NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra

 

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra: In our day to day life, we come across many queries such as 'What is your height?' How should a football player hit the ball to give a pass to another player of his team? The answer to the second question is a force. It has magnitude and direction. Such quantities are called as vector quantities. In this article, you will get NCERT solutions for class 12 maths chapter 10 vector algebra. Important topics that are going to be discussed in this chapter are vector quantities, operations on vectors, geometric properties and algebraic properties like addition, multiplication, etc. In CBSE NCERT solutions for class 12 maths chapter 10 vector algebra article, questions from all these topics are covered. Vector algebra plays an important role in physics also. Solutions of NCERT for class 12 maths chapter 10 vector algebra will build your concepts of vectors which will help you to solve the many problems of physics in competitive exams like JEE mains, Jee Advanced, BITSAT, VITEEE. Check all NCERT solutions from class 6 to 12 at a single place, which will help you to get a better understanding of concepts in a much easy way.

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

10.4 Addition 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

NCERT solutions for class 12 maths chapter 10 Vector Algebra- Exercise questions

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

Question:1 Represent graphically a displacement of 40 km, 30 \degree east of north.  

Answer:

 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.

 

Represent graphically a displacement of 40 km

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 \degree

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. 20 m/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 both magnitude as well as 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 \vec{a} and vector \vec{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 \vec{b} and Vector\vec{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 \vec{a}and vector \vec{c} have the same magnitude but different direction, they are colinear and not equal.

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

Answer:

True,  \vec a  and -\vec a  are collinear. they both are 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 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 colinear vectors with the same magnitude can have opposite  direction 

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

Question:1 Compute the magnitude of the following vectors: 

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

Answer:

Here

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

    Magnitude of \vec a

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

Question:1 Compute the magnitude of the following vectors:

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

Answer:

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}

Question:1 Compute the magnitude of the following vectors: 

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

Answer:

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

Question:2  Write two different vectors having same magnitude

Answer:

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}

Question:3 Write two different vectors having same direction.

Answer:

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

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,

\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

Question:6 Find the sum of the vectors \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

Answer:

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

Question:7 Find the unit vector in the direction of the vector \vec a = \hat i + \hat j + 2 \hat k

Answer:

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}}

Question:8 Find the unit vector in the direction of vector \vec { 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 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}}

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

Answer:

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}}

Question:10 Find a vector in the direction of vector5 \hat i - \hat j + 2 \hat k which has magnitude 8 units.

Answer:

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}}

Question:11 Show that the vectors 2 \hat i -3 \hat j + 4 \hat k and - 4 \hat i + 6 \hat j - 8 \hat k  are collinear.

Answer:

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.

Question:12 Find the direction cosines of the vector \hat i + 2 \hat j + 3 \hat k

Answer:

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 )

Question:13 Find the direction cosines of the vector joining the points A(1, 2, –3) and B(–1, –2, 1), directed from A to B.

Answer:

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 )

Question:14 Show that the vector \hat i + \hat j + \hat k  is equally inclined to the axes OX, OY and OZ.

Answer:

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.

Question:15 (1)  Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are  i + 2 j - kand  - i + j + k respectively, in the ratio 2 : 1  internally

Answer:

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}

Question:15 (2) Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are \hat i + 2 \hat j - \hat k and  - \hat i + \hat j + \hat krespectively, in the ratio 2 : 1 externally

Answer:

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

Question:16 Find the position vector of the mid point of the vector joining the points P(2, 3, 4) and Q(4, 1, –2).

Answer:

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

Question:17 Show that the points A, B and C with position vectors, \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, respectively form the vertices of a right angled triangle.

Answer:

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.

Question:18 In triangle ABC (Fig 10.18), which of the following is not true: 

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

Answer:

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.

Question:19 If are two collinear vectors, then which of the following are incorrect:
(A) \vec b = \lambda \vec a for some saclar \lambda 
(B) \vec a = \pm \vec b
(C) the respective components of \vec a \: \:and \: \: \vec bare not proportional
(D) both the vectors \vec a \: \:and \: \: \vec b have same direction, but different magnitudes.

Answer:

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).

NCERT solutions for class 12 maths chapter 10 vector algebra-Exercise: 10.3

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

Answer:

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}.

Question:2 Find the angle between the vectors \hat i - 2 \hat j + 3 \hat k \: \:and \: \: 3 \hat i - 2 \hat j + \hat k

Answer:

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}

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

Answer:

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.

 

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 

\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}}

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

\\\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.

