|\vec b|=|\vec c|$

$cos\theta_1=cos\theta_2=cos\theta_3$

$\theta_1=\theta_2=\theta_3$

Hence vector $\vec a+\vec b +\vec c$  is equally inclined to $\vec a , \vec b \: \: and \: \: \vec c$ .

Given in the question,

$\vec a , \vec b$ are perpendicular and we need to prove that $\dpi{100} ( \vec a + \vec b ) . (\vec a + \vec b ) = |\vec a ^2 | + |\vec b |^2$

LHS= $( \vec a + \vec b ) . (\vec a + \vec b ) = \vec a .\vec a+\vec a.\vec b+\vec b.\vec a+\vec b.\vec b$

$= \vec a .\vec a+2\vec a.\vec b+\+\vec b.\vec b$

$= |\vec a |^2+2\vec a.\vec b+\+|\vec b|^2$

if  $\vec a , \vec b$ are perpendicular,  $\vec a.\vec b=0$

$( \vec a + \vec b ) . (\vec a + \vec b ) = |\vec a |^2+2\vec a.\vec b+\+|\vec b|^2$

$= |\vec a |^2+2\cdot0+\+|\vec b|^2$

$= |\vec a |^2+|\vec b|^2$

= RHS

LHS ie equal to RHS

Hence proved.

Given in the question

$\theta$ is the angle between two vectors $\vec a \: \: and \: \: \vec b$

$\vec a \cdot \vec b \geq 0$

$|\vec a| | \vec b |cos\theta\geq 0$

this will satisfy when

$cos\theta\geq 0$

$0\leq\theta\leq \frac{\pi}{2}$

Hence option B is the correct answer.

$\\A ) \theta = \frac{\pi }{4} \\\\ B ) \theta = \frac{\pi }{3} \\\\ C ) \theta = \frac{\pi }{2} \\\\ D ) \theta = \frac{2\pi }{3}$

Gicen in the question

$\vec a \: \: and \: \: \vec b$  be two unit vectors and $\theta$  is the angle between them

$|\vec a|=1,\:and\:\:|\vec b|=1$

also

$|\vec a + \vec b|=1$

$|\vec a + \vec b|^2=1$

$|\vec a|^2 + |\vec b|^2+2\vec a.\vec b=1$

$1 + 1+2\vec a.\vec b=1$

$\vec a.\vec b=-\frac{1}{2}$

$|\vec a||\vec b|cos\theta =-\frac{1}{2}$

$cos\theta =-\frac{1}{2}$

$\theta =\frac{2\pi}{3}$

Then $\vec a + \vec b$ is a unit vector if $\theta =\frac{2\pi}{3}$

Hence option D is correct.

(A) 0

(B) –1

(C) 1

(D) 3

To find the value of  $\hat i ( \hat j \times \hat k ) + \hat j ( \hat i \times \hat k ) + \hat k ( \hat i \times \hat j )$

$\\\hat i ( \hat j \times \hat k ) + \hat j ( \hat i \times \hat k ) + \hat k ( \hat i \times \hat j ) \\=\hat i.\hat i+\hat j(-\hat j)+\hat k.\hat k\\=1-1+1\\=1$

Hence option C is correct.

$\\A ) 0 \\\\ B ) \pi /4 \\\\ C ) \pi / 2 \\\\ D ) \pi$

Given in the question

$\theta$ is the angle between any two vectors  $\vec a \: \:and \: \: \vec b$ and $|\vec a \cdot \vec b |=|\vec a \times \vec b |$

To find the value of $\theta$

$|\vec a \cdot \vec b |=|\vec a \times \vec b |$

$|\vec a|| \vec b |cos\theta=|\vec a|| \vec b |sin\theta$

$cos\theta=sin\theta$

$tan\theta =1$

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

Hence option D is correct.

## NCERT solutions for class 12 maths chapter wise

 chapter 1 Solutions of NCERT for class 12 maths chapter 1 Relations and Functions chapter 2 CBSE NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions chapter 3 NCERT solutions for class 12 maths chapter 3 Matrices chapter 4 Solutions of NCERT for class 12 maths chapter 4 Determinants chapter 5 CBSE NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability chapter 6 NCERT solutions for class 12 maths chapter 6 Application of Derivatives chapter 7 Solutions of NCERT for class 12 maths chapter 7 Integrals chapter 8 CBSE NCERT solutions for class 12 maths chapter 8 Application of Integrals chapter 9 NCERT solutions for class 12 maths chapter 9 Differential Equations chapter 10 Solutions of NCERT for class 12 maths chapter 10 Vector Algebra chapter 11 CBSE NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry chapter 12 NCERT solutions for class 12 maths chapter 12 Linear Programming chapter 13 Solutions of NCERT for class 12 maths chapter 13 Probability

## Benefits of NCERT solutions

• NCERT solutions are explained in a step-by-step manner, so it will be very easy for you to understand the concepts.

• NCERT Solutions for class 12 maths chapter 10 vector algebra will give you some new way to solve the problem.

• Performance in the 12th board exam plays a very important role in deciding the future, so you can get admission to a good college. Scoring good marks in the exam is now a reality with the help of these solutions of NCERT for class 12 maths chapter 10 vector algebra.

• To develop a grip on the concept, you should solve the miscellaneous exercise also. In CBSE NCERT solutions for class 12 maths chapter 10 vector algebra article, you will get a solution of miscellaneous exercise also.

Happy learning !!!