Free NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra solved by Expert Teachers as per NCERT (CBSE) Book guidelines and brought to you by Toppers Bulletin. These Vector Algebra Exercise Questions with Solutions for Class 12 Maths covers all questions of Chapter Vector Algebra Class 12 and help you to revise complete Syllabus and Score More marks as per CBSE Board guidelines from the latest NCERT book for class 12 maths. You can read and download NCERT Book Solution to get a better understanding of all topics and concepts.
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
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) Vectors
andare 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) 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.
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 .gif){kind=small}
,find the unit vector in the direction of vector
.gif){kind=small}
Answer : The given vector are and
.gif)
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 vectors
are 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 vectors
and
with magnitudes
and 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 vector
on the vector
.
Answer :
Let
and
.
Now, projection of vector
on
is given by,
Hence, the projection of vector onis 0.
Q4 : Find the projection of the vector
on the vector
.
Answer :
Let
and
.
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 : Find
and
, 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 : If
are such that
is 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, vectorsatisfying
can 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 vectors
and
]
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 vectors
form the vertices of a right angled triangle.
Answer :
Let vectors be position vectors of points A, B, and C respectively.
Now, vectors
represent the sides of ΔABC.
Hence, ΔABC is a right-angled triangle.
Q18 : 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)
Answer :
Vector
is a unit vector if
.
Hence, vector
is 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 angles
with
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 sides
and
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 which
is 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,
Let
be 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 are
externally 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.