Question:6 Find |\vec a| \: \: and\: \:| \vec b | , if ( \vec a + \vec b ). ( \vec a - \vec b )=8 \: \:and \: \: |\vec a |\: \:= 8 \: \:|\vec b |.

Answer:

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}}

Question:7 Evaluate the product ( 3\vec a - 5 \vec b ). ( 2 \vec a + 7 \vec b ).

Answer:

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

Question:8 Find the magnitude of two vectors \vec a \: \: and \: \: \vec b , having the same magnitude and such that the angle between them is  60 \degree   and their scalar product is 1/2

Answer:

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

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

Answer:

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 \sqrt{13}

Question:10 If  \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 are such that \vec a + \lambda \vec b is perpendicular to \vec c  , then find the value of \lambda

Answer:

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,

Question:11 Show that   |\vec a | \vec b + |\vec b | \vec a  is perpendicular to   |\vec a | \vec b - |\vec b | \vec a , for any two nonzero vectors  \vec a \: \: \: and \: \: \vec b.

Answer:

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.

Question:12 If \vec a . \vec a = 0 \: \: and \: \: \vec a . \vec b = 0 , then what can be concluded about the vector  \vec b?

Answer:

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

Question:13 If  \vec a , \vec b , \vec c  are unit vectors such that \vec a + \vec b + \vec c = \vec 0, find the value of \vec a . \vec b + \vec b. \vec c + \vec c . \vec a 

Answer:

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 \frac{-3}{2}

Question:14 If either vector \vec a = 0 \: \: or \: \: \vec b = 0 \: \: then \: \: \vec a . \vec b = 0 . But the converse need not be true. Justify your answer with an example 

Answer:

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.

Question:15 If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2), respectively, then find \angle ABC , [\angle ABC is the angle between the vectors \overline{BA}\: \: and\: \: \overline{BC} ] .

Answer:

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

Question:16 Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear.

Answer:

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.

Question:17 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.

Answer:

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.

Question:18 If \vec a  is a nonzero vector of magnitude ‘a’ and  \lambda  a nonzero scalar, then \lambda \vec a is unit vector if 

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

Answer:

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

Question:1 Find |\vec a \times \vec b |, if \vec a = \hat i - 7 \hat j + 7 \hat k \: \: and \: \: \vec b = 3 \hat i - 2 \hat j + 2 \hat k

Answer:

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 |\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 |\vec a \times \vec b | is 19\hat j+19\hat k

Question:2 Find a unit vector perpendicular to each of the vector \vec a + \vec b \: \: and\: \: \vec a - \vec b , where \vec a = 3 \hat i + 2 \hat j + 2 \hat k \: \:and \: \: \vec b = \hat i + 2 \hat j - 2 \hat k

Answer:

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.

Question:3 If a unit vector \vec a makes angles \frac{\pi }{3}with\hat i , \frac{\pi }{4}   with \hat j and an acute angle \theta \: \: with \hat k  then find \theta \: \:  and hence, the components of \vec a.

Answer:

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 )

Question:4 Show that ( \vec a - \vec b ) \times (\vec a + \vec b ) = 2 ( \vec a \times \vec b )

Answer:

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.

Question:5 Find \lambda and \mu  if ( 2 \hat i + 6 \hat j + 27 \hat k ) \times ( \hat i + \lambda j + \mu \hat k ) = \vec 0

Answer:

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}

Question:6  Given that  \vec a . \vec b = 0 \: \:and \: \: \vec a \times \vec b = 0 and . What can you conclude about the vectors\vec a \: \:and \: \: \vec b ?

Answer:

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 bare 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

Question:7  Let the vectors \vec a , \vec b , \vec c  be given as \vec a_1 \hat i + \vec a_2 \hat j + \vec a_3 \hat k , \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   Then show that \vec a \times ( \vec b + \vec c ) = \vec a \times \vec b + \vec a \times \vec c

Answer:

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.

Question:8 If either \vec a = \vec 0 \: \: or \: \: \vec b = \vec 0  then \vec a \times \vec b = \vec 0  . Is the converse true? Justify your answer with an example.

Answer:

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. 

Question:9 Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).

Answer:

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 \frac{\sqrt{61}}{2} square units

Question:10  Find 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.

Answer:

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}

Question:11 Let the vectors \vec a \: \: and\: \: \vec b  be such that |\vec a| = 3 \: \: and\: \: |\vec b | = \frac{\sqrt 2 }{3} , then  \vec a \times \vec bis a unit vector, if the angle between is \vec a \: \:and \: \: \vec b 

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

Answer:

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.

