NCERT Solutions for Class 12 Maths Chapter 10: Vector Algebra

Displacement 40 km, 30° East of North

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

(i) 10 kg is a measure of mass, it has no direction, it is magnitude only and therefore it is a scalar.

(ii) 2 meters North-West is a measure of velocity. It has magnitude and direction both and hence it is a vector.

(iii) 40° is a measure of angle. It has no direction, it has magnitude only. Therefore it is a scalar.

(iv) 40 Watt is a measure of power. It has no direction, only magnitude and therefore, it is a scalar.

(v) 10–19 coulomb is a measure of electric charge and it has magnitude only, therefore, it is a scalar.

(vi) 20 m/sec2 is a measure of acceleration. It is a measure of rate of change of velocity, therefore, it is a vector.

(i) Time-scalar

(ii) Distance-scalar

(iii) Force-vector

(iv) Velocity-vector

(v) Work done-scalar

(i) and have the same initial point and therefore coinitial vectors.

(ii) and have the same direction and same magnitude. Therefore and are equal vectors.

(iii) and have parallel support, so that they are collinear. Since they have opposite directions, they are not equal. Hence and are collinear but not 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.

Collinear vectors are those vectors that are 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.

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

The given vectors are:

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

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

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.

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 

Given:

The unit vector in the direction of vector is given by = / |a|

Given: Points P (1, 2, 3) and Q (4, 5, 6)

Hence, the unit vector in the direction of is

Given: Vectors 

Hence, the unit vector in the direction of (+ ) is

Let =

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

Let

= λ

where λ = 2

Hence, the given vectors are collinear.

The given vector is =

We know that the direction cosines of a vector are coefficients of 

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

Let =

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

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

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

(i) Internally:

=

(ii) Externally:

=

Position vectors of P and Q are given as:

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

Hence, ABC is a right-angled triangle.

Hence, the equation given in alternative C is incorrect.

The correct answer is C.

Option (D) is not true because two collinear vectors can have different directions and also different magnitudes.

The option (A) and option (C) are true by definition of collinear vectors.

Option (B) is a particular case of option (A).

It is given that,

Hence, the angle between the given vectors and is π/4.

The given vectors are

Let = and =

Now, projection of vector on is given by,

Hence, the projection of vector on is 0.

Let = and =

Now, projection of vectoron is given by,

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

Let θ be the angle between the vectors and

Given:

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

Hence, ΔABC is a right-angled triangle.

It is given that. = 0 and . = 0 .

Hence, vectorsatisfying. = 0can be any vector.

Since, + += 0 are unit vectors.

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

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 vectors.

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.

We have,

We have

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

. = 0

Then,

(i) Either || = 0 or || = 0, or ⊥(in case and are non – zero)

x = 0

(ii) Either || = 0 or || = 0 or ||(in case and are non – zero)

But, and cannot be perpendicular and parallel simultaneously.

Hence || = 0 or || =0.

Hence, the given result is proved.

Take any parallel non-zero vectors so that x = 0

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

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:

Hence, the area of ΔABC is √61/2 sq. units.

The area of the parallelogram whose adjacent sides are and is |x|.

Adjacent sides are given as:

Hence, the area of the given parallelogram is 15√2 sq. units.

It is given that || = 3 and || = √2/3

We know that x = ||||sin θ , where n is a unit vector perpendicular to both and and θ is the angle between and.

Now, x is a unit vector if | x | = 1

Therefore, option (B) is correct.

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 and is | x |.

Hence, the area of the given rectangle is | x | = 2 sq. units.
Therefore, option (C) is correct.