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NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra

Edited By Ramraj Saini | Updated on Sep 19, 2023 09:16 AM IST | #CBSE Class 12th

NCERT Vector Algebra Class 12 Questions And Answers

NCERT Solutions for Class 12 Maths Chapter 10 provided here. The name of this chapter is Vector Algebra. NCERT Solutions for Class 12 Maths Chapter 10 are explained in a detailed manner to help students prepare for their board exams and competitive exams. Important topics that are going to be discussed in Vector Algebra class 12 are vector quantities, operations on vectors, geometric properties, and algebraic properties like addition, multiplication, etc. Questions related to all these topics are covered in the Vector Algebra Class 12 NCERT solutions. Check all NCERT solutions from classes 6 to 12 in a single place, which will help in better understanding of concepts in a much easier way. Also, check NCERT solutions for class 12 also.

Vector Algebra not only helps to solve problems in Mathematics. It is also helps in solving problems of Class 11 and 12 Physics also. Students may be familiar with some of the concepts discussed in Vectors Class 12 as Class 11 Physics also discuss the concepts of vectors. The concepts studied in the Class 12 Maths ch 10 are also used in the upcoming chapter Three Dimensional Geometry.

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NCERT Vector Algebra Class 12 Questions And Answers PDF Free Download

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NCERT Vector Algebra Class 12 Questions And Answers - Important Formulae

For vector a represented as a = xi + yj + zk The magnitude (length) of the vector is given by:

|a| = √(x² + y² + z²)

Vector Addition:

A + B = B + A (Commutative Law)

A + (B + C) = (A + B) + C (Associative Law)

Dot Product (Scalar Product) of Vectors:

A • B = |A| |B| cos θ (Where θ is the angle between vectors A and B)

Cross Product (Vector Product) of Vectors:

A × B = |A| |B| sin θ (Where θ is the angle between vectors A and B)

Scalar Multiplication:

k(A + B) = kA + kB

Additive Identity: A + 0 = 0 + A

Free download NCERT Vector Algebra Class 12 Solutions for CBSE Exam.

NCERT Class 12 Maths Chapter 10 Question Answer (Intext Questions and Exercise)

Class 12 Maths Chapter 10 NCERT Solutions 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 degrees east of north as shown in the figure.

1626668157287

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

1626668486601

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


Class 12 Maths Chapter 10 NCERT Solutions 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 vector 5 \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 - k and - 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 k respectively, 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:

15948360623461594836059639

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 b are 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 1517995000744796 and 1517995001522472 are proportional,
Therefore, (c) is not true.

D) The vectors 1517995002319114 and 1517995003131589 can have different magnitude as well as different directions.

Therefore, (d) is not true.

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


Class 12 Maths Chapter 10 NCERT Solutions 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.


Class 12 vector algebra NCERT solutions - 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 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

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 b is 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.


Class 12 vector algebra NCERT Solutions - 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 vectors 2\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 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 =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 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 .

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+|\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

Hence option D is correct.

Vector Algebra Class 12 - Topics

The main topics that will cover while practising Vector Algebra Class 12 NCERT solutions are:

  • Concepts of Vectors
  • Position Vector, Magnitude and Direction of Vectors and Unit Vector
  • Direction Cosines and Ratios
  • Addition and Subtraction of Vectors
  • Collinear Vectors
  • Dot Product and Cross Product of Vectors
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NCERT Vector Algebra Class 12 Solutions

A total of 5 exercises are covered in Class 12 Maths chapter 10 Vector Algebra. That are four exercises and one miscellaeous exercise. The exercises are listed below. Students can doenload the solutions for Individual exercises also. There are a total of 73 questions from these 5 exercises.

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NCERT solutions for class 12 maths - Chapter wise

Key Features of NCERT Solutions for Class 12 Maths Chapter 10 – Vector Algebra

NCERT Solutions for ch 10 maths class 12 - Vector Algebra is a comprehensive guide that helps students understand the concepts of vector algebra. Some of the key features of these solutions are:

  1. Comprehensive coverage: The vectors class 12 ncert solutions cover all the topics included in the Class 12 Maths syllabus, ensuring that students are well-prepared for their exams.