Question:12 Area of a rectangle having vertices A, B, C and D with position vectors 

 - \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

Answer:

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

Question:1 Write down a unit vector in XY-plane, making an angle of 30 \degree with the positive direction of x-axis.

Answer:

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

Question:2 Find the scalar components and magnitude of the vector joining the points
       P(x_1, y_1, z_1) \: \: and \: \: Q(x_2, y_2, z_2).

Answer:

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)

Question:3 A girl walks 4 km towards west, then she walks 3 km in a direction 30 \degree east of north and stops. Determine the girl’s displacement from her initial point of departure.

Answer:

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

Question:4  If \vec a = \vec b + \vec c , then  is it true that |\vec a| =| \vec b |+| \vec c | ? Justify your answer. 

Answer:

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 |

Question:5  Find the value of x for which x ( \hat i+ \hat j + \hat k ) is a unit vector.

Answer:

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 \frac{1}{\sqrt{3}}

Question:6 Find a vector of magnitude 5 units, and parallel to the resultant of the vectors \vec a = 2 \hat i + 3 \hat j - \hat k \: \: and \: \: \vec b = \hat i - 2 \hat j + \hat k

Answer:

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

Question:7 If \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, find a unit vector parallel to the vector 2\vec a - \vec b + 3 \vec c.

Answer:

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

Question:8 Show that the points A(1, – 2, – 8), B(5, 0, –2) and C(11, 3, 7) are collinear, and find the ratio in which B divides AC.

Answer:

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.

Question:9 Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are ( 2 \vec a + \vec b ) \: \:and \: \: ( \vec a - 3 \vec b ) externally in the ratio 1: 2. Also, show that P is the mid point of the line segment RQ.

Answer:

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

Question:10 The two adjacent sides of a parallelogram are 2 \hat i - 4 \hat j + 5 \hat k \: \:and \: \: \hat i - 2 \hat j - 3 \hat k . Find the unit vector parallel to its diagonal. Also, find its area.

Answer:

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}.

Question:11 Show that the direction cosines of a vector equally inclined to the axes OX, OY  and OZ are \pm \left ( \frac{1}{\sqrt 3 } , \frac{1}{\sqrt 3 } , \frac{1}{\sqrt 3 } \right )

Answer:

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 )

Question:12 Let \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  . Find a vector \vec d which is perpendicular to both  \vec a \: \: and \: \: \vec b \: \: and \: \: \vec c . \vec d = 15

Answer:

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

Question:13 The scalar product of the vector \hat i + \hat j + \hat k with a unit vector along the sum of vectors2\hat i + 4 \hat j -5 \hat k  and \lambda \hat i + 2 \hat j +3 \hat k is equal to one. Find the value of \lambda.

Answer:

Let, the sum of vectors2\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

Question:14 If \vec a , \vec b , \vec c are mutually perpendicular vectors of equal magnitudes, show that the vector \vec a+\vec b +\vec c  is equally inclined to \vec a , \vec b \: \: and \: \: \vec c .

Answer:

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 cis 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 .

Question:15 Prove that ( \vec a + \vec b ) . (\vec a + \vec b ) = |\vec a ^2 | + |\vec b |^2 , if and only if  \vec a , \vec b are perpendicular, given \vec a \neq 0 , \vec b \neq 0

Answer:

Given in the question,

\vec a , \vec b are perpendicular and we need to prove that ( \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.

Question:16 Choose the correct answer If \theta is the angle between two vectors \vec a \: \: and \: \: \vec b  , then \vec a \cdot \vec b \geq 0 only when
     \\A ) 0 < \theta < \frac{\pi }{2} \\\\ \: \: \: \: B ) 0 \leq \theta \leq \frac{\pi }{2} \\\\ \: \: \: C ) 0 < \theta < \pi \\\\ \: \: \: D) 0 \leq \theta \leq\pi

Answer:

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}

Answer:

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.

Question:18 The value of  \hat i ( \hat j \times \hat k ) + \hat j ( \hat i \times \hat k ) + \hat k ( \hat i \times \hat j ) is

 (A) 0

(B) –1

(C) 1

(D) 3

Answer:

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.

Question:19  Choose the correct. If  \theta is the angle between any two vectors  \vec a \: \:and \: \: \vec b , then  |\vec a \cdot \vec b |=|\vec a \times \vec b | when \theta
is equal to

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

Answer:

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

NCERT solutions for class 12 subject wise

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.  

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