# CLASS 12 MATHS CHAPTER 10-VECTOR ALGEBRA

### Solve The Following Questions.

Question1. Represent graphically a displacement of 40 km, 30° east of north.

Solution :
Displacement 40 km, 30° East of North

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

Question2. Check the following measures as scalars and vectors:

(i) 10 kg

(ii) 2 meters north-west

(iii) 40°

(iv) 40 Watt

(v) 10–19 coulomb

(vi) 20 m/sec2

Solution :
(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 us 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.

Question3. Classify the following as scalar and vector quantities:

(i) time period

(ii) distance

(iii) force

(iv) velocity

(v) work done

Solution :
(i) Time-scalar

(ii) Distance-scalar

(iii) Force-vector

(iv) Velocity-vector

(v) Work done-scalar

Question4. In the adjoining figure, (a square) identify the following vectors:

(i) Coinitial

(ii) Equal

(iii) Collinear but not equal

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

(ii)  and  have 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.

Question5. 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.

Solution :
(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.

Exercise 10.2

Solve The Following Questions.

Question1. Compute the magnitude of the following vectors:

Solution :
The given vectors are:

Question2. Write two different vectors having same magnitude.

Solution :

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

Question3. Write two different vectors having same direction.

Solution :

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

Question4. Find the values of x and y so that the vectors are equal

Solution :

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.

Question5. Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (–5, 7).

Solution :

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

Question6. Find the sum of the vectors:

Solution :
Given:

Question7. Find the unit vector in the direction of the vector

Solution :

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

Question8. Find the unit vector in the direction of the vector  where P and Q are the points (1, 2, 3) and (4, 5, 6) respectively.

Solution :

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

Hence, the unit vector in the direction of  is

Question9. For given vectors  find the unit vector in the direction of  +

Solution :
Given: Vectors

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

Question10. Find the vector in the direction of vector  which has magnitude 8 units.

Solution :
Let  =

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

Question11. Show that the vectors  are collinear.

Solution :
Let

= λ

where λ = 2

Hence, the given vectors are collinear.

Question12. Find the direction cosines of the vector

Solution :
The given vector is  =

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

Question13. Find the direction cosines of the vector joining the points A (1, 2, –3) and B (–1, –2, 1) directed from A to B.

Solution :

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

Question14. Show that the vector  is equally inclined to the axes OX, OY and OZ.

Solution :
Let

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

Question15. Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are  and – respectively, in the ratio 2 : 1

(i) internally

(ii) externally.

Solution :

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

P and Q in the ratio m: is given by

(i) Internally:

(ii) Externally:

=

Position vectors of P and Q are given as:

Question16. Find the position vector of the mid-point of the vector joining the points P (2, 3, 4) and Q (4, 1, – 2).

Solution :

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

Question17. Show that the points A, B and C with position vectors   respectively form the vertices of a right angled triangle.

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

Hence, ABC is a right-angled triangle.

Question18. In triangle ABC (Fig. below), which of the following is not true:

Solution :

Hence, the equation given in alternative C is incorrect.

Question19. If  and  are two collinear vectors, then which of the following are incorrect:

(A)  = λ for some scalar λ

(B)  = ±

(C) The respective components of  and  are proportional.

(D) Both the vectors  and  have same direction, but different magnitudes.

Solution :
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).

Exercise 10.3

Solve The Following Questions.
Question 1. Find the angle between two vectors  and  with magnitude √3 and 2 respectively having  . = √6

Solution :

It is given that,

Hence, the angle between the given vectors and is π/4.
Question 2. Find the angle between the vectors  .

Solution :

The given vectors are

Question 3. Find the projection of the vector  on the vector

Solution :
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

Solution :
Let  =  and

Now, projection of vectoron 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.
Solution :

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

Question 6. Find

Solution :

Question 7. Evaluate the product

Solution :

Question 8. Find the magnitude of two vectors  and having the same magnitude such that the angle between them is 60° and their scalar product is 1/2.

Solution :

Let θ be the angle between the vectors and

Question 9. Find  if for a unit vector .

Solution :

Question 10. If  are such that + λ is perpendicular to  then find the value of λ

Solution :
Given:

Question 11. Show that  is perpendicular to  for any two non-zero vectors  and

Solution :

Question 12. If and . = 0 and . = 0 , then what can be concluded about the vector ?

Solution :

It is given that. = 0 and . = 0 .

Hence, vectorsatisfying. = 0can be any vector.
Question 13. If ,  and are unit vectors such that + += 0 find the value of

Solution :
Since, + += 0 are unit vectors.

Question 14. If either vector . But the converse need not be true. Justify your answer with an example.
Solution :

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 ]
Solution :

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.

