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NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra

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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.

Also Refer,

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.

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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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

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

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.


As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.


Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.


Believe in Yourself! You can make anything happen


All the very best.

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects  and  we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

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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

Bio Medical Engineer

The field of biomedical engineering opens up a universe of expert chances. An Individual in the biomedical engineering career path work in the field of engineering as well as medicine, in order to find out solutions to common problems of the two fields. The biomedical engineering job opportunities are to collaborate with doctors and researchers to develop medical systems, equipment, or devices that can solve clinical problems. Here we will be discussing jobs after biomedical engineering, how to get a job in biomedical engineering, biomedical engineering scope, and salary. 

4 Jobs Available
Data Administrator

Database professionals use software to store and organise data such as financial information, and customer shipping records. Individuals who opt for a career as data administrators ensure that data is available for users and secured from unauthorised sales. DB administrators may work in various types of industries. It may involve computer systems design, service firms, insurance companies, banks and hospitals.

4 Jobs Available
Ethical Hacker

A career as ethical hacker involves various challenges and provides lucrative opportunities in the digital era where every giant business and startup owns its cyberspace on the world wide web. Individuals in the ethical hacker career path try to find the vulnerabilities in the cyber system to get its authority. If he or she succeeds in it then he or she gets its illegal authority. Individuals in the ethical hacker career path then steal information or delete the file that could affect the business, functioning, or services of the organization.

3 Jobs Available
Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

3 Jobs Available
Geothermal Engineer

Individuals who opt for a career as geothermal engineers are the professionals involved in the processing of geothermal energy. The responsibilities of geothermal engineers may vary depending on the workplace location. Those who work in fields design facilities to process and distribute geothermal energy. They oversee the functioning of machinery used in the field.

3 Jobs Available
Remote Sensing Technician

Individuals who opt for a career as a remote sensing technician possess unique personalities. Remote sensing analysts seem to be rational human beings, they are strong, independent, persistent, sincere, realistic and resourceful. Some of them are analytical as well, which means they are intelligent, introspective and inquisitive. 

Remote sensing scientists use remote sensing technology to support scientists in fields such as community planning, flight planning or the management of natural resources. Analysing data collected from aircraft, satellites or ground-based platforms using statistical analysis software, image analysis software or Geographic Information Systems (GIS) is a significant part of their work. Do you want to learn how to become remote sensing technician? There's no need to be concerned; we've devised a simple remote sensing technician career path for you. Scroll through the pages and read.

3 Jobs Available
Geotechnical engineer

The role of geotechnical engineer starts with reviewing the projects needed to define the required material properties. The work responsibilities are followed by a site investigation of rock, soil, fault distribution and bedrock properties on and below an area of interest. The investigation is aimed to improve the ground engineering design and determine their engineering properties that include how they will interact with, on or in a proposed construction. 

The role of geotechnical engineer in mining includes designing and determining the type of foundations, earthworks, and or pavement subgrades required for the intended man-made structures to be made. Geotechnical engineering jobs are involved in earthen and concrete dam construction projects, working under a range of normal and extreme loading conditions. 

3 Jobs Available
Cartographer

How fascinating it is to represent the whole world on just a piece of paper or a sphere. With the help of maps, we are able to represent the real world on a much smaller scale. Individuals who opt for a career as a cartographer are those who make maps. But, cartography is not just limited to maps, it is about a mixture of art, science, and technology. As a cartographer, not only you will create maps but use various geodetic surveys and remote sensing systems to measure, analyse, and create different maps for political, cultural or educational purposes.

3 Jobs Available
Budget Analyst

Budget analysis, in a nutshell, entails thoroughly analyzing the details of a financial budget. The budget analysis aims to better understand and manage revenue. Budget analysts assist in the achievement of financial targets, the preservation of profitability, and the pursuit of long-term growth for a business. Budget analysts generally have a bachelor's degree in accounting, finance, economics, or a closely related field. Knowledge of Financial Management is of prime importance in this career.

4 Jobs Available
Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Underwriter

An underwriter is a person who assesses and evaluates the risk of insurance in his or her field like mortgage, loan, health policy, investment, and so on and so forth. The underwriter career path does involve risks as analysing the risks means finding out if there is a way for the insurance underwriter jobs to recover the money from its clients. If the risk turns out to be too much for the company then in the future it is an underwriter who will be held accountable for it. Therefore, one must carry out his or her job with a lot of attention and diligence.

