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# NCERT Solution for Class 12 Mathematics Chapter 10

NCERT Solution for Class 12 Mathematics Chapter 10, “Vector Algebra,” is an important study material designed to help students understand vector algebra’s fundamental concepts and principles. Swastik Classes, a leading coaching institute, has developed comprehensive NCERT solutions that provide step-by-step explanations and solved examples to help students develop a deeper understanding of the subject. The chapter covers the scalar and vector products of two and three-dimensional vectors, triple scalar product, and vector triple product. With the help of Swastik Classes’ NCERT solutions, students can improve their problem-solving skills and gain the confidence to tackle complex vector algebra problems. These solutions are also helpful for students who are preparing for competitive exams like JEE, NEET, and other entrance exams. Overall, Swastik Classes’ NCERT Solution for Class 12 Mathematics Chapter 10 is an essential resource for students who want to excel in mathematics and build a strong foundation in vector algebra.

## Answers of Mathematics NCERT solutions for class 12 Chapter 10 Vector

Exercise 10.1

Question 1:

Represent graphically a displacement of 40 km, 30° east of north.

Here, vector represents the displacement of 40 km, 30° East of North.

Question 2:

Classify the following measures as scalars and vectors.

1. 10 kg
2. 2 metres north-west
3. 40°
4. 40 Watt
5. 10–19 coulomb
6. 20 m/s2

(i) 10 kg is a scalar quantity because it involves only magnitude.

(ii) 2 meters north-west is a vector quantity as it involves both magnitude and direction.

(iii) 40° is a scalar quantity as it involves only magnitude.

(iv) 40 watts is a scalar quantity as it involves only magnitude.

(v) 10–19 coulomb is a scalar quantity as it involves only magnitude.

(vi) 20 m/s2 is a vector quantity as it involves magnitude as well as direction.

Question 3:

Classify the following as scalar and vector quantities.

1. time period
2. distance
3. force
4. velocity
5. work done

(i) Time period is a scalar quantity as it involves only magnitude.

(ii) Distance is a scalar quantity as it involves only magnitude.

(iii) Force is a vector quantity as it involves both magnitude and direction.

(iv) Velocity is a vector quantity as it involves both magnitude as well as direction.

(v) Work done is a scalar quantity as it involves only magnitude.

Question 4:

In Figure, identify the following vectors.

(i) Co initial (ii) Equal (iii) Collinear but not equal

(i) Vectors and are co initial because they have the same initial point

(ii) Vectors and are equal because they have the same magnitude and direction.

(iii) Vectors and are collinear but not equal. This is because although they areparallel, their directions are not the same.

Question 5:

Answer the following as true or false.

(i) and are collinear.

(ii) Two collinear vectors are always equal in magnitude.

(iii) Two vectors having same magnitude are collinear.

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

(i) True.

Vectors and are parallel to the same line.

(ii) False.

Collinear vectors are those vectors that are parallel to the same line.

(iii) False.

Two vectors having the same magnitude need not necessarily be  parallel to the same line.

(iv) False

Only if the magnitude and direction of two vectors are the same, regardless of the positions of their initial points the two vector are said to be equal.

Exercise 10.2

Question 1:

Compute the magnitude of the following vectors:

The given vectors are:

Question 2:

Write two different vectors having same magnitude.

Hence, are two different vectors having the same magnitude. The vectors are different because they have different directions.

Question 3:

Write two different vectors having same direction.

The direction cosines of are the same. Hence, the two vectors have the same direction.

Question 4:

Find the values of x and y so that the vectors are equal

The two vectors will be equal if their corresponding components are equal.

Hence, the required values of x and y are 2 and 3 respectively.

Question 5:

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

The vector with the initial point P(2,1) and terminal point Q(–5,7) can be given by,

Hence, the required scalar components are–7 and 6 while the vector components are and

Question 6:

Find the sum of the vectors .

The given vectors are .

Question 7:

Find the unit vector in the direction of the vector

The unit vector in the direction of vector is given by

Question 8:

Find the unit vector in the direction of vector , where P and Q are the points (1, 2, 3) and (4, 5, 6),respectively.

The given points are P (1, 2, 3) and Q (4, 5, 6).

Hence, the unit vector in the direction ofis

Question 9:

For given vectors, and , find the unit vector in the direction of the vector

The given vectors are and .

Hence, the unit vector in the direction of is

.

Question 10:

Find a vector in the direction of vector which has magnitude 8units.

Hence, the vector in the direction of vector which has magnitude 8 units is given by,

Question 11:

Show that the vectors are collinear.

.

Hence, the given vectors are collinear.

