O Math 10 2075

or, 23 cos θ + 1 sin θ = 1
2 2

or, sin 60° cos θ + cos 60° sin θ = 1
2

or, sin (60° + θ) = sin 30°
This value gives θ = – 30° which is not included in the given range.

or, sin (60° + θ) = sin (180° – 30°)

or, 60° + θ = 150°
∴ θ = 90°.

Again sin (60° + θ) = sin (360° + 30°)

or, 60° + θ = 390°
∴ θ = 330°

Hence, θ = 90°, 330°.

Example 7: Solve : sin 3x + sin x = sin 2x (0° ≤ x ≤ 180°)

Solution:

sin 3x + sin x = sin 2x

or, 2 sin 3x + x cos 3x – x = sin 2x
2 2

or, 2 sin 2x cos x – sin 2x = 0 Trigonometry

or, sin 2x (2 cos x – 1) = 0

Either, sin 2x = 0 ... ... ... (1)

or, 2 cos x – 1 = 0 ... ... ... (2)

From (1), sin 2x = 0

or, sin 2x = sin 0°

or, 2x = 0°

∴ x = 0°
Again, sin 2x = 0 = sin 180°
or, 2x = 180°
∴ x = 90°
and sin 2x = sin 360°
or, 2x = 360°
∴ x = 180°

From (2), 2 cos x = 1

or, cos x = 1
2

or, cos x = cos 60°

∴ x = 60°

Hence, x = 0°, 60°, 90°, 180°.

Trigonometric Equations / 197

Exercise 5.6

1. Find the value of θ. (0° ≤ θ ≤ 180°)

a. cos θ = 21 b. sin θ =1 c. cos 2θ = 23 d. sin 3θ =1
2. Solve the following equations. (0° 2≤
x ≤ 90°)

a. sin x = cos 2x b. sin 3x = cos 7x

c. cos 2x = sin 3x d. tan x = cot 5x

3. Solve the following trigonometric equations. (0° ≤ A ≤ 180°)

a. sin2A = sin A b. cos2A = cos A

c. 2 cos2A = 3 sin A d. 1 + cos A = 2 sin2A

4. Solve the following equations. (0° ≤ θ ≤ 180°)

a. 7 cos2θ + 3 sin2θ = 4 b. 5 sin2θ + 3 cos2θ = 4

c. cos2θ = 3 sin2θ + 4 cos θ d. 4 cos2θ = 4 – 7 sin2θ

e. 4 sec2θ – 7 tan2θ = 3 f. tan2θ – 3 sec θ + 3 = 0

g. cot θ = 2 cos θ g. 2 sin θ = tan θ

5. Solve the following equations. (0° ≤ θ ≤ 360°)

a. sin θ + cos θ = 2 b. cos θ + sin θ = 1
2
c. cos θ + sin θ = 1 d. cos θ + 3 sin θ = 2

e. 3 cos θ + sin θ = 3 f. 3 sin θ – cos θ = 2

g. cos θ – 3 sin θ = 2 h. 4 sin θ + 3 cos θ = 5

i. 2 sin θ + 3 cos θ = 3 j. 3 sin θ + cos θ = 1

6. Solve the following equations. (0° ≤ θ ≤ 360°)

a. 2 sin2θ = cos θ + 1 b. 2 cos2θ + sin θ = 2

c. 2 cos θ + 2 sec θ = 5 d. 2 sin θ + cot θ – cosec θ = 0

e. cos2θ = 3 sin2θ + 4 cos θ f. tan2θ + (1 – 3) tan θ – 3 = 0

g. ( 3 + 1) tan θ – 3 = tan θ h. cot2θ + 3+ 1 cot θ + 1 = 0
3
7. Solve the following equations. (0° ≤ θ ≤ 360°)

a. sin 2θ = 3 tan θ cos 2θ b. cos θ + cos 2θ + cos 3θ = 0

c. sin θ + sin 2θ + sin 3θ = 0 d. sin 5θ + sin 3θ = cos 2θ – cos 6θ

8. Solve the following equations.
1
a. sin32q + cos 2q = 4 (0° ≤ θ ≤ 90°)
(0° ≤ θ ≤ 180°)
b. tan θ + tan 2θ + tan 3θ = 0

c. tan4θ + tan6θ = tan3θ sec2θ (0° ≤ θ ≤ 360°)

d. If 2 sin x sin y = 23, cot x + cot y = 2, find x + y.
e. If 4 cos A cos B = 1 and tan A + tan B = 2, find the value of A + B.

f. If 2 cos x cos y = 1 and tan x + tan y = 2, find the value of (x + y).



198 / Optional Mathematics Class 10

5.7 Heights and Distances

Introduction

Trigonometry is extensively used in many branches of science. In this section we use trigonometry
to measure the distance between two points and the height of the tower, building, tree, lamp post etc.
If we look at on object above the horizontal line, the angle formed between the horizontal line and
the line of sight is called the Angle of Elevation.

If we look an object below the horizontal line, the angle formed between the horizontal line and the
line of sight is called the Angle of Depression.

Y P Horizontal Line M

Angle of Depression

Line of Sight Line of Sight

Angle of Elevation

O Horizontal Line A Q Object R

If an object Y (aeroplane) is above the horizontal line OA, then the anlge between the horizontal line Trigonometry
and sight line i.e. AOY is called angle of elevation. Similarly, if an object R (cat) is below the horizontal
line PM, then the angle made by horizontal line and sight line i.e. MPR is called the angle of depression.

We have already solved some of the problems of heights and distances in compulsory mathematics
of class ten. Here we deal with the further problems related to heights and distances.

Example 1: A straight palm tree 30m high is broken by the wind but not completely separated and its
upper part meets the ground at an angle of 60°. Find the distance of the point where the top of the
tree meets the ground from the foot and also the height at which the tree is broken.

Solution: M

Let MN be the height of the tree and it is
broken at P. Let Q be the point on the ground at
which the top of the tree meets. So, MP = PQ.
Here, MN = 30m and PQN = 60°.

Suppose MP = PQ = x m, then PN = (30 – x) m. P
From right angled triangle PQN,

sin 60° = PN
PQ

or, 23 = 30 – x
x
60°
or, 3x = 60 – 2x
NQ

or, 3 + 2 x = 60

Heights and Distances / 199

or, x= 60
3+2

or, x = 60 + 2
1.732
60
or, x = 3.732 = 16.08 m.

Now, PN = (30 – x) m = (30 – 16.08) m = 13.92 m.

Again, tan 60° = PN
QN

or, 3 = 13.92 = 13.92
QN 1.732

∴ QN = 8.04 m.

Hence, the required distance is 8.04m and the height at which the tree broken is 13.92m.
Example 2: Two men are on the opposite side of a tower of 80m high. They observed the angle of elevation

of the top of the tower and found to be 30° and 60°. Find the distance between them.

Solution:

Let AB be the tower of height 80m and C and D be the position of two men such that ACB
= 60° and ADB = 30°.

From ∆ABD, From ∆ACD, A

tan 30° = AB tan 60° = AB 60°
BD CB
or, 13 80 CB
= BD or, 3 = 80
CB
30°
∴ BD = 80 3m or, CB = 80 m
3 D

Hence, the distance between them

CD = CB + BD = 80 + 80 3 = 80 + 240 = 320 = 184.75m
3 3 3
Example 3: A person on the top of a house 45m high observes the angle of depressions of two objects

on the same side of the horizontal plane from the foot of the house to be 30° and 45°. Find the

distance between the first object from the house and also the distance between two objects.

Solution: M A

Let AB be the house, C and D be two points on the level 30° 45°
of B where MA be the horizontal line so,
MAD = ADB = 30° and MAC = ACB = 45°. 45m

Here, AB = 45m, BC = ? and DC = ?

From right angled triangle ACB, 30° 45°

tan 45° = AB D C B
or, BC
∴
1 = 45
BC

BC = 45m

200 / Optional Mathematics Class 10

Again, from right angled ∆ADB, tan 30° = AB
DB
or, 13
= 45
BD

or, BD = 45 3

or, BD = 45 × 1.732

or, BD = 77.94

Hence, DC = BD – BC = 77.94m – 45m = 32.94m.

Example 4: The shadow of a tower standing on level ground is found to be 40 meter longer when the
sun's altitude is at 30° than when its is at 60°. Find the height of the tower.

Solution:

Let AB be the height of the tower and BC = d be the the A

shadow when the sun's altitude is 60°. BD is the shadow of the

tower when the sun's altitude is 30°. From question, CD = 40m.
AB
From ∆ABC, tan 60° = BC
or, 3 = AB
d 60° 30°
B
or, AB = 3d ... ... ... (i) C 40m D

Again from ∆ABD, AB
BD
tan 30° = Trigonometry

or, 13 = AB
BC + CD

or, 13 = d 3d
+ 40

or, 3d = d + 40

or, 3d – d = 40

or, 2d = 40

or, d = 20

Now, the height of the tower is AB = 3d = 3 × 20 = 20 3 m

Example 5: The angles of elevation of the top of a tower observed from 27m and 75m away from its foot

on the same side are found to be complementary. Find the height of the tower. A
Solution:

Let AB be the tower, C and D are the points of observation.

From question, BC = 27m, BD = 75m. q 75m C 90°– q B
ADB and ACB are complementary. 27m
So,if ADB = q then ACB = 90° – q. D

Heights and Distances / 201

From right angled triangle ADB,

tan q = AB = AB ... ... ... (i)
BD 75

Agin from right anlged triangle ACB,

tan (90° – q) = AB
CB
AB
or, cot q = 27 ... ... ... (ii)

Multiplying equations (i) and (ii), we get

tan q cot q = AB × AB
75 27
AB2
or, 1 = 2025

or, AB2 = 2025

or, AB = 2025 = 45

Hence, the height of the tower is 45m.
Example 6: Two poles elected at two opposite sides of a road are one longer than the other by 24m. The

angle of elevation of the top of longer pole from a point midway between the road is 60° and the
angle of depression of the same point from the top of the other is 30°. Find the length of the poles
and the distance of the mid point from either pole.

Solution: A

Here B and D are the opposite sides of road and AB x + 24
and CD are the poles erected. BE = ED, AEB = 60° and
CED = 30°.

From question, AB = CD + 24m. C
Let CD = x m then AB = (x + 24)m
x
From ABE, tan 60° = AB 60° 30°
BE D
BE
or, 3 = x + 24
BE As, BE = ED
x + 24 = 3x
or, BE = x + 24 ... ... ... (i)
3 3
or, x + 24 = 3x
Again from ∆CDE, tan 30° = CD or, 3x – x = 24
ED or, 2x = 24
or, x = 12m
or, 13 = x
ED

or, ED = 3x ... ... ... (ii)

Hence, the length of the poles are CE = x m = 12m and AB = (x + 24)m = (12 + 24)m = 35m.
The distance of the mid-point from either side of the road is ED = 3x = 3 × 12m = 12 3m.

202 / Optional Mathematics Class 10

Example 7: From the top of a vertical column of 60m, the angle of depressions of the top and the bottom
of a house on the same plane of the column are respectively 30° and 60°. Find the height of the
house and its distance from the column.

Solution: F 30° A
E 30°
Let AB be the vertical column and ED, a house and 60°
AB = 60m. FA be the horizontal line and draw EC//FA.
Here, FAE = AEC = 30° and FAD = ADB = 60°, C
ED = ? and BD = ? If ED = x m,

AC = AB – BC = AB – ED = (60 – x)m.

From right angled ∆ABD,

tan 60° = AB
BD
60°
or, 3 = 60 B
BD D

∴ BD = 60 m = 20 3m
3

Again, from right angled DAEC

AC Hence, height of the house = ED = BC
tan 30° = EC
AC = AB – AC
or, 1 20 3
3 = (As BD = EC) = (60 – 20) m = 40m Trigonometry

∴ AC = 20 m Distance between the column and the house = BD = 20 3m.

Example 8: A flagstaff stands on the top of a tower. The angles subtended by the tower and flagstaff at
a point 60 metres away from the foot of the tower and 45° and 15° respectively. Find the length
of the flagstaff.