3 Jobs Available
Finance Executive
3 Jobs Available
Operations Manager

Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.

3 Jobs Available
Investment Director

An investment director is a person who helps corporations and individuals manage their finances. They can help them develop a strategy to achieve their goals, including paying off debts and investing in the future. In addition, he or she can help individuals make informed decisions.

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
Transportation Planner

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

3 Jobs Available
Plumber

An expert in plumbing is aware of building regulations and safety standards and works to make sure these standards are upheld. Testing pipes for leakage using air pressure and other gauges, and also the ability to construct new pipe systems by cutting, fitting, measuring and threading pipes are some of the other more involved aspects of plumbing. Individuals in the plumber career path are self-employed or work for a small business employing less than ten people, though some might find working for larger entities or the government more desirable.

2 Jobs Available
Construction Manager

Individuals who opt for a career as construction managers have a senior-level management role offered in construction firms. Responsibilities in the construction management career path are assigning tasks to workers, inspecting their work, and coordinating with other professionals including architects, subcontractors, and building services engineers.

2 Jobs Available
Urban Planner

Urban Planning careers revolve around the idea of developing a plan to use the land optimally, without affecting the environment. Urban planning jobs are offered to those candidates who are skilled in making the right use of land to distribute the growing population, to create various communities. 

Urban planning careers come with the opportunity to make changes to the existing cities and towns. They identify various community needs and make short and long-term plans accordingly.

2 Jobs Available
Highway Engineer

Highway Engineer Job Description: A Highway Engineer is a civil engineer who specialises in planning and building thousands of miles of roads that support connectivity and allow transportation across the country. He or she ensures that traffic management schemes are effectively planned concerning economic sustainability and successful implementation.

2 Jobs Available
Environmental Engineer

Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems. 

2 Jobs Available
Naval Architect

A Naval Architect is a professional who designs, produces and repairs safe and sea-worthy surfaces or underwater structures. A Naval Architect stays involved in creating and designing ships, ferries, submarines and yachts with implementation of various principles such as gravity, ideal hull form, buoyancy and stability. 

2 Jobs Available
Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
Veterinary Doctor
5 Jobs Available
Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
Speech Therapist
4 Jobs Available
Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth. 

4 Jobs Available
Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
Audiologist

The audiologist career involves audiology professionals who are responsible to treat hearing loss and proactively preventing the relevant damage. Individuals who opt for a career as an audiologist use various testing strategies with the aim to determine if someone has a normal sensitivity to sounds or not. After the identification of hearing loss, a hearing doctor is required to determine which sections of the hearing are affected, to what extent they are affected, and where the wound causing the hearing loss is found. As soon as the hearing loss is identified, the patients are provided with recommendations for interventions and rehabilitation such as hearing aids, cochlear implants, and appropriate medical referrals. While audiology is a branch of science that studies and researches hearing, balance, and related disorders.

3 Jobs Available
Hospital Administrator

The hospital Administrator is in charge of organising and supervising the daily operations of medical services and facilities. This organising includes managing of organisation’s staff and its members in service, budgets, service reports, departmental reporting and taking reminders of patient care and services.

2 Jobs Available
Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

4 Jobs Available
Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
Videographer
2 Jobs Available
Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

2 Jobs Available
Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

5 Jobs Available
Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

3 Jobs Available
Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
Linguist

Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning). 

Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian

2 Jobs Available
Public Relation Executive
2 Jobs Available
Travel Journalist

The career of a travel journalist is full of passion, excitement and responsibility. Journalism as a career could be challenging at times, but if you're someone who has been genuinely enthusiastic about all this, then it is the best decision for you. Travel journalism jobs are all about insightful, artfully written, informative narratives designed to cover the travel industry. Travel Journalist is someone who explores, gathers and presents information as a news article.

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
QA Manager
4 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Production Manager
3 Jobs Available
Merchandiser
2 Jobs Available
QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

2 Jobs Available
Metallurgical Engineer

A metallurgical engineer is a professional who studies and produces materials that bring power to our world. He or she extracts metals from ores and rocks and transforms them into alloys, high-purity metals and other materials used in developing infrastructure, transportation and healthcare equipment. 

2 Jobs Available
Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

4 Jobs Available
QA Manager
4 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
ITSM Manager
3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
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