Question 12:

Find the direction cosines of the vector

Hence, the direction cosines of are

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.

The given points are A (1, 2, –3) and B (–1, –2, 1).

Hence, the direction cosines of are

Question 14:

Show that the vector is equally inclined to the axes OX, OY, and OZ.

Therefore, the direction cosines of

Now, letα, β, and γ be the angles formed by with the positive directions of x, y, and z axes.

Then, we have

Hence, the given vector is equally inclined to axes OX, OY, and OZ.

*Question 15:

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are respectively, in the ration2:1internallyexternally

The position vector of point R dividing the line segment joining two points P and Q in the ratio m: n is given by:

Internally:

Externally:

Position vectors of P and Q are given as:

ThepositionvectorofpointRwhichdividesthelinejoiningtwopointsPandQ internally in the ratio 2:1 is given by,

The position vector of point R which divides the line joining two points P and Q externally in the ratio 2:1 is given by,

*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).

The position vector of mid-point R of the vector joining points P (2, 3, 4) and Q (4, 1, –2) is given by,

*Question 17:

Show that the points A, B and C with position vectors, ,

respectively form the vertices of a right-angled triangle.

Position vectors of points A, B, and C are respectively given as:

Hence, ABC is a right-angled triangle.

*Question 18:

In triangle ABC which of the following is not true:

A.

B.

C.

D.

On applying the triangle law of addition in the given triangle, we have:

From equations (1) and (3), we have:

Hence, the equation given in alternative C is incorrect. The correct answer is C.

*Question 19:

If  are two collinear vectors, then which of the following are incorrect:

A.  , for some scalar λ

B.

C. the respective components of are proportional

D. both the vectors have same direction, but different magnitudes

If are two collinear vectors, then they are parallel. Therefore, we have:

(For some scalar λ)

If λ = ±1, then.

Thus, the respective components of are proportional. However, vectors can have different directions.

Hence, the statement given in D is incorrect.

Exercise 10.3

Question 1:

Find the angle between two vectors and with magnitudes and 2, respectively having .

It is given that,

Hence, the angel between the given vectors and is

Question 2:

Find the angle between the vectors

The given vectors are and

Also, we know that

Question 3:

Find the projection of the vector on the vector

Let and .

Now, projection of vector on is given by,

Hence, the projection of vector on is 0.

Question 4:

Find the projection of the vector on the vector

Let  and .

Now, projection of vector on is given by,

Question 5:

Show that each of the given three vectors is a unit vector:

Also, show that they are mutually perpendicular to each other.

Thus, each of the given three vectors is a unit vector.

Hence, the given three vectors are mutually perpendicular to each other.

Question 6:

Find and  , if .

Question 7:

Evaluate the product

Question 8:

Find the magnitude of two vectors , having the same magnitude and such that the angle between them is 60° and their scalar product is

Let θ be the angle between the vectors It is given that

We know that

Question 9:

Find , if for a unit vector

Question 10:

If are such that is perpendicular to , then find the value of λ.

Hence, the required value of λ is 8.

Question 11:

Show that is perpendicular to  , for any two nonzero vectors

Hence,  and are perpendicular to each other.

Question 12:

If, then what can be concluded about the vector ?

It is given that.

Hence, vector satisfying can be any vector.

*Question 13:

If and are unit vectors such that find the value of

Consider the given vectors,

Given

So,

[Distributivity of scalar product over addition]

…..(1)

Next

….(2)

And,

…..(3)

From (1), (2) and (3),

[Scalar product is commutative]

Hence the value is

*Question 14:

If either vector , then . But the converse need not be true. Justify your answer with an example.

We now observe that:

Hence, the converse of the given statement need not be 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 ∠ABC. [∠ABC is the angle between the vectors and

The vertices of OABC are given as A(1,2,3), B(–1,0,0), and C (0,1,2). Also, it is given that ∠ABC is the angle between the vectors and .

Now, it is known that:

.

*Question 16:

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

The given points are A (1, 2, 7), B (2, 6, 3), and C (3, 10, –1).

Hence, the given points A, B, and C are collinear.

*Question 17:

Show that the vectors form the vertices of a right-angled triangle.

Let vectors be position vectors of points A, B, and C respectively.

Now, vectors represent the sides of OABC.

Hence, OABC is a right-angled triangle.

*Question 18:

If is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λis unit vector if

(A) λ = 1

(B) λ = –1

(C)

(D)

Vector is a unit vector if

Hence, vector is a unit vector if The correct answer is D.

Exercise 10.4

Question 1:

Find if and

We have,

and

Question 2:

Find a unit vector perpendicular to each of the vector and , where and .

