NCERT Solutions for Class 12 Science Math Chapter 4 Vector Algebra are provided here with simple step-by-step explanations. These solutions for Vector Algebra are extremely popular among Class 12 Science students for Math Vector Algebra Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 12 Science Math Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class Class 12 Science Math are prepared by experts and are 100% accurate.

Page No 428:

Question 1:

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

Answer:

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

Page No 428:

Question 2:

Classify the following measures as scalars and vectors.

(i) 10 kg (ii) 2 metres north-west (iii) 40°

(iv) 40 watt (v) 10–19 coulomb (vi) 20 m/s2

Answer:

(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.

Page No 428:

Question 3:

Classify the following as scalar and vector quantities.

(i) time period (ii) distance (iii) force

(iv) velocity (v) work done

Answer:

(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.

Page No 428:

Question 4:

In Figure, identify the following vectors.

(i) Coinitial (ii) Equal (iii) Collinear but not equal

Answer:

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

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

(iii) Vectorsand are collinear but not equal. This is because although they are parallel, their directions are not the same.

Page No 428:

Question 5:

Answer the following as true or false.

(i) andare 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.

Answer:

(i) True.

Vectors andare parallel to the same line.

(ii) False.

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

(iii) False.

It is not necessary for two vectors having the same magnitude to be parallel to the same line.

(iv) False.

Two vectors are said to be equal if they have the same magnitude and direction, regardless of the positions of their initial points.



Page No 440:

Question 1:

Compute the magnitude of the following vectors:

Answer:

The given vectors are:

Page No 440:

Question 2:

Write two different vectors having same magnitude.

Answer:

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

Page No 440:

Question 3:

Write two different vectors having same direction.

Answer:

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

Page No 440:

Question 4:

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

Answer:

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.

Page No 440:

Question 5:

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

Answer:

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

Page No 440:

Question 6:

Find the sum of the vectors.

Answer:

The given vectors are.

Page No 440:

Question 7:

Find the unit vector in the direction of the vector.

Answer:

The unit vector in the direction of vector is given by.

Page No 440:

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.

Answer:

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

Hence, the unit vector in the direction of is

.

Page No 440:

Question 9:

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

Answer:

The given vectors are and.

Hence, the unit vector in the direction of is
a+ba+b=i^+k^2=12i+12k.

Page No 440:

Question 10:

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

Answer:

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

Page No 440:

Question 11:

Show that the vectorsare collinear.

Answer:

.

Hence, the given vectors are collinear.

Page No 440:

Question 12:

Find the direction cosines of the vector

Answer:

Hence, the direction cosines of

Page No 440:

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:

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

Hence, the direction cosines of are

Page No 440:

Question 14:

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

Answer:

Therefore, the direction cosines of

Now, let α, β, and γbe 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.

Page No 440:

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 ration 2:1

(i) internally

(ii) externally

Answer:

The position vector of point R dividing the line segment joining two points

P and Q in the ratio m: n is given by:

  1. Internally:

  1. Externally:

Position vectors of P and Q are given as:

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

(ii) 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,



Page No 441:

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:

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

Page No 441:

Question 17:

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

Answer:

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



AB2+CA2=35+6=41=BC2


Hence, ABC is a right-angled triangle.

Page No 441:

Question 18:

In triangle ABC which of the following is not true:

A.

B.

C.

D.

Answer:

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.

Page No 441:

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

Answer:

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.

The correct answer is D.



Page No 447:

Question 1:

Find the angle between two vectorsandwith magnitudesand 2, respectively having.

Answer:

It is given that,

Now, we know that.

Hence, the angle between the given vectors andis.

Page No 447:

Question 2:

Find the angle between the vectors

Answer:

The given vectors are.

Also, we know that.

Page No 447:

Question 3:

Find the projection of the vectoron the vector.

Answer:

Letand.

Now, projection of vectoronis given by,

Hence, the projection of vector onis 0.

Page No 447:

Question 4:

Find the projection of the vectoron the vector.

Answer:

Letand.

Now, projection of vectoronis given by,

Page No 447:

Question 5:

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

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

Answer:

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

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



Page No 448:

Question 6:

Findand, if.

Answer:

Page No 448:

Question 7:

Evaluate the product.

Answer:

Page No 448:

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.

Answer:

Let θ be the angle between the vectors

It is given that

We know that.

Page No 448:

Question 9:

Find, if for a unit vector.

Answer:

Page No 448:

Question 10:

Ifare such thatis perpendicular to, then find the value of λ.

Answer:

Hence, the required value of λ is 8.

Page No 448:

Question 11:

Show that is perpendicular to, for any two nonzero vectors

Answer:

Hence, andare perpendicular to each other.

Page No 448:

Question 12:

If, then what can be concluded about the vector?

Answer:

It is given that.