Solution:

Let DB be the height of the tower and AD be the flagstaff on the top of the tower. C be a point on

the ground 60m away from the foot B of the tower i.e. BC = 60m. ACD = 15° and DCB = 45°. From

right angled ∆DBC, DB
BC
tan 45° = A

or, 1 = DB
60

∴ DB = 60 m. D
As DBCD is a right angled isosceles triangle, DB = BC = 60m

Again, from right angled ∆ABC,

tan 60° = AB
BC
AB
or, 3 = 60 15°

∴ AB = 60 3 m = 60 × 1.732 m = 103.92 m. C 45° 60m B

Heights and Distances / 203

Hence, the length of the flagstaff is AD

= AB – BD

= (103.92 – 60) m = 43.92m.

Example 9: A man observes that at a point due east of a certain tower its angle of elevation is 60°. He

then walks 100 feet due south on a horizontal plane and finds an angle of elevation is 45°. Find

the height of the tower and its original distance from him.

Solution: A

Let AB be the tower, C be a point on the ground such that ACB x ft N
E
= 60°. Again CD⊥BC, CD = 100 ft. and ADB = 45°. (It is a three B
60° C
dimensional figure in which ∆ABC, ∆ABD and ∆BCD are right angled
45°
triangles) AB = ? and BC = ? Let AB = x feet, then from right angled
D
∆ABD, tan 45° = AB Again, from right angled DABC
BD
AB
or, 1 = x tan 60° = BC 100ft
BD
x
∴ BD = x ft. or, 3 = BC (As BD = EC)

Now, from ∆BCD, ∴ BC = x ft
3
BD2 = BC2 + CD2

or, x2 = x 2

3 + (100)2

or, x2 = x2 + 10000
3
or, 3x2 = x2 + 30000

or, 2x2 = 30000

or, x2 = 15000

∴ x = 15000

x = 50 6 ft. 50 6
2
Hence, the height of the tower (AB) = 50 6 ft and the required distance (BC) = ft = 50 2 ft.

Exercise 5.7

1. a. The altitude of the sun is 45° and the shadow of a tower is 170m. Find the height of the tower.

b. A 25m long ladder is resting on the house. The ladder forms an angle of 60° with the ground.
At what height of the house does the ladder reach?

c. From the roof of a house, the angle of depression of a puppy is found to be 30°. If the height of
the house is 12m, how far is the puppy from the house?

d. A flagstaff is placed at one corner of a level rectangular play ground 40m long and 30m wide.
If the angle of elevation of the top of the flagstaff from the opposite corner of the play-ground
is 13º, calculate the height of the flagstaff upto two decimal places (Take tan 13º = 0.231).

204 / Optional Mathematics Class 10

2. a. A tree, 12 metres high, is broken by the wind in such a way that its top touches the ground and Trigonometry
makes an angle of 60º with the ground. At what height from the bottom the tree is broken by
the wind?

b. A 45m high tree is broken by the wind but not completely separated and its upper part meets
the ground at an angle of 30°. At what height is the tree broken? Also find the upper broken
part of the tree.

c. A vertical pole is struck by a speed car and breaks into two parts but not completely, the top of
pole strike the ground at an angle of 30° and at a distance of 20 feet from the foot of the pole.
Find height of the pole before it was broken.

3. a. Two men are on the opposite sides of a 60m high tower. They observed the angles of elevation
of the tower and found to be 30° and 60°. Find the distance between them.

b. Two men are on the opposite side of a pole. They observed the angles of elevation of the top of
the pole as 60° and 45°. If they are 80m apart each other, find the height of the tower.

c. The angles of elevation of the top of a light house in the mid sea from the two boats are 30°
and 45° respectively. If the boats are on the opposite side of the light house and they are 200m
apart, find the height of the lighthouse.

4. a. The angle of elevation of a tower was observed to be 60° from a point. On walking 200 meter
away from the point, it was found to be 30°. Find the height of the tower.

b. The angle of elevation of the top of a tower from a point was observed to be 45º. On walking
30m away from that point it was found to be 30º. Find the height of the tower.

c. From the aeroplane flying vertically over a straight road, the angles of depression of two
consecutive km stones on the same side are 45º and 60º. Find the height of the aeroplane.

d. From the top of a tower 100 meters high the measures of the angles of depression of two object
due east of the tower are found to be 45º and 60º. Find the distance between the objects.

5. a. The shadow of a tower when the angle of elevation of the sun is 45° is found to be 5m longer
when it is 60°. Find the height of the tower.

b. The shadow of a tower on the ground is found to be 45m longer than the sun's altitude of the
tower is 45° than it is 60°. What will be the height of the tower and the length of the shadow
when the sun's altitude is 60°.

c. When the sun's altitude is 45°, the shadow of a tower standing on a plain is 60m. How much
will the shadow of the tower be longer when the sun's altitude is 30°?

d. Determine the height of a mountain if the elevation of its top at an unknown distance from the
base is 30° and at a distance 10km further off from the mountain, along the same line, the angle
of elevation is 15°. [Use tan 15° = 0.27]

6. a. The angles of elevation of the top of a tower observed from two points on the same plane and
same side of the tower to be complementary. If the points are at a distance of 20m and 45m
from the foot of the tower, find the height of the tower.

Heights and Distances / 205

b. The angles of elevation of the top of a tower as observed from two points at the distances of
144m and 121m from the foot of the tower are found to be complementary. Find the height of
the tower.

c. The angles of elevation of the top of a tower as observed from the distance of 36m and 16m
from the bottom of the tower are found to be complementary. Find the height of the tower.

7. a. Two poles stand on either side of a road. At the mid-way between the two posts the angles of
inclination of their tops are 30º and 60º. Find the length of the shorter post if the other post is
15 metres long.

b. Two poles erected on two opposite sides of a road are one longer than the other by 10m. The
angle of the elevation of the top of the longer pole from a point mid-way between the road is
60° and the angle of depression of the same point on the road from the top of the other is 45°.
Find the length of the poles and the distance of the mid-point from either pole.

c. Two lamp posts are of equal height. A boy standing midway between them observes the angle
of elevation of either post to be 30º. After walking 15 meters towards one of them, he observes
its elevation to be 45º, find the height of posts and the distance between them.

d. Two posts are 120 metres apart and the height of one is double that of the other. From the
middle point of the line joining their feet an observer finds the angular elevation of their tops
to be complementary. Find the height of the shorter post.

8. a. From the top of 21m high cliff, the angles of depression of the top and the bottom of a tower
are observed to be 45° and 60° respectively. Find the height of the tower.

b. The angle of depression of the top of a 70 metres tall building from the top of a tower is 45º
and the angle of elevation of the top of the tower from the foot of the building is 60º. Find the
height of the tower.

c. From the foot of a house the angle of elevation of the top of a temple is 45° and from the
top of the house the angle of elevation of the top of the temple is 15°. If the height of the
house is 40m, find the height of the temple and distance from the house to the temple.
(tan 15° = 0.268)

9. a. The angle of depression of the top of a 5m high pole observed from the top of a house is 60°
and the angle of elevation of the top of the pole from the foot of the house is 30°. Find the
height of the house.

b. The angle of depression of the bottom of a pole of height 20m when observed from the top of
a tower is found to be 60º and the angle of depression of the bottom of the tower from the top
of the pole is 22º. Find the height of the tower and the distance between the pole and the tower.
(tan 22º = 0.4)

c. From the foot of a column the angle of elevation of the top of a tower 45º and from the top of
the column the angle of depression of the bottom towards the tower is 30º. A man walks 10
meters from the bottom of the column towards the tower. He notices the angle of elevation of
its top to be 60º. Find the heights of the column and the tower.

206 / Optional Mathematics Class 10

d. A bird is sitting on the top of a tree which is in front of a house. The angles of elevation and the
depression of the bird from the bottom and the top of a house are 60° and 30° respectively. If the
height of the tree is 24m, find the height of the house and distance between the house and the tree.

10. a. A statue stands on the top of a column. The angles subtended by the column and statue at a
point 100m away from the foot of the column are 45° and 15° respectively. Find the height of
the statue.

b. The angle of elevation of the top of an incomplete house as observed from a point on the
ground is 30°. If the height of the complete house is 30m, how high must the house be raised
so that the angle of elevation of the top of the complete house as observed from the same point
on the ground is 45°?

c. A pole is surmounted on its top by a flagstaff. The angles of elevation of the top and the bottom
of the flagstaff as observed from a point 30 meters away from the foot of the pole are found to
be 45° and 30° respectively. Find the height of the flagstaff.

d. PQ is a vertical pole with its foot Q in a level ground. A point R divides PQ such that PR : RQ
= 7 : 3. If parts PR and RQ subtend equal angles at a point 20m from the foot of the pole, find
the height of the pole.

e. A pole is divided by a point in the ratio 1:9 from the ground. If both parts of the pole subtend
equal angles at a point on ground level 20m away from its bottom, find the height of the
pole.

11. a. A man observes that at a point due south of a certain tower its angle of elevation is 60º. The Trigonometry
then walks 60 metres due west on a horizontal plane and observes that the angle of elevation
is 30º. Find the height of the tower and his original distance from it.

b. The angle of elevation of the top of a tower at a place A due south of it is θ and at a place due
west of A and at a distance s from it is a. Show that the height of the tower is s tan q tan a .
tan2q – tan2a

c. From the top of a light house, the angles of depression of two ships on the opposite sides of it

are observed to be θ and β. If the height of the light house be h meter and the line joining the

ships passes through the foot of the light house, show that the distance between the ships is

h(tan q + tan b) metres.
tan q tan b

12. a. The angle of elevation of an aeroplane from a point on the ground is 45°. After 15 second flight,

the elevation changes to 30°. If the aeroplane is flying at a height of 3000m, find the speed of

the plane.

b. A tower in a city is 150m high and a hotel at the city centre is 20m high. The angle of elevation
of the top of tower at the top of the hotel is 5°. A building of h metre height is situated on the
straight road connecting the tower with the city centre at a distance of 1.2 km from the tower.
Find the value of h if the top of the hotel, the top of the building and top of the tower are in a
straight line. Also, find the distance of the tower from city centre. [Use tan 5° = 0.0875]


Heights and Distances / 207

6. VEsdCafsTdfaORS

Curriculum Contents Included Periods

a. Scalar product of vectors (Condition of perpendicular of vectors)
b. Vector geometry
– Midpoint theorem, Section formula

– Line joining midpoints of two sides of a triangle is parallel to the third side

and half of it. →g (→a →b →c )

– Position vector of centroid of a triangle, = 1 + +
3
– Relation between a line joining vertex and midpoint of the base and base of

an isosceles triangle 17

– Quadrilateral joining the midpoints of the sides of a quadrilateral is a
parallelogram.

– Diagonals of a parallelogram bisect each other.

– Diagonals of a rectangle are equal.

– The diagonals of a rhombus bisect each other at right angles.

– Angle of semi-cercle is a right angle.

– The midpoint of hypotenus of a right angled triangle is equidistant from its
vertices.

Curriculum

On the completion of this unit, students will be able to
find the scalar product of two vectors and also find the anlge between them and state the condition

of parallelism and orthogonality.
apply triangle law, parallelogram law and polygonal law of vector addition to prove theorems.
obtain the section formulae and mid-point formula by vector method and also prove some

geometrical theorems vectorically.
obtain the centroid formula by vector method and solve related problems.

Specification Grid

Contents Knowledge Understanding Application Higher ability Total Total
Vectors Each of 1 mark Each of 2 marks Each of 4 marks Each of 5 marks Quesitons Marks

1 2 – 13 10

208 / Optional Mathematics Class 10

6 VECTORS

6.1 Dot Product of Vectors

Introduction

Suppose the position vectors of P and Q are →OP = x1 and Y Q
y1
O→Q = x2 X' O PC
y2 . then from triangle law of vector addition Y' A BX

O→Q = →OP + →PQ

or, →PQ = O→Q – →OP = x2 – x1 = x2 – x1
y2 y1 y2 – y1

∴ →PQ = O→Q – →OP

Sum and Difference of Two Vectors

The addition (or subtraction) of two vectors is the vector obtained by adding (or subtracting) the

corresponding components of the given two vectors. C

If →AB = a and →BC = c are two vectors then the sum a+c c
b d b+d d

of given two vectors is →AB + →BC = a + c = a+c = →AC. A a B
b d b+d b
a–c –c
If →BD = – →BC = – c = – c be a negative vector of →BC then b–d –d
d – d

the difference →AB – →BC = →AB + (–→BC)

= →AB + →BD = a + – c = a–c = A→D. D Vectors
b – d b–d

Product of Two Vectors

Two vectors can be multiplied in two ways.
1. Vector product of two vectors or cross product and
2. Scalar product of two vectors or dot product.