We have,

and

Hence, the unit vector perpendicular to each of the vectors and is given by the relation,

Question 3:

If a unit vector makes an angles with with and an acute angle θ with then find θ and hence, the compounds of

Let unit vector have (a1, a2, a3) components.

Since  is a unit vector, .

Also, it is given that makes an angles with with and an acute angle θ with then we have:

Hence, and the components of are

Question 4:

Show that

Question 5:

Find λ and µ if

On comparing the corresponding components, we have:

Hence,

Question 6:

Given that and  . What can you conclude about the vectors ?

Then,

Either or  , or

Either  or , or . But, and  cannot be perpendicular and parallel simultaneously.

Hence, or .

*Question 7:

Let the vectors given as . Then show that

We have,

On adding (2) and (3), we get:

Now, from (1) and (4), we have:

Hence, the given result is proved.

*Question 8:

If either or , then Is the converse true? Justify your answer with an example.

Take any parallel non-zero vectors so that.

It can now be observed that:

Hence, the converse of the given statement need not be true.

*Question 9:

Find the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and

C (1, 5, 5).

The vertices of triangle ABC are given as A (1, 1, 2), B (2, 3, 5), and

C (1, 5, 5).

The adjacent sides and of OABC are given as:

Hence, the area of OABC

*Question 10:

Find the area of the parallelogram whose adjacent sides are determined by the vector

The area of the parallelogram whose adjacent sides are is Adjacent sides are given as:

Hence, the area of the given parallelogram is .

*Question 11:

Let the vectors and be such that and then, is a unit vectors, if the angle between and is

(A) (B) (C) (D)

It is given that and

We know that , where is a unit vector perpendicular to both. and and θ is the angle between and

Now is a unit vector if

Hence, is a unit vector if the angle between and is . The correct answer is B.

*Question 12:

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

and  respectively is

(A) (B) 1 (C) 2 (D) 4

The position vectors of vertices A, B, C, and D of rectangle ABCD are given as:

The adjacent sides and of the given rectangle are given as:

Now, it is known that the area of a parallelogram whose adjacent sides are is .

Hence, the area of the given rectangle is The correct answer is C.

Miscellaneous Solutions

Question 1:

Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of x-axis.

If is a unit vector in the XY-plane, then

Here, θ is the angle made by the unit vector with the positive direction of the x-axis. Therefore, for θ = 30°:

Hence, the required unit vector is .

Question 2:

Find the scalar components and magnitude of the vector joining the points

.

The vector joining the points can be obtained by,

Hence, the scalar components and the magnitude of the vector joining the given points are respectively

and .

Question 3:

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

Let O and B be the initial and final positions of the girl respectively. Then, the girl’s position can be shown as:

Now, we have:

By the triangle law of vector addition, we have:

Hence, the girl’s displacement from her initial point of departure is

Question 4:

Now, by the triangle law of vector addition, we have .

It is clearly known that represent the sides of OABC.

Also, it is known that the sum of the lengths of any two sides of a triangle is greater than the third side.

Hence, it is not true that .

Question 5:

Find the value of x for which is a unit vector.

is a unit vector

Hence, the required value of x is

Question 6:

Find a vector of magnitude 5 units, and parallel to the resultant of the vectors

We have,

Let be the resultant of

Hence, the vector of magnitude 5 units and parallel to the resultant of vectors is

Question 7:

If , find a unit vector parallel to the

Vector .

We have,

Hence, the unit vector along is

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.

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

Thus, the given points A, B, and C are collinear.

Now, let point B divide AC in the ratio . Then, we have:

On equating the corresponding components, we get:

Hence, point B divides AC in the ratio

Question 9:

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

It is given that

.

It is given that point R divides a line segment joining two points P and Q externally in the ratio 1: 2. Then, on using the section formula, we get:

Therefore, the position vector of point R is . Position vector of the mid-point of RQ=

Hence, P is the mid-point of the line segment RQ.

Question 10:

The two adjacent sides of a parallelogram are and . Find the unit vector parallel to its diagonal. Also, find its area.

Adjacent sides of a parallelogram are given as: and Then, the diagonal of a parallelogram is given by .

Thus, the unit vector parallel to the diagonal is

Hence, the area of the parallelogram is square units.

*Question 11:

Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are

Let a vector be equally inclined to axes OX, OY, and OZ at angle α. Then, the direction cosines of the vector are cos α, cos α, and cos α.

Hence, the direction cosines of the vector which are equally inclined to the axes are

*Question 12:

Let and . Find a vector which is perpendicular to both and , and.

Let

.