  2. Simplified language: The vectors maths class 12 solutions are written in simple language, making it easy for students to understand the concepts of vector algebra.

  3. Detailed explanations: The solutions provide detailed explanations of concepts, which help students to understand the fundamental principles of vector algebra.

  4. Step-by-step approach: The algebra chapters of class 12 solutions follow a step-by-step approach, which helps students to understand the solution process in a structured way.

  5. Illustrated solutions: The solutions are accompanied by diagrams and illustrations, which help students to visualize the solution process and understand the concepts better.

  6. Conceptual clarity: The solutions aim to develop the conceptual clarity of students, rather than just providing them with the final answers. This helps students to build a strong foundation in the subject.

  7. Previous years' question papers: The solutions also include the solutions to previous years' question papers, and miscellaneous exercise chapter 10 class 12, which help students to get a better understanding of the exam pattern and types of questions asked in the exams.

NCERT solutions for class 12 - Subject Wise

NCERT Solutions Class Wise

What are Vectors and Scalars

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.

Benefits of NCERT vector algebra class 12 solutions

  • NCERT solutions for maths chapter 10 class 12 are explained in a step-by-step manner, so it will be very easy for you to understand the concepts.

  • NCERT Solutions for maths chapter 10 class 12 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 NCERT Class 12 Maths solutions chapter 10 vector algebra article, you will get a solution of miscellaneous exercise also.

NCERT Books and NCERT Syllabus

Frequently Asked Question (FAQs)

1. What is the weightage of the chapter vector algebra for CBSE board exam?

Two chapters combined vector algebra & three-dimensional geometry has 17% weightage in 12th board maths final exam. So the vector algebra class 12 NCERT solutions can be practised to get a good score in the CBSE board exam. Students can solve problems related to the NCERT syllabus of vector algebra using NCERT solutions and NCERT Exemplar solutions

2. What are the important topics in chapter vector algebra?

Some basic concepts of vectors, addition and multiplication of a vector by a scalar, vector joining two points, scalar (or dot) product of two vectors, projection of a vector on a line and vector (or cross) product of two vectors are the important topics of this chapter.    

3. Where can I find solutions to the exercises for NCERT Solutions for class 12 math chapter 10?

Choosing appropriate study material is essential for chapter 10 class 12 Maths, as it aids in the efficient resolution of textbook problems. It requires considerable patience to choose the appropriate reference guide from the various options available in the market. Careers360 provides solutions for both chapter-wise and exercise-wise problems in PDF format. Students can use this resource to clear their doubts instantly while solving problems. For ease students can study vector algebra class 12 pdf both online and offline.

4. Where can I find the complete solutions of NCERT for class 12 maths?

A Here you will get the detailed NCERT solutions for class 12 maths  by clicking on the link.

5. Which is the best book for CBSE class 12 Maths?

NCERT textbook is the best book for CBSE class 12 maths. You don't need to buy any supplementary book. All you need to do is rigorous practice of all the questions given in the NCERT textbook.

6. Which are the most difficult chapters of NCERT Class 12 Maths syllabus?

Most of the students consider integration and applications of integrations as the most difficult chapters in CBSE class 12 maths but with the regular practice of NCERT questions you will be able to have a strong grip on them also.

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  • Passing NIOS in October 2024 will make you eligible for NIT admissions in 2025 . NIT admissions are based on your performance in entrance exams like JEE Main, which typically happen in January and April. These exams consider the previous year's Class 12th board results (or equivalent exams like NIOS).

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  • Focus on WAT and PI: If you receive a shortlist, prepare extensively for the Written Ability Test (WAT) and Personal Interview (PI). These stages assess your communication, soft skills, leadership potential, and suitability for the program.

  • Work Experience (if applicable): If you have work experience, highlight your achievements and how they align with your chosen IIM Bangalore program.

Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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