1. Vector Product of Two Vectors (Cross Product)

When two vectors are multiplied and the resulting value is a vector then the
product is known as a Vector Product. In this process the resultant obtained is a vector →a ×→b
→b
in nature. If →a = a1 and →b = a2 be two vectors then their vector product is denoted by →a
b1 b2
→a × →b and is defined as

→a × →b = | →a | | →b | sin q η→ = ab sin q η→ where η→ is the unit vector normal to the plane of →a and →b.

Dot Product of Vectors / 209

2. Scalar Product of Two Vectors (Dot Product)

When two vectors are multiplied and the resulting value is a scalar then the product is known as

Scalar Product. In this process, we get a scalar value is resultant. Dot product of two vectors is equal to the

product of their lenghts mltiplied by the cosine of the angle between them. B

Let O→A = →a and →OB = →b be the two vectors and AOB = θ. If →
b
BM⊥OA then OM is known as projection of OB on OA.

Here, OM = projection of →b on →a q

OM = OB cos θ = |→OB| cos θ = |→b| cos θ ... ... (1) OM → A
a
|→b | cos q
The scalar product of vector →a and →b is denoted by →a.→b and is
defined as →a.→b = magnitude of →a × projection of →b on →a

∴ →a . →b = | →a | | →b | cos q
There are the following conditions of the scalar product of →a and →b .

Name Parallelism Negativism Perpendicularity
Figure →a
→u →v →q →q
→b →p →p
→u →v
→a →b → →
→u →v
→a →b →q p →q p
Unlike parallel vectors
Name of Like parallel vectors Perpendicular or
vector Angle between →u and →v is
Angle Angle between →a and →b is 180° i.e. q = 180° Orthogonal vectors
0° i.e. q = 0° →u . →v = | →u | | →v | cos q
Condition →a . →b = | →a | | →b | cos q →u . →v = | →u | | →v | cos 180° Angle between →p and →q is
→a . →b = | →a | | →b | cos 0° →u . →b = | →u | | →b | .(– 1)
→a . →b = | →a | | →b | .1 = ab →u . →v = – | →u | | →v | = – uv 90° i.e. q = 90°
→a . →b = | →a | | →b | = ab
→p . →q = | →p | | →q | cos q
→p . →q = | →p | | →q | cos 90°
→p . →q = | →p | | →q | .0 = 0
→p . →q = 0

Scalar Product of Standard Unit Vector

A unit vector along its positive axis is called standard unit vector. There are

two types of standard unit vectors. B
a. Standard unit vector along X-axis is denoted by →i . In the
figure O→A = →i = 1 . Therefore |O→A| = |→i | = 1. →j
0 denoted by →j .
b. Standard unit vector along Y-axis is In the O →i A

figure →OB = →j = 0 . Therefore |→OB| = |→j | = 1.
1

210 / Optional Mathematics Class 10

Since, →i and →i are parallel to each other, q = 0°.
→i .→i = |→i |.|→i | cos 0° = 1 × 1 × 1 = 1.
\ →i .→i = 1. Similarly, →j .→j = 0.
Again, since →i and →j are perpendicular to each other, so q = 90°.
→i .→j = |→i |.|→j | cos 90° = 1 × 1 × 0 = 0.
\ →i .→j = 0. Similarly, →j .→i = 0.

Scalar Product of Two Vectors in Terms of Their Components

Let M→A = →a = x1 and M→B = →b = x2 be two vectors and θ be the angle between →a and →b .
y1 y2

Then, woer, have→a→a. →b. →b==x1xyx112 →.i .→yxi 22+=x1(yx21→→ii .→+j y1→j ).(x2→i + y2→j ) A
+ y1x2→j .→i + y1y2→j .→j
→a
or, →a . →b = x1 x2 × 1 + x1y2 × 0 + y1x2 × 0 + y1y2 × 1 θ

M →b B

[  →i .→i = 1 = →j .→j and →i .→j = 0 = →j .→i ]
∴ →a . →b = x1 x2 + y1 y2

Properties of Scalr Product of Vectors

If →a , →b and →c are any three vectors, then the scalar product of the vectors holds the following

properties.

1. →a . →b = →b . →a (Commutative law)

2. →a .( →b + →c ) = →a . →b + →a . →c and ( →a + →b ). →c = →a . →c + →b . →c (Distributive law)

3. If →a . →b = 0, then →a ⊥ →b

4. If →a . →b = ab, then →a and →b are like parallel.

5. If →a . →b = – ab, then →a and →b are unlike parallel. Vectors

6. →a . →a = ( →a )2 = a2

7. ( →a + →b ) = a2 + 2 →a . →b + b2

8. ( →a – →b ) = a2 – 2 →a . →b + b2

9. ( →a + →b )( →a – →b ) = a2 – b2

Example 1: Find the scalar product of the following pair of vectors:

a. →a = 3 and →b = –2 b. →a = 5 →i – 3 →j and →b = 2 →i + 6 →j
4 5
Solution:
3 –2
a. Here, given two vectors are →a = 4 and →b = 5
Now, →a . →b = x1x2 + y1y2

= 3 × (– 2) + 4 × 5 = – 6 + 20 = 14

b. Here →a = 5→i – 3→j = 5 and →b = 2→i + 6→j = 2
– 3 6

Dot Product of Vectors / 211

Now, →a.→b = x1x2 + y1y2
= 5 × 2 + (– 3) × 6

= 10 – 18 = – 8

Example 2: If →a = 3 and →b = –2 then find the dot product of the vectors (2 →a + →b ) and ( →a – 3 →b ).
1 5
Solution:
Here, →a = 3 →b = – 2
1 and 5.

So, 2 →a + →b = 2 3 – 2 6 – 2 6 – 2 4
1 +5= 2 + 5 = 2+5 = 7

and →a – 3 →b = 3 – 3 –2 = 3 – –6 = 3+6 = 9
1 5 1 15 1 – 15 – 14

Now, (2 →a + →b ).( →a – 3 →b ) = 49
7 . – 14

= x1x2 + y1y2 = 4 × 9 + 7 × (–14) = 36 – 98 = – 62

Angle between Two Vectors

Let →PQ = →a and →PR = →b be two vectors and q be the angle between them, then

→a . →b = | →a | | →b | cos q →a = x1 Q
y1
or, cos θ = →a . →b
| →a | | →b |

\ q = cos–1 →a . →b θ →b = x2 R
| →a | | →b | y2
P

When two vectors are given in component form,

Let →PQ = →a = x1 and →PR = →b = x2 are given two vectors in components form, then the scalar
y1 y2

product of →a and →b is →a . →b = x1x2 + y1y2 , | →a | = x12 + y12 and | →b | = x22 + y22

cos θ = →a . →b = x1x2 + y1y2
| →a | | →b | x12 + y12 x22 + y22

\ q = cos–1 x1 x2 + y1 y2
x12 + y12 x22 + y22

Example 3: Find the angle between the following pairs of vectors:

a. →a = 5 and →b = 1 b. →a = 4 →i – 3 →j and →b = 3 →i + 7 →j
3 4

Solution:

a. Here, given two vectors are →a = 5 and →b = 1
3 4

212 / Optional Mathematics Class 10

So, | →a | = 52 + 32 = 25 + 9 = 34 b. Here →a = 4 →i – 3 →j = 4 and →b = 3 →i + 7 →j = 3
| →b | = 12 + 42 = 1 + 16 = 17 – 3 7
And →a . →b = x1x2 + y1y2
So, | →a | = 42 + (–3)2 = 16 + 9 = 5 and
= 5×1+3×4 | →b | = 32 + 72 = 9 + 49 = 58
And →a . →b = x1x2 + y1y2

= 5 + 12 = 17 = 4 × 3 + (–3) × 7

→a . →b 17 = 12 – 21 = – 9
→a | | →b 34 × 17
Now, cos θ = | | = →a . →b – 9
→a | | →b × 58
Now, cos θ = | | = 5

or, cos θ = 2× 17 17 = 1 or, cos θ = – 9 = – 9
17 × 2 5 × 7.62 38.1

or, cos θ = cos 45°

∴ θ = 45° or, cos θ = – 0.236
or, cos θ = cos 104°
∴ θ = 104°

Example 4: If →a = 3 and →b = 0 then prove that ( →a + →b ) and (2 →a – →b ) are perpendicular to each
other. 1 5

Solution:

Here, →a = 3 and →b = 0
1 5.

So, →a + →b = 3 + 0 3+0 3
1 5 = 1+5 = 6

and 2 →a – →b =2 3 – 0 = 6 – 0 6–0 6
1 5 2 5 = 2–5 = –3

Now, ( →a + →b ).(2 →a – →b ) = 3 6 Vectors
6 . –3

= x1x2 + y1y2 = 3 × 6 + 6 × (–3) = 18 – 18 = 0
Hence, ( →a + →b ) and (2 →a – →b ) are perpendicular to each other.

Example 5: Find the value of k so that the pair of vectors k →i – 2 →j and 4 →i + (k + 3) →j are perpendicular.

Solution:

Let the given vectors are →a = k→i – 2→j = k and →b = 4→i + (k + 3)→j = 4
– 2 k + 3
As →a and →b are perpendicular to each other

→a . →b = 0

or, x1x2 + y1y2 = 0
or, k × 4 + (– 2) × (k + 3) = 0

or, 4k – 2k – 6 = 0

Dot Product of Vectors / 213

or, 2k = 6
∴ k = 3
Hence, the required value of k is 3.

Example 6: Prove that the vectors – 4 →i + 5 →j and 8 →i – 10 →j are parallel to each other.

Solution:

Let →a = – 4→i + 5→j = – 4 and →b = 8→i – 10→j = 8
5 –10

Magnitude of →a = |→a| = (– 4)2 + 52 = 41 and Alternately, – 4
→a = – 4 →i + 5 →j 5
Magnitude of →b = |→b| = 82 + (–10)2 = 164 = 2 41 and Let = and
So, →a . →b = x1x2 + y1y2
→b = 8 →i – 10 →j = 8
–10
8 – 4
= – 4 × 8 + 5 × (– 10) Now, →b = –10 =–2 5 = – 2 →a

= – 32 – 50 = – 82. \ →b = – 2 →a and →a and →b are unlike

Now, cos q = | →a . →b | = – 82 41 = – 1 parallel vectors.
→a | | →b 41 × 2

or, cos θ = cos 180º
∴ θ = 180º
Hence, the given two vectors are parallel in opposite direction.

Example 7: For what value of p, two vectors 2 →i – 3 →j and 4 →i + p →j are parallel to each other?

Solution:

Here, the given vectors are 2→i – 3→j and 4→i + p→j are parallel to each other.

For two parallel vectors →a and →b, we have Alternately,
→a.→b = ± |→a| |→b| The given two vectors are 2→i – 3→j and 4→i + p→j .
or, x1x2 + y1y2 = ± x12 + y12 x22 + y22
Now, for given two vectors, For any two parallel (like/unlike) vectors, their

2 × 4 + (– 3) × p = ± 22 + (–3)2 42 + p2 corresponding components are proportional.

or, 8 – 3p = ± 13 16 + p2 i.e. x1 = y1
Squaring both sides, we get x2 y2
2 – 3
(8 – 3p)2 = (± 13 16 + p2)2 or, 4 = p
or, 64 – 48p + 9p2 = 13(16 + p2)
or, 2p = – 12

or, 64 – 48p + 9p2 = 208 + 13p2 \ p = – 6

or, 13p2 – 9p2 + 48p + 208 – 64 = 0 Hence, the value of p is – 6.

or, 4p2 + 48p + 144 = p

or, 4(p2 + 12p + 36) = 0

214 / Optional Mathematics Class 10

or, p2 + 12p + 36 = 0

or, (p + 6)2 = 0

\ p = – 6
Example 8: If (4 →a + →b )2 = (4 →a – →b )2, prove that →a and →b are perpendicular to each other.

Solution:
Here, (4 →a + →b )2 = (4 →a + →b )2
or, 16 →a 2 + 8 →a . →b + →b 2 = 16 →a 2 – 8 →a . →b + →b 2
or, 8 →a . →b + 8 →a . →b = 0
or, 16 →a . →b = 0
\ →a . →b = 0. Hence, →a and →b are perpendicular to each other.

Example 9: Find the angle between two vectors →a and →b if | →a | = 3, | →b | = 3 2 and →a . →b = 9.