Since is perpendicular to both , we have:

Also, it is given that:

On solving (i), (ii), and (iii), we get:

Hence, the required vector is.

*Question 13:

The scalar product of the vector with a unit vector along the sum of vectors

and is equal to one. Find the value of

Scalar product of with this unit vector is 1.

Hence, the value of λ is 1.

*Question 14:

If are mutually perpendicular vectors of equal magnitudes, show that the vector is equally inclined to

Since  are mutually perpendicular vectors, we have

It is given that:

Let vector be inclined to  at angles respectively. Then, we have:

Now, as , .

Hence, the vector is equally inclined

*Question 15:

Prove that , if and only if are perpendicular, given

*Question 16:

If θ is the angle between two vectors and , then only when

(A) (B) (C) (D)

Let θ be the angle between two vectors and.

Then, without loss of generality, and are non-zero vectors so that

Hence when The correct answer is B.

*Question 17:

Let and be two unit vectors and θ is the angle between them. Then is a unit vector if

(A) (B)

(C) (D)

Let and  be two unit vectors and θ be the angle between them.

Then,

Now, is a unit vector if

Hence, is a unit vector if  .  The correct answer is D.

*Question 18:

The value of is

(A) 0

(B) –1

(C) 1

(D) 3

*Question 19:

If θ is the angle between any two vectors and , then when θ is equal to

(A) 0 (B) (C) (D)

## Conclusion

Swastik Classes’ NCERT Solution for Class 12 Mathematics Chapter 10, “Vector Algebra,” is a comprehensive study material designed to help students understand the fundamental concepts and principles of vector algebra. The solutions provide step-by-step explanations and solved examples that help students to develop a deeper understanding of the subject. The chapter covers a range of topics including scalar and vector products of two and three-dimensional vectors, triple scalar product, and vector triple product. With the help of these solutions, students can improve their problem-solving skills and gain the confidence to tackle complex vector algebra problems. Swastik Classes’ NCERT solutions are designed in accordance with the latest CBSE syllabus, making them useful for students preparing for board exams or competitive exams like JEE and NEET. Overall, Swastik Classes’ NCERT Solution for Class 12 Mathematics Chapter 10 is an excellent resource for students who want to excel in mathematics and build a strong foundation in vector algebra.

The finest thing the NCERT answers Chapter 10 vector algebra is that it conveys tough parts in vernacular and easy language so that students of all intellect levels can understand them.

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One of the top IIT JEE coaching institutes is Swastik Classes. Shobhit Bhaiya and Alok Bhaiya, pioneering mentors of IIT JEE Coaching Classes, started Swastik Classes in Anand Vihar. Over the last 15 years, they have educated and sent over 2000+ students to IITs and 5000+ students to different famous universities such as BITS, NITs, DTU, and NSIT. When it comes to coaching programmes for IIT JEE, Swastik Classes is the top IIT JEE Coaching in Delhi, favoured by students from all over India.

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## FAQS on NCERT Solutions for Class 12 Maths Chapter 10 – Vector Algebra

### What is meant by vector algebra?

an algebra in which the components involved might be vectors and the assumptions and rules are based on vector behaviour.

5 exercises

### What are 3 types of vectors?

• Null Vector or Zero Vector

When the magnitude of a vector is 0 and the starting point and terminus of the vector are the same, the vector is said to be a Zero Vector. PQ, for example, is a line segment in which the coordinates of point P and point Q are the same. The symbol for a zero vector is 0. There is no fixed orientation for the zero vector.

• Vector of a unit

When the magnitude of a vector is 1 unit in length, it is said to be a unit vector. If x is a vector of magnitude x, then the unit vector is denoted by x in the vector’s direction and has a magnitude equal to 1.

However, two-unit vectors cannot be equal since their directions may differ.

• Travelling position

A position vector is defined as a point X in the plane. It just indicates the current location. Assume OX is a point in a plane with respect to its origin. If O is used as the reference origin and X is an arbitrary point in the plane, the vector is referred to as the point’s position vector.

### What is an equal vector?

When the magnitude and direction of two or more vectors are the same, they are said to be equal.

### What is the triangular law of addition?

The triangle rule of vector addition asserts that when two vectors are represented as two sides of a triangle with the same order of magnitude and direction, the magnitude and direction of the resulting vector is represented by the third side of the triangle.

### What is the resultant of vector?

The vector sum of two or more vectors is the outcome. It’s the outcome of combining two or more vectors. When you put the displacement vectors A, B, and C together, you obtain vector R. Vector R may be found using an appropriately drawn, scaled vector addition diagram, as illustrated in the picture.