Question 16. Show that the points A (1, 2, 7), B (2, 6, 3) and C (3, 10, –1) are collinear.

Solution :

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.

Solution :

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

Hence, ΔABC is a right-angled triangle.
Question 18. If is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λ is unit vector if

(A) λ = 1

(B) λ = –1
(C) a = | λ |
(D) a = 1/|λ|
Solution :

Therefore, option (D) is correct.

Exercise 10.4

Solve The Following Questions.

Question 1. Find | x | if

Solution :

We have,

Question 2. Find a unit vector perpendicular to each of the vectors

Solution :

We have

Question 3. If a unit vector  makes an angle π/3 with   and an acute angle θ with  then find θ and hence, the components of .

Solution :
Let unit vector  have (a1a2a3) components.

Question 4. Show that

Solution :

5. Find λ and μ if
Solution :

Question 6. Given that  . = 0 and  x = 0 What can you conclude about the vectors and ?

Solution :

. = 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.

Question 7. Let the vectors ,, be given as  then show that

Solution :

We have,

Hence, the given result is proved.

Question 8. It either  = 0 and  = 0 then  x = 0 Is the converse true? Justify your answer with an example.

Solution :

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

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).

Solution :

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.
Question 10. Find the area of the parallelogram whose adjacent sides are determined by the vectors

Solution :

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

Hence, the area of the given parallelogram is 15√2 sq. units.
Question 11. Let the vectors  and  such that || = 3 and || = √2/3  then x is a unit vector, if the angle between  and  is:
(A) π/6
(B) π/4
(C) π/3
(D) π/2

Solution :

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.
Question 12. Area of a rectangle having vertices A, B, C and D with position vectors  respectively is:
(A) 1/2
(B) 1
(C) 2
(D) 4

Solution :

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.

### Solve The Following Questions.

Question 1. Write down a unit vector in XY-plane making an angle of 30° with the positive direction of x-axis.

Solution :
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°:

Question2. Find the scalar components and magnitude of the vector joining the points

Solution :
Given points are

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

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.

Solution :

Let O and B be the initial and final positions of the girl respectively.

Then, the girl’s position can be shown as:

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

Solution :

Question 5. Find the value of x for which x () is a unit vector.

Solution :
x () is a unit vector if |x ()| = 1

Hence, the required value of x is ± 1/√3.

Question 6. Find a vector of magnitude 5 units and parallel to the resultant of the vectors

Solution :
Given: Vectors

Let  be the resultant of  and

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

Question 7. If  find a unit vector parallel to the vector

Solution :
Given: Vectors

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.

Solution :

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

Hence, point B divides AC in the ratio 2 : 3.

Question 9. Find the position vector of a point R which divides the line joining the two points P and Q whose position vectors are (2 + ) and (- 3) externally in the ratio 1 : 2. Also, show that P is the middle point of line segment RQ.

Solution :

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

Solution :

Adjacent sides of a parallelogram are given as:

Then, the diagonal of a parallelogram is given by  + .

Hence, the area of the parallelogram is 11√5 square units.

Question 11. Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are

Solution :

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  Find a vector  which is perpendicular to both  and and . = 15

Solution :

Question 13. The scalar product of the vector  with a unit vector along the sum of vectors  is equal to one. Find the value of λ.

Solution :

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 ,, and.

Solution :

Since,, andare mutually perpendicular vectors, we have

. =  = . = 0

It is given that:

| = | | = | |

Let vector  + +  be inclined to ,, andat angles θ1 ,θ2 and θ3 respectively.

Then, we have:

Hence, the vector (  + + ) is equally inclined to,, and.

Question 15. Prove that if and only if , are perpendicular given .

Solution :

Question 16. Choose the correct answer:

If θ is the angle between two vectors  and  then .≥0  only when:

Solution :

Let θ be the angle between two vectors and.

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

It is known that .. = | |||cosθ

Therefore, option (B) is correct.

Question 17. Choose the correct answer:

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

(A) θ = π/4

(B) θ = π/3

(C) θ = π/2

(D) θ = π/3

Solution :

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

Then, || = || = 1

Now, + is a unit vector if | + | = 1

Hence, +  is a unit vector if θ = 2π/3.

Therefore, option (D) is correct.

Question 18. Choose the correct answer:

The value of  is:

(A) 0

(B) -1

(C) 1

(D) 3

Solution :

Therefore, option (C) is correct.

Question 19. If θ  be the angle between any two vectors  and , then  when θ is equal to:

(A) 0

(B) π/4

(C) π/2

(D) π

Solution :

Let θ be the angle between two vectors  and .

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

Therefore, option (B) is correct.

Get 30% off your first purchase!

X
error: Content is protected !!
Scroll to Top