Solution:
Here, we have | →a | = 3, | →b | = 3 2 and →a . →b = 9

By using formula, →a . →b
| →a | | →b
Now, cos θ = | = 3 9 2= 1
×3 2

or, cos θ = cos 45º

∴ θ = 45º
Hence, the angle between →a and →b is 45º.

Example 10: If →a + →b + →c = 0, |→a| = 6, |→b| = 10 and |→c| = 14, find the angle between and →a and →b.

Solution: Vectors
Here, →a + →b + →c = 0
or, →a + →b = – →c

Squaring both sides, we have or, →a.→b = 30
(→a + →b)2 = (–→c)2 or, |→a| |→b| cos θ = 30
or, →a 2 + 2→a.→b + →b 2 = →c 2
or, |→a| 2 + 2→a.→b + |→b| 2 = |→c| 2 or, 6 × 10 × cos θ = 30
or, 62 + 2→a.→b + 102 = 142
or, 2→a.→b = 196 – 36 – 100 or, cos θ = 30 = 1
60 2
or, 2→a.→b = 60
or, cos θ = cos 60°
Hence, the angle between →a and →b is 60º.
∴ θ = 60°

Examplbee1tw1e:enG→aiveanntdha→bt →a + 2 →b and 5 →a – 4 →b are orthogonal. If →a and →b unit vectors, find the angle
.

Solution:
Since, →a + 2 →b and 5 →a – 4 →b are orthogonal, their dot product is zero.

Dot Product of Vectors / 215

i.e. ( →a + 2 →b ).(5 →a – 4 →b ) = 0

or, 5a2 – 4 →a . →b + 10 →b . →a – 8b2 = 0

or, 5 × 1 – 4 →a . →b + 10 →a . →b – 8 × 1 = 0 [ →a . →b = →b . →a ]

or, 6 →a . →b = 8 – 5 = 3

or, →a . →b = 3 q = 1
or, | →a |.| →b | 6 2

cos

or, 1 × 1 × cos q = 1
2

or, cos q = cos 60°

\ q = 60°
Hence, the angle between →a and →b is 60°.

Example 12: Prove that the vertices A(9, 2), B(7, 5), C(1, 1) and D(3, – 2) are the vertices of a rectangle.

Solution:

Here, O→A = 9 , →OB = 7 , →OC = 1 and O→D = 3 A D
2 5 1 – 2

Now, →AB = →OB – O→A = 7 – 9 = – 2
5 2 3

→BC = →OC – →OB = 1 – 7 = – 6
1 5 – 4

→DC = →OC – O→D = 1 3 – 2 B C
1 – – 2 = 3

and A→D = O→D – O→A = 3 – 9 – 6 O
– 2 2 = – 4

Again, →AB.→BC = – 2 – 6
3 . – 4 = (– 2) × (– 6) + 3 × (– 4) = 12 – 12 = 0
From the above vectors representing sides of quadrilateral ABCD, we have →AB = →DC, →BC = A→D and

→AB and →BC are perpendicular to each other. i.e. ABC = 90°. Hence, ABCD is a rectangle.

Example 13: If P(2, 3), Q(– 2, 1), and R(0, – 1) are the vertices of DPQR, find the measure of PQR.

Solution:

Here, the vertices of DPQR are P(2, 3), Q(– 2, 1), and R(0, – 1), then

→QP = →OP – O→Q = 2 – – 2 = 2+2 = 4 and
3 1 3–1 2

→QR = →OR – O→Q = 0 – – 2 = 0+2 2
– 1 1 – 1 – 1 = – 2
Now, |→QP | = x2 + y2 = 42 + 22 = 16 + 4 = 20 = 2 5

and |→QR| = x2 + y2 = 22 + (– 2)2 = 4 + 4 = 8 = 2 2

216 / Optional Mathematics Class 10

So, if q be the angle between →QP and →QR then
→QP .→QR
cos q = |→QP | |→QR| = x1x2 + y1y2 = 4 ×2 +2 × (– 2) = 8 –4 = 1 = 0.316
|→QP | |→QR| 2 5× 2 2 4 10 10

cos q = cos 71.58°

\ q = 71.58°

Hence, the measure of PQR is 71.58°.

Exercise 6.1A

1. Find the scalar product of the following pairs of vectors.

a. →a = 6 and →b = 2 b. m→ = 86 and →n = 3
– 2 7 –3

c. →a = 2→i – 3→j and →b = 6→i + 4→j d. →u = 4→i – 5→j and →v = 3→i + 5→j

e. →a = 2→i and →b = 3→i – 5→j f. →p = (5, – 2) and →q = (2, 5)

2 . aIf. →a =→a . →92b , →b = –3 1 , →c = 52b. and→b→d. →c= ––  23 , find c. →d . →a

d. a2 e. c2 f. d2

3. Iaf. →a =→a +12 and →b = – 2 find the scalar product between the following vectors.
→b and →a – →b3 , then b. 2 →a + →b and 3 →a

c. 2 →a + 3 →b and 4 →a – →b d. →a – 5 →b and 2 →a – →b

4. Prove that the following pairs of vectors are perpendicular to each other. Vectors

a. 35 and – 5 b. 75 and –1 251
3
c. 4 →i – 3 →j and 6 →i + 8 →j d. 2 →i + 5 →j and 5 →i – 2 →j

e. 34 →i – 1 →j and 6 →i + 16 →j f. 32 →i + →j and 5 →i – 10 →j
2 3

5. a. If →a = 2 and →b = – 3 are two vectors, then prove that →a + →b and 23 →a – 3 →b are perpendicular
1 4

to each other.

a. If →u = 3 and →v = 1 are two vectors, then prove that 2 →u + →v and 12 →u – 19 →v are
– 2 5

perpendicular to each other.

6. Find the value of k so that the following pairs of vectors are perpendicular to each other.

2k 2 k+1 6
a. – 5 and k – 1 b. 3 and – 3k

Dot Product of Vectors / 217

c. k →i – 3 →j and 4 →i – (k – 1) →j d. 5 →i + (k + 1) →j and 2 →i – 5 →j

7. Prove that the following pairs of vectors are parallel to each other.

24 –6 9
a. – 3 and – 6 b. 4 and – 6
c. 2 →i + 4 →j and 6 →i + 12 →j d. 6 →i – 3 →j and – 4 →i + 2 →j

e. 23 →i – 1 →j and 15 →i – 5 →j f. – 1 →i + 5 →j and 3 →i – 15 →j
4 2 3 6 2

8. For what value of k will the following pairs of vectors parallel?

a. –34 and 6 b. –86 and k––123
k
c. 3 →i – →j and 6 →i + k →j d. 2 →i + 6 →j and (k + 1) →i + 9 →j

e. 32 →i – 1 →j and 2 →i + k →j f. 21 →i – 43 →j and (k – 1) →i + 3 →j
2
A
9. a. From the given figure find →BA.→BC. P
4 80°
42

b. From the given figure find →PQ .→RQ. 40°

B7 C A Q Q.N. 9(b) R

Q.N. 9(a) 3 100°

c. In the given figure, →AC.→AB = 12. Find the value of x. 35°

Bx C
Q.N. 9(c)
10. a. If ( →a + →b )2 = ( →a – →b ) 2, prove that →a and →b are perpendicular
to each other.

b. If (3 →a + 2 →b )2 = (3 →a – 2 →b ) 2, prove that →a and →b are orthogonal vectors.

c. If →p and →q are orthogonal vectors, prove that (2 →p + 3 →q )2 = (2 →p – 3 →q )2.

11. a. If | →a | = 5 and | →b | = 4 and →a . →b = 10, find the angle between →a and →b .

b. If | →p | = 4 and | →q | = 2 2 and →p . →q = 4 6, then find the angle between →p and →q .

c. If →a . →b = 3 3, | →a | = 2 and | →a | = 3, find the angle between →a and →b .

d. Find the angle between two vectors →a and →b if | →a | = 5, | →b | = 3 and →a . →b = 9.

Exercise 6.1B

1. If →a = 2→i – 3→j , →b = →i + 2→j and →c = 4→i + 3→j , then prove the following relations.

a. →a .( →b + →c ) = →a . →b + →a . →c b. ( →a + →b )2 = a2 + 2 →a . →b + b2

c. ( →a – →b )2 = a2 – 2 →a . →b + b2 d. 3( →a . →b ) = (3 →a ). →b = →a .(3 →b )

2. Find the angle between the following pairs of vectors.

a. →a = 3 and →b = – 3 b. →p = 72 and →q = – 3
4 0 3
c. →a = 2→i – →j and →b = →i + 2→j d. →a = 3→i + 4→j and →b = 5→i – 12→j

218 / Optional Mathematics Class 10

e. m→ = 2→i – →j and →n = – 3→j f. →a = (5 3) and →b = (4 – 4) Vectors
3. a. Given that the vectors 2 →a + →b and 4 →a – 5 →b are orthogonal. If →a and →b are unit vectors, find

the angle between →a and →b .
b. If →p and →q are unit vectors such that 4 →p – 2 →q and 5 →p + 4 →q are orthogonal to each other, find

the angle between →p and →q .
c. If →u + 2 →v and 4 →u – →v are perpendicular to each other, where →u and →v are unit vectors. Find

the angle between →u and →v .
4. a. If →a + →b + →c = 0, |→a| = 6, |→b| = 7 and |→c| = 127, find the angle between and →a and →b.
b. If →a + →b + →c = 0, |→a| = 8, |→b| = 6 and |→c| = 2 37, find the angle between and →a and →b.
c. If →p + →q + →r = 0, |→p| = 6, |→q| = 7 and |→c| = 106, find the angle between and →p and →q.

5. a. Prove that the points A(2, 2), B(4, 6) and C(10, – 2) are the vertices of a right angled triangle.

b. Show that the vertices A(2, – 2), B(8, 4), C(5, 7) and D(– 1, 1) are the vertices of a rectangle.

6. a. Show that the points A(7, 2), B(4, 5), C(1, 2) and D(4, – 1) are the vertices of a square.

b. Show that the points P(2, 1), Q(– 1, – 1), R(2, – 3) and S(5, – 1) are the vertices of a rhombus.

7. a. If the vertices of the ∆ABC are A(2, 6), B(– 3, 3) and C(5, 1) show that ∆ABC is a right angled
triangle.

b. If the vertices of the DPQR are P(0, 2), Q(2, 5) and R(8, 1), show that DPQR is a right angled
triangle.

c. Prove that the vectors →a = 3→i – 5→j , →b = 5→i + 3→j and →c = 8→i – 2→j form an isosceles right angled
triangle.

8. a. If A(2, 2), B(4, 4) and C(7, 1) are the given three points, find the measure of ABC by vector
method.

b. If P(3, 2), Q(6, 3) and R(3, 6) are the given three points, find the measure of PRQ by vector
method.

c. If A(1, – 2), B(6, 3) and C(5, – 1) are the given three points, find the measure of CAB by
vector method.

9. a. If A(– 1, 3), B(3, 1) and C(1, 7) are the given three points, prove that CAB = 2ACB by
vector method.

b. If P(1, 2), Q(6, 4) and R(3, 7) are the given three points, prove that PQR = PRQ by vector
method.

10. a. If O be the origin and O→A = (5, 1), →OB = (1, – 1) and →OC = (11, 4), prove that A, B and C are
collinear.

b. If P, Q, R and S are four points having co-ordinates (2, 1), (– 3, 4), (1, – 5) and (6, – 8), prove
that →PQ and →RS are parallel.


Dot Product of Vectors / 219

6.2 Formulae of Vector Geometry

We can study different properties and relations relating to geometry with the help of vectors. Such a
study is known as the Vector Geometry.