Hence, vectorsatisfyingcan be any vector.

Page No 448:

Question 13:

If are unit vectors such that , find the value of .

Answer:

It is given that .

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

Page No 448:

Question 14:

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

Answer:

We now observe that:

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

Page No 448:

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 vectorsand]

Answer:

The vertices of ΔABC 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 vectorsand.

Now, it is known that:

.

Page No 448:

Question 16:

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

Answer:

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.

Page No 448:

Question 17:

Show that the vectorsform the vertices of a right angled triangle.

Answer:

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

Now, vectorsrepresent the sides of ΔABC.

Hence, ΔABC is a right-angled triangle.

Page No 448:

Question 18:

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

(A) λ = 1 (B) λ = –1 (C)

(D)

Answer:

Vectoris a unit vector if.

Hence, vectoris a unit vector if.

The correct answer is D.



Page No 454:

Question 1:

Find, if and.

Answer:

We have,

and

Page No 454:

Question 2:

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

Answer:

We have,

and

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

Page No 454:

Question 3:

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

Answer:

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

Since is a unit vector, .

Also, it is given that makes angleswith with , and an acute angle θ with

Then, we have:

Hence, and the components of are.

Page No 454:

Question 4:

Show that

Answer:

Page No 454:

Question 5:

Find λ and μ if .

Answer:

On comparing the corresponding components, we have:

Hence,

Page No 454:

Question 6:

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

Answer:

Then,

(i) Either or, or

(ii) Either or, or

But, and cannot be perpendicular and parallel simultaneously.

Hence, or.

Page No 454:

Question 7:

Let the vectors given as. Then show that

Answer:

We have,

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

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

Hence, the given result is proved.

Page No 454:

Question 8:

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

Answer:

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.

Page No 454:

Question 9:

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

C (1, 5, 5).

Answer:

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

C (1, 5, 5).

The adjacent sidesand of ΔABC are given as:

Area of ΔABC

Hence, the area of ΔABC



Page No 455:

Question 10:

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

Answer:

The area of the parallelogram whose adjacent sides are is.

Adjacent sides are given as:

Hence, the area of the given parallelogram is.

Page No 455:

Question 11:

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

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

Answer:

It is given that.

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.

Page No 455:

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)

Answer:

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:

AB×BC=2

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.



Page No 458:

Question 1:

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

Answer:

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.

Page No 458:

Question 2:

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

.

Answer:

The vector joining the pointscan be obtained by,

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

Page No 458:

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.

Answer:

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

.

Page No 458:

Question 4:

If, then is it true that? Justify your answer.

Answer:

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

It is clearly known that represent the sides of ΔABC.

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.

Page No 458:

Question 5:

Find the value of x for whichis a unit vector.

Answer:

is a unit vector if.

Hence, the required value of x is.

Page No 458:

Question 6:

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

.

Answer:

We have,

Letbe the resultant of.

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

Page No 458:

Question 7:

If, find a unit vector parallel to the vector.

Answer:

We have,

Hence, the unit vector alongis

Page No 458:

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:

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

Page No 458:

Question 9:

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

Answer:

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.

Page No 458:

Question 10:

The two adjacent sides of a parallelogram areand .

Find the unit vector parallel to its diagonal. Also, find its area.

Answer:

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

Area of parallelogram ABCD =

Hence, the area of the parallelogram issquare units.

Page No 458:

Question 11:

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

Answer:

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.

Page No 458:

Question 12:

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

Answer:

Let.

Sinceis perpendicular to bothand, we have:

Also, it is given that:

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

Hence, the required vector is.

Page No 458:

Question 13:

The scalar product of the vectorwith a unit vector along the sum of vectors and is equal to one. Find the value of.

Answer:

Therefore, unit vector alongis given as:

Scalar product ofwith this unit vector is 1.

Hence, the value of λ is 1.

Page No 458:

Question 14:

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

Answer:

Sinceare mutually perpendicular vectors, we have

It is given that:

Let vector be inclined to at angles respectively.

Then, we have:

Now, as, .

Hence, the vectoris equally inclined to.



Page No 459:

Question 15:

Prove that, if and only if are perpendicular, given.

Answer:

Page No 459:

Question 16:

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

(A) (B)

(C) (D)

Answer:

Let θ be the angle between two vectors and.

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

It is known that.

Hence, when.

The correct answer is B.

Page No 459:

Question 17:

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

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

Answer:

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

Then, .

Now, is a unit vector if.

Hence, is a unit vector if.

The correct answer is D.

Page No 459:

Question 18:

The value of is

(A) 0 (B) –1 (C) 1 (D) 3

Answer:

The correct answer is C.

Page No 459:

Question 19:

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

(A) 0 (B) (C) (D) π

Answer:

Let θ be the angle between two vectors and.

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

Hence, when θisequal to.

The correct answer is B.



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