Here is a NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra. This solutions covers all questions of Maths Chapter 10 Vector Algebra Class 12 as per CBSE Board guidelines from the latest NCERT book for class 12 maths. Following topics and sub-topics in Class 12 Maths Chapter 10 Vector Algebra are covered.
10.1 Introduction
10.2 Some Basic Concepts
10.3 Types of Vectors
10.4 Addition of Vectors
10.5 Multiplication of a Vector by a Scalar
10.5.1 Components of a vector
10.5.2 Vector joining two points
10.5.3 Section formula
10.6 Product of Two Vectors
10.6.1 Scalar (or dot) product of two vectors
10.6.2 Projection of a vector on a line
10.6.3 Vector (or cross) product of two vectors.

Vector Algebra NCERT Solutions – Class 12 Maths

NCERT Solutions for Class 12 Maths Chapter 10

Exercise 10.1 : Solutions of Questions on Page Number : 428
Q1 : Represent graphically a displacement of 40 km, 30° east of north.
Answer :
Here, vector represents the displacement of 40 km, 30° East of North.


Q2 : 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.


Q3 : 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.


Q4 : 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.


Q5 : 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.


Exercise 10.2 : Solutions of Questions on Page Number : 440


Q1 : Compute the magnitude of the following vectors:

Answer :
The given vectors are:


Q2 : 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.


Q3 : Write two different vectors having same direction.
Answer :

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


Q4 : 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.


Q5 : 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


Q6 : Find the sum of the vectors.
Answer :
The given vectors are.


Q7 : Find the unit vector in the direction of the vector.
Answer :
The unit vector in the direction of vector is given by.


Q8 : 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.


Q9 : For given vector  and ,find the unit vector in the direction of vector 

Answer : The given vector are and  

Hence the unit vector in the direction

.


Q10 : 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,


Q11 : Show that the vectorsare collinear.
Answer :

Hence, the given vectors are collinear.


Q12 : Find the direction cosines of the vector
Answer :

Hence, the direction cosines of


Q13 : 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


Q14 : 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.


Q15 : 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:
Internally:

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,


Q16 : 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,


Q17 : 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:


Q18 : 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.


Q19 : 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.


Exercise 10.3 : Solutions of Questions on Page Number : 447


Q1 : 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.


Q2 : Find the angle between the vectors
Answer :
The given vectors are.

Also, we know that.


Q3 : Find the projection of the vectoron the vector.
Answer :
Letand.
Now, projection of vectoronis given by,

Hence, the projection of vector onis 0.


Q4 : Find the projection of the vectoron the vector.
Answer :
Letand.
Now, projection of vectoronis given by,


Q5 : 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.


Q6 : Findand, if.
Answer :


Q7 : Evaluate the product.
Answer :


Q8 : 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.


Q9 : Find, if for a unit vector.
Answer :


Q10 : Ifare such thatis perpendicular to, then find the value of λ.
Answer :

Hence, the required value of λ is 8.


Q11 : Show that is perpendicular to, for any two nonzero vectors
Answer :

Hence, andare perpendicular to each other.


Q12 : If, then what can be concluded about the vector?
Answer :
It is given that.
Hence, vectorsatisfyingcan be any vector.


Q13 : If are unit vectors such that , find the value of .
Answer :
It is given that .

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


Q14 : 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.


Q15 : 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:


Q16 : 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.


Q17 : 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.


Q18 : 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.


Exercise 10.4 : Solutions of Questions on Page Number : 454


Q1 : Find, if and.
Answer :
We have,
and


Q2 : 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,


Q3 : 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


Q4 : Show that

Answer :


Q5 : Find λ and μ if .
Answer :

On comparing the corresponding components, we have:

Hence,


Q6 : 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.


Q7 : 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.


Q8 : 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


Q9 : 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


Q10 : 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


Q11 : 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.


Q12 : Area of a rectangle having vertices A, B, C, and D with position vectors and respectively is
(A) (B) 1
(C) 2 (D)4
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:


Exercise Miscellaneous : Solutions of Questions on Page Number : 458


Q1 : 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.


Q2 : 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.


Q3 : 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.


Q4 : 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.


Q5 : Find the value of x for whichis a unit vector.
Answer :
is a unit vector if.

Hence, the required value of x is.


Q6 : 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


Q7 : If, find a unit vector parallel to the vector.
Answer :
We have,

Hence, the unit vector alongis


Q8 : 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


Q9 : 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.


Q10 : 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


Q11 : Let and. Find a vector which is perpendicular to both and, and.
Answer :
Let.
Since Is perpendicular to both and, we have:
Also, it is given that:
On solving (i), (ii), and (iii), we get:
Hence, the required vector is.


Q12 : The scalar product of the vector with a unit vector along the sum of vectors and is equal to one. Find the value of.
Answer :
Therefore, unit vector along is given as:
Scalar product of with this unit vector is 1.
Hence, the value of λ is 1.


Q13 : If are mutually perpendicular vectors of equal magnitudes, show that the vector is equally inclined to and.
Answer :
Sincere 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 to.


Q14 : Prove that, if and only if are perpendicular, given.
Answer :


Q15 : 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.


Q16 : 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.


Q17 : The value of is
(A) 0 (B) –1 (C) 1 (D) 3
Answer :

The correct answer is C.


Q18 : 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.