Midpoint Formula
If →a and →b are position vectors of two points A and B respectively and M is the middle point of
1 ( →a →b ).
the line segment AB, then the position vector of M is 2 +

Proof:

Le O be the origin and AB is a line segment. A
Position vector of A = O→A = →a and
Position vector of B = →OB = →b . a M
Let M be the midpoint of AB, hence A→M = M→B.
i.e. A→O + O→M = M→O + →OB O m
i.e. O→M – M→O = – A→O + →OB
i.e. O→M + O→M = O→A + →OB B
i.e. 2O→M = O→A + →OB
b

i.e. O→M = 1 ( →a + →b )
2

Hence, O→M = →a + →b is the position vector of midpoint M of AB.
2

Section Formula: Internal Division

If →a and →b be the position vectors of the points A and B respectively and →p be the position vector

of point P which divides AB in the ratio m:n internally, then →p = m→b + n→a
m+n
Proof:
Here, O→A = →a , →OB = →b , →OP = →p and AP:PB = m:n Y

S o, or, nAPB→AP P= m →PB b B
n
n
=m
or, n( A→O + →OP ) = m( →PO + →OB ) P
or, n(– O→A + →OP ) = m(– →OP + →OB )
pm
or, n(– →a + →p ) = m(– →p + →b )
or, – n →a + n →p = – m →p + m →b A
or, n →p + m →p = m →b + n →a
a

X' O X
Y'

or, →p (m + n) = m →b + n →a

Hence, →p = m →b + n →a which is the position vector of P.
m + n

220 / Optional Mathematics Class 10

Section Formula: External Division

If →a and →b are the position vectors of points A and B respectively, and →q is the position vector of

a point Q which divides AB in the ratio m:n externally, then →q = m→b – nn→a .
m –
Proof:
Here, O→A = →a , →OB = →b , O→Q = →q and AQ:BQ = m:n

So, ABQQ = m
n

or, n A→Q = m →BQ Y
Q
or, n( A→O + O→Q ) = m( →BO + O→Q )

or, n(– O→A + O→Q ) = m(– →OB + O→Q ) q n
b m
or, n(– →a + →q ) = m(– →b + →q )
B
or, – n →a + n →q = – m →b + m →q
A

or, m →q – n →q = m →b – n →a a

or, →q (m – n) = m →b – n →a X' O X
Y'

Hence, →q = m →b – n →a which is the position vector of Q.
m – n

Centroid Formula

The position vector of the centroid of a triangle is one third of the sum of the position vectors of
its vertices.

Let O be the origin and G the centroid of DABC and AD is median. So, D is the midpoint of BC.

O→D = →OB + →OC [ From midpoint formula] Vectors
2

The centroid of a triangle divides each median in the ratio 2:1 A
from its vertex.

So, AG:GD = 2:1.

Now, O→G = 2 × O→D +1 × →OA G
2 +1 D
B C
O
2× →OB + →OC + O→A →OB →OC O→A
= 2 3
= + +

3

Hence, O→G = O→A + →OB + →OC which is the required result.
3

Formulae of Vector Geometry / 221

Example 1: If the position vectors of A and B are 2 →i + 3 →j and 6 →i + →j respectively, find the position
vector of the middle point M of AB.

Solution: B
Here, O→A = 2 →i + 3 →j and →OB = 6 →i + →j
O→A →OB 2→i + 3→j M
The position vector of mid-point M of AB is O→M = + A
2 O 6→i + →j
2→i 3→j + 6→i →j
or, O→M = 4→i + 2 +
∴ O→M = +
2→j

Example 2: If the position vectors of A and B are 5→i + 3→j and 2→i – 3→j respectively, find the position
vector of a point which divides AB in the ratio 2:3

a. internally b. externally

Solution:
Here, O→A = 5→i + 3→j and →OB = 2→i – 3→j

a. Let point M divide AB in the ratio 2:3 internally then the position vector of M is

O→M = 2 × →OB + 3 × O→A 3→j ) →→ A
or, O→M = 2(2→i – 32→j ) + 3(5→i + 5i +3j
or, O→M = 2
4→i – 6→j +21+5→i3 + 9→j
= 19→i + 3→j O M
5 5
O→M 19 →i 3 →j 2→i – 3→j 3
∴ = 5 + 5
B

b. Let point N divide AB in the ratio 2:3 externally then the position vector of N is N

O→N = 2 × →OB – 3 × O→A
or, O→N =
or, O→N = 2(2→i –23–→j3) – 3(5→i + 3→j ) 2

4→i – 6→j –21–5→i3 – 9→j = – 11→i – 15→j 5→i + 3→j A 3
–1 – 1 O
∴ O→N = 11→i + 15→j
→→

2i –3j B

Example 3: In the given figure, →OA = →p and →OB = →q . If A→M = 4 →AB , find O→M. M

Solution: O →q B
Here, O→A = →p, →OB = →q and A→M = 4 →AB . →p A
In DABM, O→M = O→A + A→M [ Triangle law of vector addition]
or, O→M = O→A + 4 →AB
or, O→M = O→A + 4( →OB – O→A ) = O→A + 4 →OB – 4 O→A = 4 →OB – 3 O→A
\ O→M = 4→q – 3→p

222 / Optional Mathematics Class 10

Alternatively

Here, O→A = →p, →OB = →q and A→M = 4 →AB . So, AB = 13.
BM
O→M O→A
Hence, →OB = 1 × 1 + 3 ×
+ 3
O→M + 3→p
or, →q = 4

\ O→M = 4→q – 3→p

Example 4: In D+L4M→j ,Nfi,n→dOLO→N=.2→i + 3→j , O→M = →i – 2→j and position vector of centroid G of DLMN is
O→G 3→i
=

Solution: = →OL + O→M + O→N
3
As, G is the centre of DLMN, O→G
or, 3 O→G = →OL + O→M + O→N

or, O→N = 3 O→G – →OL – O→M

or, O→N = 3(3→i + 4→j ) – (2→i + 3→j ) – (→i – 2→j )

or, O→N = 9→i + 12→j – 2→i – 3→j – →i + 2→j

\ O→N = 6→i + 11→j .
Example 5: If ABCD is a parallelogram in which →AB = 7→i + 2→j and →BC = →i + 5→j , find the →AP and →CQ

where P and Q are points of trisection of diagonal BD. C

Solution:

Here, →AB = →DC = 7→i + 2→j and →BC = A→D = →i + 5→j as opposite D P
sides of given parallelogram. Q

As P divides BD in the ratio BP:PD = 1:2 = m:n, B Vectors

Considering →AB = 7→i + 2→j as →a and A→D = →i + 5→j as →b, we have →AP = m→b + n→a A
m + n
1(→i 5→j ) + 2(7→i 2→j )
or, →AP = + 1+2 +

or, →AP = 15→i + 9→j
3
∴ →AP = 5→i + 3→j

Again Q divides BD in the ratio BQ:QD = 2:1 = m:n, considering →CB = – →i – 5→j as →a and →CD = – 7→i – 2→j as →b.

→QC = m→b + n→a
m+n
7→i 2→j ) + 1(–→i 5→j ) 14→i 4→j →i 5→j
or, →CQ = 2(– – 2+1 – = – – 3 – –

or, →CQ = – 15→i – 9→j = – 5→i – 3→j
3

Formulae of Vector Geometry / 223

Example 6: In the given figure, AD, BE and CF are the medians of A E
the triangle ABC; prove that →AD + →BE + →CF = 0. F

Solution:

In ∆ABC, D, E and F are midpoints of BC, CA and AB respectively. B D C
Here, A→D + →BE + →CF
= ( →AB + →BD ) + ( →BC + →CE ) + ( →CA + →AF )

= →AB + 1 →BC + →BC + 1 →CA + →CA + 1 →AB
2 2 2

= 3 →AB + 3 →BC + 3 →CA Alternately,
2 2 2
Using mid-point formula,
3 →AB →BC →CA )
= 2 ( + + A→D = 1 ( →AB + →AC ),
2
= 0 (As →AB + →BC + →CA = 0). →BE 1 ( →BC →BA )
= 2 + and

→CF = 1 ( →CA + →CB )
2
Now, A→D + →BE + →CF

= 1 ( →AB + →AC ) + 1 ( →BC + →BA ) + 1 ( →CA + →CB )
2 2 2
1 ( →AB →AC →BC →BA →CA →CB )
= 2 + + + + +

= 1 ( →AB – →CA + →BC – →AB + →CA – →BC )
2
1
= 2 × 0 = 0

Example 7: PQRSTU is a regular hexagon; show that PU
→PR + →PS + →UP + →TP = 3 →PQ .

Solution: Q T
R S
PQRSTU is a regular hexagon and joins UR. Then by
the property of a regular hexagon, 2 →PQ = →UR P U
Q T
Now, from DPRU, →UP + →PR = →UR
Again, from DPST, →TP + →PS = →TS R S
So, →PR + →PS + →UP + →TP
= ( →UP + →PR ) + ( →TP + →PS )
= →UR + →TS
= 2 →PQ + →PQ ( →TS = →PQ )
= 3 →PQ
Hence, →PR + →PS + →UP + →TP = 3 →PQ .

224 / Optional Mathematics Class 10

Exercise 6.2A

1. a. If the position vectors of two points A and B are 3→i + 2→j and 5→i – 6→j respectively, find the

position vector of mid-point M of AB.
b. If the position vector of M and N are 9→i + 3→j and 3→i – 5→j respectively, find the position vector

of point P such that M→P = →PN .
c. If the position vectors of two points A and B are – 2→i + 3→j and – 3→i + 4→j respectively, find the

position vector of point C where B is the midpoint of AC.
d. If the position vector of the midpoint of the line segment EF is 4→i + 7→j and the position vector

of the point E is – 3→i – 4→j , find the position vector of the point F.
2. a. If 5→i + 3→j and 5→i – 2→j are the position vectors of points A and B respectively, find the position

vector of point P which divides AB in the ratio 1:4 internally.
b. In ∆ABC, →AB = 2→i + 3→j and →BC = 4→i – 4→j . If M divides AC in the ratio 1:4 internally, find B→M.

c. If the line AB is divided by a point M in the ratio 1:2 and the position vector of points A and
M are 2→i + 6→j and 3→i + →j , find the position vector of B.

3. a. If the co-ordinates of the points A and B are (– 1, – 2) and (5, 1) respectively, find the position
vector of the point C which divides AB.

i. internally in the ratio 2:1 and ii. externally in the ratio 5:2

b. Find the position vectors of the points dividing the join of points with position vectors
2→a – 3→b and 3→a – 2→b in the ratio 2:3

i. internally and ii. externally

4. a. In the given figure O→A = →u, →OB = →v and B C Vectors
→BA = 3 →BC , find the value of →OC in terms of C B
→u and →v.

b. Find →OC if →AC = 3 →AB where the given O Q.N. 4(a) A O Q.N. 4(b) A
figure, O→A = →a, →OB = →b. O

c. In the given figure O→A = →a and →OB = →b. If DA B Q.N. 4(c) C
→AC = 3 →AB , find →OC . C P
d.
In the given figure, →CA = 2→a and →AB = →b. A Q.N. 4(d) B p
If DC = 2BC, find the value of →DC .

e. In the diagram, →PQ = →p and →QS = →q. If QS =
SR
31, express →PR in terms of →p and →q. R Q.N.4(e) S q Q

Formulae of Vector Geometry / 225

15 3
5. a. If the position vectors of the vertices A, B and C of triangle ABC are 1 , – 1 and 6

respectively, find the position vector of the centriod of triangle ABC.

25
b. The position vectors of vertices Q and R of triangle PQR are – 8 and – 1 respectively. If the
5
position vector of the centroid G of DPQR is , find the position vector of the vertex P.
c. D is the midpoint of the side BC of DABC. – 5 position vectors of A and D are 3→i + 2→j and
The
3→i + 5→j respectively, find the position vector of centroid G of ∆ABC.

d. In DABC, O→A = 3→i – 5→j , →OB = – 7→i + 4→j and the position vector of centroid G is 2→i + →j . Find
→OC .

6. a. Prove that →PQ + →QR + →RS + →SP = 0 in the given quadrilateral PQRS.
R

b. In the given diagram, prove that S C

→AB + →BC + →CD + →DE + →EA = 0. QD B

Exercise 6.2B Q.N. 6(a) Q.N.6(b)

P A
E

1. a. If ABCD is s parallelogram in which →AB = 2→i – 7→j and →BC = 5→i – →j , find A→M and →CN

where M and n are the points of trisection of diangonal BD.
b. PQRS is a parallelogram in which →PQ = 3→i – 2→j . If A and B are the points of trisection of

diagonal QS and →PA = 4→i + 2→j , find →QR and →RB .
c. If ABCD is a parallelogram in which →AB = 3→i – 4→j and A→D = 3→i + →j . Find B→M and →BN

where M and N are the points of trisection of diagonal AC.
d. If ABCD is a parallelogram in which →AB = 6→i + 3→j and →BC = 4→i – 7→j and the diangonals

intersect at M. If P is the mid-point of AM and Q is the midpoint of CM, Find →DP and →BQ .

2. a. In the given figure, D and E are the mid- A
b.
points of BC and CA respectively. If B Q.N. 2(a) D E
position vectors of A, B and C are →a, →b and P
→c respectively, find →AE and →DE . T
In the given figure, →p, →q and →r are the C R

position vectors of P, Q and R respectively. Q Q.N. 2(b) S
If QS : SR = 2:1 and PT : TR = 3:1, find →ST
and →TR .

226 / Optional Mathematics Class 10

3. a. The position vectors of three points A, B and C are →a + →b, 2→a – →b and 2→b – →a respectively. If
AM : MB = AN : NC = 1:2, then show that →BC = 3M→N.

b. The position vectors of four points A, B, C and D are 2→a + 4→c, 5→a + 3 3→b + 4→c, – 2 3→b + →c
2→a →c 2
and + respectively, show that AB is parallel to CD and CD = 3 AB.

4. a. OABC is a parallelogram. Points P and Q divide OC and CQ B
BC such that CP:PO = CQ:QB = 1:3. If O→A = →a and P
→OC = →c, find the vector which represents vector →PQ and
prove that →PQ and →OB are parallel vectors. OA

b. In the adjoining figure, OAB is a triangle. X is the point on O
OA such that OX:XA = 3:1. Y is the midpoint of OB and
Z is the mid-point of XB. If O→A = →a and →OB = →b, find O→X , X Y
O→Y and →OZ in terms of →a and →b. Also show that →YZ is A Z
parallel to O→A , and find the ratio of the lengths YZ:OA.
B

c. ABCD is a parallelogram. Points P and Q are on AB and AD such that AP:PB = AQ:QD = 1:3
and M and N are the midpoints of CD and BC respectively, prove that M→N = 2 →QP

5. a. In DOAB, →OB = →u, →OP = →v and →BA = 5 B O

→PA , prove that O→A = 1 (5→v – →u). P
4 Q.N. 5(a) A
b. In DOAC, →OB = →p, O→A = →q and
→OC = 4→p – 3→q, prove that →BC = 3 →AB . A B Q.N. 5(b) C

O

6. a. If A, B and C are the mid-points of sides QR, P A Vectors
PR and PQ of DPQR respectively, prove that CB FO E
→AP + →BQ + →CR = 0.

b. In DABC, the medians AD, BE and CF Q A RB DC
intersect at O, prove that D→O + →EO + →FO = 0.
Q.N. 6(a) Q.N. 6(b)

7. a. If ABCDE is a regular pentagon, prove that: AD C
→AB + →AC + →BC + →DC + →ED + →AE = 3 →AC
BE A B
b. In the given figure, ABCD is a C Q.N. 7(a) D O
parallelogram. If O→A = →a, →OB = →b,
→OC = →c find O→D . Q.N. 7(b)

b. If AC and BD are the diagonals of a parallelogram ABCD, prove that: →AC + →DB = 2 →AB .


Formulae of Vector Geometry / 227

6.3 Theorems on Vector Geometry

We have many theorems on plane geometry. Those theorems can also be proved by the vector

goemetry. Some of them are included in the curriculum of addtional mathematics class 10.

Theorem 1: A

The line joining mid-points of two sides of a trianlge is parallel P Q
to the third side and half of it.

Given : In triangle ABC, P and Q are the mid-points of sides AB

and AC respectively. B C

To Prove: PQ//BC and PQ = 1 BC.
Proof: 1. 2
→BA = 2 →PA , →AC = 2 A→Q [ P and Q are mid-points of AB and AC]

2. →PQ = →PA + A→Q A ]
3. →BC = →BA + →AC
4. →BC = 2 →PA + 2 A→Q = 2( →PA + A→Q ) [Triangle law of vector addition Q
P
A ]

[Triangle law of vector addition C
B

[From 1 and 3]

5. →BC = 2 →PQ [From 2 and 4]

or, →PQ = 1 →BC
2
1
Hence, PQ//BC and PQ = 2 BC as equal vectors are parallel and equal in length.

Theorem 2:

The line joining vertex and mid-point of the base of an isosceles triangle is perpendicular to the

base. A
Given : ABC is an isosceles triangle with AB = AC and AD is

median of DABC.

To Prove: AD⊥BC.

Proof: 1. A→D = 1 →AB + →AC A B D C
2
[Mid-point theorem B D C]

2. →BC = →BA + →AC A ]
= →AC – →AB
[Triangle law of vector addition C
B

3. A→D . →BC = 1 →AB + →AC . →AC – →AB
2
1 →AC 2 – →AB 2
= 2

= 1 (AC2 – AB2) = 1 × 0 = 0 [ AC = AB]
2 2
4. A→D ⊥ →BC [ A→D . →BC = 0]

Hence, AD⊥BC.

228 / Optional Mathematics Class 10

Theorem 3:

The quadrilateral joining the mid-points of the adjacent sides of any quadrilateral is a

parallelogram. C

Given : ABCD is a quadrilateral and P, Q, R and S are the midpoints of D R
AB, BC, CD and DA respectively.

To Prove: PQRS is a parallelogram. SQ

Construction: Join AC. C

Proof: 1. →AC = →AB + →BC [Triangle law of vector addition AB ] A
= 2 →PB + 2 →BQ
P B

[ P and Q are mid-points of AB and AC respectively]

= 2 →PB + →BQ = 2 →PQ

2. Similarly, →AC = 2 →SR [Same as 1]

3. →PQ = →SR [From 1 and 2]

4. PQ//SR, PQ = SR [ Equal vectros are parallel and equal in length]

5. PQRS is a parallelogram. [From 4]

Alternately: If E, F, G and H are the mid-points of the sides AB, BC, CD and D G C
O F
Given: B
E
DA respectively.

To Prove: EFGH is a parallelogram. H

Construction: Let O be the origin.

Proof: 1. →OE = 1 O→A + →OB , →OF = 1 →OB + →OC , A
2 2

O→G = 1 →OC + O→D , O→H = 1 O→A + O→D [Mid-point theorem] Vectors
2 2

2. →EF = →OF – →OE = 1 →OB + →OC – 1 O→A + →OB
2 2

= 1 →OB + →OC – O→A – →OB = 1 →OC – O→A
2 2

3. H→G = O→G – O→H = 1 →OC + O→D – 1 O→A + O→D
2 2

= 1 →OC + O→D – O→A – O→D = 1 →OC – O→A
2 2

4. →EF = H→G [From 2 and 3]

5. EF//HG and EF = HG [ Equal vectors are parallel and equal in length]

6. EFGH is a parallelogram. [From 5]

Theorems on Vector Geometry / 229

Theorem 4:

The diagonals of a parallelogram bisect each other. A D

Given: ABCD is a parallelogram and AC and BD are its diagonals.

To Prove: AC and BD bisect each other. NM

Construction: Let M and N are mid-points of AC and BD respectively. C
B
Proof: →BA + →BC
1. B→M A
= 1
2 [Mid-point theorem M ]
BC

2. →BN = 1 →BD [As N is the mid-point of BD]
2
AD
3. →BD = →BA + →BC
[Parallelogram law of vector addition N ]
4. →BN = 1 →BA + →BC BC
2
5. B→M = →BN [From 2 and 3]

[From 1 and 4]

6. M and N coincide. [B→M = →BN ]

Hence, mid-points of AC and BD coincide. i.e. AC and BD bisect each other.

Alternately,

Given: PQRS is a parallelogram, PR and QS are the diagonals. S R
P Q
To Prove: PR and QS bisect each other.

Construction: Let O be the origin. O
Proof:
1. →PQ = →SR

[Vectors representing opposite sides of parallelogram]

2. →PQ = O→Q – →OP , →SR = →OR – →OS

3. O→Q – →OP = →OR – →OS [From 1 and 2]

or, O→Q + →OS = →OP + →OR

4. O→Q + →OS = →OP + →OR [Dividing both sides of 3 by 2]
2 2

i.e. Position vector of mid-point of QS

= position vector of mid-point of PR

5. Mid-points of QS and PR are same. [From 4]

Hence, PR and QS bisect each other.

230 / Optional Mathematics Class 10

Theorem 5:

The diagonals of a rectangle are equal.

Given: ABCD is a rectangle and AC and BD are its diagonals. A D
To Prove: AC = BD C

Proof: 1. →AC = →AB + →BC

or, →AC 2 = ( →AB + →BC )2

= →AB 2 + 2 →AB . →BC + →BC 2 B

= →AB 2 + 2 × 0 + →BC 2 [ABC = 90°, so →AB . →BC = 0]

= →BC 2 + →AB 2

2. →BD = →BC + →CD

or, →BD 2 = ( →BC + →CD )2

= →BC 2 + 2 →BC . →CD + →CD 2

= →BC 2 + 2 × 0 + →CD 2 [BCD = 90°, so →BC . →CD = 0]

= →BC 2 + →CD 2

= →BC 2 + →AB 2 [| →AB | = | →CD |, so →AB 2 = →CD 2]

3. →AC 2 = →BD 2 [From 2 and 3]

or, | →AC | = | →BD |

\ AC = BD

Hence, diagonals of rectangle are equal.

Theorem 6:

The diagonals of a rhombus are perpendicular to each other. Vectors

Given: PQRS is a rhombus and PR and QS are its diangonals. P S
R
To Prove: PR⊥QS

Proof: 1. →PR = →PQ + →QR [Triangle law of vector addition]
= – →QP + →QR = →QR – →QP
Q
2. →QS = →QR + →RS
= →QR + →QP [ →RS = →QP ]
3. →PR . →QS = →QR – →QP →QR + →QP
= →QR 2 – →QP 2
= →QR 2 – →QR 2 = 0 [| →QR | = | →QP |, sides of a rhombus]
4. →PR ⊥ →QS [ →PR . →QS = 0 in 3]

or, PR⊥QS

Hence, the diagonals of rhombus are perpendicular to each other.

Theorems on Vector Geometry / 231

Theorem 7: C

The angle at the circumferece of a semicircle is a right angle.

Given: ACB is a semicircle with centre O and diameter AB. B OA
To Prove: ACB = 90°

Construction: Join OC.

Proof: 1. A→O = →OB [Vectors representing radii of semicirle in same direction]

2. →AC = A→O + →OC [Triangle law of vector addition O C

3. →BC = →BO + →OC [Triangle law of vector addition B ]

= – →OB + →OC = – A→O + →OC A
C

]

O



4. →AC . →BC = A→O + →OC . – A→O + →OC

= →OC 2 – A→O 2 = →OC 2 – →OC 2 [Since | A→O | = | →OC |, A→O 2 = →OC 2]

5. →AC ⊥ →BC or, ACB = 90° [ →AC . →BC = 0]

Hence, the angle at the circumference of a semicircle is a right angle.

Theorem 8:

The mid-point of the hypotenuse of a right angled triangle is equidistant from each vertex.

Given: ABC is a right angled triangle with ABC = 90°. M is the A

midpoint of htypotenuse AC.

To Prove: BM = CM = AM M
Proof: 1. →BA = B→M + M→A
M

[Triangle law B C]

2. →BC = B→M + M→C A B C

[Triangle law B M]

= B→M + A→M [ M is the midpoint of AC, so M→C = A→M]
= B→M – M→A

3. →BA . →BC = 0 [ ABC = 90°]

4. B→M + M→A . B→M – M→A = 0 [From 1, 2 and 3]

or, B→M 2 – M→A2 = 0

or, B→M 2 = M→A2

or, |B→M| = |M→A|

or, BM = MA [ M is the mid-point of AC]
5. AM = CM

6. AM = BM = CM [From 4 and 5]

232 / Optional Mathematics Class 10

Example 1: In the given figure, B is the mid-point of BC and AD⊥BC. A
DC
Prove that ABC is an isosceles triangle.

Given: In ∆ABC, BD = DC and AD⊥BC.

To Prove: ∆ABC is an isosceles triangle.

Proof: 1. →BC = →BA + →AC [Triangle law of vector addition A B

]
BC
= →AC – →AB

2. A→D = 1 →AB + →AC [Mid-point theorem A ]
2 B D
3. A→D . →BC = 0 C

or, 12 →AB + →AC . →AC – →AB = 0
or, 21 →AC 2 – →AB 2 = 0
or, →AC 2 – →AB 2 = 2 × 0 = 0
or, →AC 2 = →AB 2 or, | →AC | = | →AB |

4. AC = AB [From 3]

Hence, ∆ABC is an isosceles triangle.

Example 2: If the diagonals of a quadrilateral bisect each other, prove that it is a parallelogram.

Given: ABCD is a quadrilateral. The diagonals AC and BD bisect D C
each other. i.e. AO = OC, BO = OD
b a
To Prove: ABCD is a parallelogram. Vectors
Construction: Let A→O = →OC = →a and →BO = O→D = →b. O B

ab

A

Proof: 1. A→D = A→O + O→D D
= →a + →b
[Triangle law of vector addition O]
A

2. →BC = →BO + →OC C
= →b + →a
[Triangle law of vector addition O ]
B

3. A→D = →BC [From 1 and 2]

4. AD//BC, AD = BC [Equal vectors are parallel and equal in length]

5. ABCD is a parallelogram. [Being a pair of opposite sides equal and parallel in 4]

Theorems on Vector Geometry / 233

Example 3: Prove that the parallelogram with equal diagonals is a rectangle.

Given: ABCD is a parallelogram and AC = BD.

To Prove: ABCD is a rectangle.

Proof: 1. →AC = →AB + →BC [Triangle law of vector addition] A D
B C
2. →BD = →BC + →CD

= →BC – →DC = →BC – →AB [ →AB = →DC ]
3. →AC 2 = →BD 2

or, →AB + →BC 2 = →BC – →AB 2

or, →AB 2 + 2 →AB . →BC + →BC 2 = →BC 2 – 2 →BC . →AB + →AB 2

or, 2 →AB . →BC = – 2 →AB . →BC [ →BC . →AB = →AB . →BC ]

or, 4 →AB . →BC = 0

or, →AB . →BC = 0

4. →AB ⊥ →BC [ →AB . →BC = 0]

Hence, ABCD is a rectangle as a parallelogram with an angle 90°.
Example 4: ABCD is a parallelogram and P is the point of intersection of the diagonals. If →a, →b, →c, →d

and →p are the position vectors of A, B, C, D and P respectively, prove that →a + →b + →c + →d = 4→p.

Given: ABCD is a parallelogram and diagonals AC and A
BD intersect at point P. Also O→A = →a, →OB = →b, D

P

→OC = →c, O→D = →d and →OP = →p. B
C

To Prove: →a + →b + →c + →d = 4→p O

Proof: 1. P is the mid-point AC and BD. [Diangonals of parallelogram bisect each other]
O→A →OC [Mid-point theorem]
2. →OP = +
2
or, 2 →OP = O→A + →OC

or, 2→p = →a + →c

3. →OP = →OB + O→D [Mid-point theorem]
or, 2 →OP 2 →OB +
O→D
=

or, 2→p = →b + →d

4. 2→p + 2→p = →a + →b + →c + →d

\ →a + →b + →c + →d = 4→p. Hence the result.

234 / Optional Mathematics Class 10

Example 5: If the diagonals of a parallelogram are at right angles to each other prove that it is a rhom-
bus.

Given: ABCD is a parallelogram, its diagonals AC and BD are perpendicular to each other.

To Prove: ABCD is a rhombus.

Construction: Let →AB = →DC = →a and →BC = A→D = →b A D

Proof: 1. →AC = →AB + →BC = →a + →b [Triangle law of vector addition]

2. →BD = →BC + →CD = →b – →a [Triangle law of vector addition] O
3. →AC . →BD = 0 C
4. (→a + →b) (→b – →a) = 0 [AC⊥BD] B
[From 1, 2 and 3]

or, (→b)2 – (→a)2 = 0

or, (→b)2 = (→a)2

or, |→b|2 = |→a|2

or, |→b| = |→a|

∴ | →BC | = | →AB |

Hence, AB = BC and ABCD is a rhombus.
Example 6: Prove the pythagorus theorem.

Given: In DABC, ABC = 90°.
To Prove:
AC2 = AB2 + BC2 A
Proof: 1. B
→AC = →AB + →BC [Triangle law of vector addition]

2. →AC 2 = →AB + →BC  2 [Squaring both sides of 1] Vectors

or, →AC 2 = →AB 2 + 2. →AB . →BC + →BC 2 C

3. | →AC | 2 = | →AB | 2 + 2 × 0 + | →BC | 2 [AB⊥BC]

5. AC2 = AB2 + BC2 Proved.

Example 7: In the given figure PS = SR = QS. Prove that DPQR is a right angled triangle.

Given: PS = SR = QS. P
To Prove: DPQR is a right angled triangle.

Proof: 1. | →PS | = | →SR | = | →QS | [ PS = SR = QS] S

2. →QP = →QS + →SP = →QS – →PS [Triangle law of vector addition] R
Q

Theorems on Vector Geometry / 235

3. →QR = →QS + →SR [Triangle law of vector addition]

= →QS + →PS [ →PS = →SR ]

4. →QP . →QR = ( →QS – →PS ) ( →QS + →PS ) [From 2 and 3]

= →QS 2 – →PS 2

= →QS 2 – →QS 2 = 0 [ | →PS | = | →QS | from 1]

5. →QP ⊥ →QR [ →QP . →QR = 0]

or, ®PQR = 90°

Hence, DPQR is a right angled triangle. A B
Example 8: In the figure, ABCD is a trapezium where AB//DC. P

and Q are the mid-points of AD and BC respectively. Prove P Q
C
that →PQ = 1 ( →AB + →DC ).
2

D

Given: In trapezium ABCD, AB//DC, P and Q are mid points of AD and BC respectively.

To Prove: →PQ = 1 ( →AB + →DC )
Proof: 1. 2
→PQ = →PA + →AB + →BQ
[Polygon law of vector addition]

2. →PQ = →PD + →DC + →CQ [Polygon law of vector addition]

3. 2 →PQ = →PA + →AB + →BQ + →PD + →DC + →CQ [Adding 1 and 2]

3. 2 →PQ = →PA + →AB + →BQ – →PA + →DC – →BQ [ →PD = – →PA and →CQ = – →BQ ]

or, 2 →PQ = →AB + →DC

\ →PQ = 1 ( →AB + →DC )
2

Example 9: Prove that the medians of an isosceles triangle drawn from the extremes of the base are

equal.

Given: DPQR is an isosceles triangle with PQ = PR, QX and RY are the medians.
To Prove:
QY = RX. P

Proof: 1. Q→Y = →QP + →PY [Triangle law of vector addition]

= →QP + 1 →PR [Y is the midpoint of PR] X Y
2 Q R
2. Q→Y 2 = →QP 1 →PR
+ 2 2 [Squaring both sides of 1]

= →QP 2 + 2. →QP . 1 →PR + 1 →PR 2
2 4
→QP 2 →QP . →PR 1 →PR
= + + 4 2

236 / Optional Mathematics Class 10

= →QP 2 + →QP . →PR + 1 →QP 2 [As PQ = PR, →QP 2 = →PR 2]
4
54+→Q→PPR2 →QP . →PR
= + [Triangle law of vector addition]

3. →XR = →XP

= 1 →QP + →PR [Y is the midpoint of PR]
2

4. →XR 2 = 1 →QP + →PR 2 [Squaring both sides of 1]
2

= 1 →QP 2 + 2.21 →QP . →PR + →PR 2
4

= 1 →QP 2 + →QP . →PR + →PR 2
4

= 1 →QP 2 + →QP . →PR + →QP 2 [As PQ = PR, →QP 2 = →PR 2]
4

= 5 →QP 2 + →QP . →PR
4

5. Q→Y 2 = →XR 2 [From 2 and 4]
or, |Q→Y| = | →XR | i.e. QY = RX Proved.

Exercise 6.3

1. In the given figure, →AP = →PB and →AC = 2 →PQ then prove A
that →BQ = →QC .
P

2. In the given figure, PQ = PR and PA⊥QR then prove that A B Q.N.1 Q P
is the midpoint of QR. C

3. If D and E are the mid-points of sides AB and AC of ABC A Q Q.N.2 A R Vectors
DE
respectively, then show that: →BE + →DC = 3 →BC .
2

4. In the given figure, A, B and C are the mid-points of the sides B Q.N.3 CA P
PQ, QR and RP of DPQR respectively and O is any point,
prove that: O→A + →OB + →OC = →OP + O→Q + →OR C

B
QR
Q.N.4

5. In the adjoining figure, ABCD is a parallelogram. Let L O
B LC
and M be the middle points of BC and CD respectively.
Express →AL and A→M in terms of →AB and A→D and show
→AL A→M 3 →AC M
that + = 2 . A Q.N.5 D

6. In the adjoining figure, ABCD is a parallelogram. L and B LC

M are the mid-points of BC and CD respectively; prove M
that : →BD = 2L→M. A Q.N.6 D

Theorems on Vector Geometry / 237

7. If M and N are the mid-points of AD and BC of the quadrilateral ABCD, prove that
→AB + →DC = 2M→N.
CD

8. ABCD is a parallelogram and G is the point of intersection G O
A
of its diagonals. If O is any point prove that: B
O→A + →OB + →OC + O→D = 4 O→G A D

9. If P and Q are the mid-points of diagonals AC and BD of

quadrilateral ABCD respectively, then prove that: QP

→AB + A→D + →CB + →CD = 4 →PQ . BC

10. In the given figure, PQRS is a parallelogram. M and N P Q
are two points on the diagonal SQ. If SM = NQ, prove N

that PMRN is a parallelogram. MC
R R
11. In the given figure, P, Q, R and S are the mid-points S Q.N.10 Q

of the sides AB, BC, CD and DA respectively of the D B

quadrilateral ABCD. Prove that PQRS is a parallelogram. A PB S P

12. In the given figure, P, Q, R and S are the mid-points of the QS A Q.N.11
sides AB, AD, CD and BC respectively of the quadrilateral

ABCD. Prove that PR and QS bisect each other. DRC

Q.N.12

13. The diagonals of a rhombus bisect each other at right angles. Prove this statement.

14. If the diagonals of a quadrilateral bisect each other at right angles, prove that it is a rhombus.

15. Prove that the triangle having two equal medians is an isosceles triangle.

16. In the given figure O be the centre of a circle, OC be the O BR
perpendicular drawn through centre O to the chord AB. PS O
Prove vectorically that AC = BC. AC
Q.N.17
17. In the adjoining figure, O be the centre of a semicircle. Q.N.16
Prove that: RS2 = PS.SQ (Using vector method) R

18. If the bisector of vertical angle of any triangle is P Q
S
perpendicular to the base then prove that the triangle is
Q.N.18
an isosceles. Q
[Hints : →PS . →QR = 0
or, →PS .( →QP + →PR ) = 0
or, →PS . →QP + →PS . →PR = 0
or, →PS . →PR = →PS . →PQ

or, PS.PR cos SPR = PS.PQ cos SPQ

∴ [PR = PQ] 

238 / Optional Mathematics Class 10

7. TRANSsFdafOsdfRa MATION

Curriculum Contents Included Periods
17
a. Combination of transformations
– (Single combination of any two of reflection, rotation, translation and

enlargement)
b. Transformation using matrices

Objectives

On completion of this unit, students will be able to:
find the image of an object under the given combined transformations.

verify the relation "if two lines of reflection are parallel then one reflection followed by another
reflection is equivalent to translation".

verfy the relation "if two lines of reflection are intersected then one reflection followed by
another reflection is equivalent to the rotation".

verify the ralation "two reflections is equivalent two single rotation".

transform the object by using matrices.

Teaching Materials B'' A" Y C'

DABC is rotated through 90° about O(0, 0) 5 (4,4)
to get the image DA'B'C'. DA'B'C' is further
rotated through 180° about the point M(0, 3) 4
to get the image DA"B"C". C'' 3 (0,3) A' B'

\ DABC (4,4), 270° DA"B"C" 2

1

X' -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 X Transformation

-1
-2 A C

-3

-4

-5 B

Y'

Specification Grid

Contents Knowledge Understanding Application Higher ability Total Total
Transformation Each of 1 mark Each of 2 marks Each of 4 marks Each of 5 marks Quesitons Marks

1 – 1 1 3 10

 / 239

7 TRANSFORMATION

Review

We discussed various types of transformations in class IX. They are (a) Translation (b) Reflection
(c) Rotation and (d) Enlargement. The results of the various transformations mentioned above are given
below.

Combination of Transformations

Study the following figures of transformations.

Two Translations Two Reflections
Y Y

B' y=x

T

B T'

X' O X
T''
B''

X' O X
Y'
fig. (i) Y' fig. (ii)

Two Rotations Two Enlargements
Y Y

A'' A' C''
X'
X C'
A C

X' O fig. (iv) X
Y'
Y' fig. (iii)

a. In figure (i), triangle B is translated through vector 4 and B' is again translated through 2
to get the final image B". 4 – 6

b. In figure (ii), triangle T is reflected on the line y = x to get image T' and T' is again reflected
on the X–axis to get the image T". T" is the final image of triangle T.

240 / Optional Mathematics Class 10

c. In figure (iii), triangle A is rotated through 90º about O(0, 0) to get A' and A' is further rotated
through 90º about O(0, 0) to get A". A" is the final image of A.

d. In figure (iv), triangle C is enlarged with centre O(0, 0) and scale factor 2 and C' is again
enlarged with centre O and scale factor 2 to get final image C".

In each of the above cases, two transformations are of the same form i.e. reflection and reflection etc.
Besides these, the reflected objects can further be rotated or rotated objects can further be enlarged and so on.

Definition

When an object has been transformed, its image can further be transformed again to get a new image.
The single transformation which maps the given object to the final image of the above two transformations
is called a combination of transformation.
If a transformation T1 maps a point P to point P' and another transformation T2 maps the point P' to the
point P" then the combination of transformations T1 and T2 denoted by T2T1 (or T2°T1) maps the point P to P".
The transformations T2T1 and T1T2 have different meanings. T2T1 is the transformation in which T1 has the first
effect and then T2 whereas T1T2 is the transformation in which T2 has the first effect and then T1. In general,
T2T1 ≠ T1T2 and the combination of transformation T with itself is denoted by T°T or TT or T2.

7.1 Combination of Translations

Review

Translation is one of the transformations in which an

object is moved through a vector called translation vector. In the T'
given figure triangle T is translated through translation vector

5
4 . The object and the image in translation are congruent.

If P(x, y) be a point on the plane and P' be the image of P T
a
under the translation through translation vector b , then

P(x, y) → P'(x + a, y + b).

Two Translations Y Transformation

A point or an object once translated can further be translated.

Let P(1, 3) be a point on the plane. If P' be the image of X' P(1,3) P'(4,5)
3 O P''(6,0) X
Y'
the point P under the translation T1 = 2 and P" be the image of
2

point P' under the translation T2 = – 5 . Then the co-ordinates
of points P' and P" are P'(1 + 3, 3 + 2) i.e. P'(4, 5) and

P"(4 + 2, 5 – 5) i.e. P"(6, 0).

Combination of Translations / 241

32 5 5
Also, T1 + T2 = 2 + – 5 = – 3 and the point P(1, 3) translated by – 3 gives the image
32
(1 + 5, 3 – 3) i.e. (6, 0) which is P". Hence the translation of P by T1 = 2 followed by T2 = – 5 is
325
equivalent to single translation T2 + T1 = T2°T1 = 2 + – 5 = – 3 .

ac
Hence, if the translation T1 = b is followed by another translation T2 = d then the combined

a c a+c
translation T2°T1 (or T2T1) is described by b + d = b + d . Also the combined translations T2T1 and T1T2

a+c
have the same effect i.e. T2T1 = T1T2 = b + d .

ac

P(x, y) b P'(x + a, y + b) d P"(x + a + c, y + b + d)

Example 1: 2 –5
Solution: T1 = and T = are two translation vectors. Find T1T2(7, 2).
1 2 3

2 –5 2 –5 2 – 5 –3
Here, T1 = 1 and T2 = 3 then T1T2 = 1 + 3 = 1 + 3 = 4

–3
Now, T1T2(7, 2) = (7 – 3, 2 + 4) [Since 4 is the translation vector T1T2]

= (4, 6)

Hence, the image of (7, 2) after the translation by T1T2 is (4, 6).

Example 2: The image of a point (2, 3) is the point (7, – 4) under a translation a followed by another
3
b

translation vector 5 . Find the values of a and b.
Solution:

a 3
Here, the given point is (2, 3) and the final image is (7, – 4). T1 = b and T2 = 5

3 a 3+a
Now, T1 followed by T2 is T2T1 = 5 + b = 5 + b

So, P(2, 3) T2T1 P'(2 + 3 + a, 3 + 5 + b)

i.e. P'(5 + a, 8 + b)

From question P' is (7, – 4).

So, (5 + a, 8 + b) = (7, – 4)

Hence, 5 + a = 7 and 8 + b = – 4

or, a = 7 – 5 or, b = – 4 – 8

= 2 = – 12

242 / Optional Mathematics Class 10

Example 3: If P(– 1, 3), Q(– 4, 1) and R(– 2, 0) be the vertices of DPQR and T1 = 22
and T = be
3 2 – 6

two translations then find the image of DPQR under the combined translation T2T1 , show the

translation on the graph. Y
Solution:

22 P(-1,3)
Here, T1 = 3 and T2 = – 6 then

2 2 = 2+2 = 4 Q(-4,1) P'(3,0)
T2T1 = 3 + – 6 3–6 – 3
X' R(-2,0)O X
Now, the images of P, Q and R under the combined Q(0,-2)

translation T2T1 are

P(– 1, 3) → P'(– 1 + 4, 3 – 3) i.e. P'(3, 0) R'(2,-3)

Q(– 4, 1) → Q'(– 4 + 4, 1 – 3) i.e. Q'(0, – 2) Y'
R(– 2, 0) → R'(– 2 + 4, 0 – 3) i.e. R'(2, – 3)

The translation is shown in the above figure.

Exercise 7.1A

23
1. a. If T1 be the translation by the vector 1 and T2 the translation by the vector – 3 , find the

translation vector T1T2.
–3 1

b. If Ta be the translation by the vector 4 , Tb be the translation by the vector – 5 and Tc be the
3

translation vector 5 , find the following translation vectors.

i. TaTb ii. TbTc iii. TcTa

23
c. If T1 be the translation by the vector – 3 and T1T2 be the combined translation vector 6 , find

the translation vector T2.
55

d. If T2 be the translation by the vector 0 and T1T2 be the combined translation vector 6 , find

the translation vector T1. Transformation
32

2. a. If T1 be the translation by the vector 2 and T2 be the translation by the vector 4 , find the

image of A(– 2, 1) under the translation T1T2.

3
b. Find the image of the point (3, 8) under the translation T1 = – 2 followed by the translation

2
T2 = –3 .
2 3 – 4
3. If T1 be the translation 2 , T2 be the translation – 4 , T3 be the translation 2 and T4 be the

1
translation – 3 , find the images of the points by the following transformations:

Combination of Translations / 243

a. T1T2(2, 4) b. T2T1(1, 5) c. T2T3(– 1, – 4)

d. T3T1(2, 5) e. T4T2(–1, 5) f. T2T4(a, b)

g. T2T2(6, – 2) h. T32(6, 6) i. T43(– 2, 3)
a
4. a. The image of point (2, 5) is the point (6, – 4) under the translation followed by a translation
3 b

– 4 . Find the value of a and b. 1

b. The image of point M(– 6, 7) is the point (3, 2) under the translation T1 = –2 followed by a
a
translation T2 = b . Find T2.

Exercise 7.1B

–2 0
1. a. A(2, 4), B(3, 5), C(0, 3) are the vertices of a triangle ABC. If T1 = 1 and T2 = 2 are two

translations, find the co-ordinates of the image of ∆ABC under the combined transformation

T2T1. Show the translation on a graph.
12

b. A(–2, 3), B(3, –2) and C(2, 4) are the vertices of ∆ABC. If T1 = 2 and T2 = – 6 are two
translations, find the co-ordinates of the image of ∆ABC under the combined transformation

T1T2. Draw the translation on a graph.

c. For the ∆ABC with vertices A(– 2, 3), B(– 4, 1) and C(– 3, 5), find the co-ordinates of the

image ∆A'B'C' after translated through 3 . Again find the image ∆A"B"C" of ∆A'B'C' after
2

the translation through translation vector 3 . Present ∆ABC, ∆A'B'C' and ∆A"B"C" on the
– 4
same graph.

2. a. If a ∆ABC with vertices A(– 1, 1), B(1, 1) and C(1, 4) is mapped onto ∆A'B'C' by a translation

T and co-ordinates of A'(4, – 2). Find the co-ordinates of B' and C' and draw it on graph.

b. If P(3, 5), Q(4, 1), R(1, – 1) and S(– 3, 1) are the vertices of a quadrilateral PQRS. Quadrilateral
PQRS is mapped onto quad.P'Q'R'S' by a tanslation and co-ordinates of Q' is (1, – 3). find the
co-ordinates of remaining vertices of quad.P'Q'R'S'. Show the translation on a graph.

3. a. A(– 2, – 1), B(2, 3), C(8, – 1) and D(2, – 5) are the vertices of a kite, find its image under T1⸰T2
– 3 4

where T1 = 5 and T2 = – 4 are the translation vectors.
b. O(0, 0), A(2, 0), B(3, 1) and C(1, 1) are the vertices of a quadrilateral. This quadrilateral is

translated by the vector →AB to O', A', B' and C'. Again the quadrilateral O'A'B'C' is further
translated by vector →BC to O"A"B"C". Find the co-ordinates of the vertices of both images

and draw all the three quadrilaterals in the same graph paper.


244 / Optional Mathematics Class 10

7.2 Combination of Reflections T
T'
Review

In reflection, when an object is reflected on a given line, the line
is called the axis of reflection. The object and image are congruent. In
the figure the triangle T is reflected on the line y = x. The object and the
image are congruent to each other.

If P(x, y) be a point on the plane and P' be the image of P
under the reflection about various axis then,

1. Reflection about X-axis, P(x, y) X-axis P'(x, –y)

2. Reflection about Y-axis, P(x, y) Y-axis P'(–x, y)

3. Reflection about y = x, P(x, y) y = x P'(y, x)

4. Reflection about y = –x, P(x, y) y = –x P'(–y, –x)

5. Reflection about x = h, P(x, y) x = h P'(2h – x, y)

6. Reflection about y = k, P(x, y) y = k P'(x, 2k – y)

Two Reflections

A point or an object once reflected, can further be reflected. There are two cases for the combination
of two reflections.

Case I : When the axes of reflections are parallel.

Case II : When the axes of reflections intersect at a point.

Case I. When the axes of reflections are parallel Y

Let P be a point on the plane and AB and CD

two axes of reflections which are parallel. If P1 be the AC

reflection of P about AB and P2 be the reflection of P1 M P1 N P2
B DX
about CD and PP1P2 intersect AB and CD at M and N

respectively as shown in the figure. P2 is the image of P Transformation

under the combined reflection. P

Here, P→M = M→P1 and P→1N = N→P2 X' O
Now, P→P2 = P→P1 + →P1P2

=2→MP1 + 2→P1N Y'

= 2(→MP1 + →P1N)

=2M→N

Hence, if the axes of two reflections are parallel, then the reflection followed by reflection is a
translation. The distance of translation is twice the distance between the axes of reflections and the direction
is from the first axis of reflection to that of the second.

Combination of Reflections / 245

Example 1: Find the image of a point (2, – 5) under the reflection in X-axis followed by reflection in

the line y = 3.
Solution:

The image of point A(2, – 5) under the reflection in X-axis is A'(2, 5).

The image of point A'(2, 5) under the reflection in the line y = 3 is A"(2, 2 × 3 – 5) i.e. A"(2, 1).

Example 2: If R1 is the reflection about Y– axis and R2, the reflection about x = 4, find the image of

∆ABC under the combined reflections R2R1 where the vertices of ∆ABC are A(– 3, 1), B(– 2, 3) and

C(– 1, 2). Show that R2R1 is a translation.
Solution:

The vertices of ∆ABC are A(– 3, 1), B(– 2, 3) and C(– 1, 2).

Reflecting about Y–axis, Y x=4
P(x, y) → P'(– x, y)
x=0

So for reflection R1, B(-2,3) B'(2,3) B''(6,3)
A(– 3, 1) → A'(3, 1)
B(– 2, 3) → B'(2, 3) C(-1,2) C'(1,2) C''(7,2)
C(– 1, 2) → C'(1, 2) A''(5,1)
Reflecting about x = 4, P(x, y) →P'(2 × 4 – x, y) A(-3,1) A'(3,1)
X
X' O R

Now for reflection R2, Y'
A'(3, 1) → A"(2 × 4 – 3, 1) i.e. A"(5, 1)

B'(2, 3) → B"(2 × 4 – 2, 3) i.e. B"(6, 3)

C'(1, 2) → C"(2 × 4 – 1, 2) i.e. C"(7, 2)

Hence, image of ∆ABC under the combined reflection R2R1 is ∆A"B"C" where A"(5, 1), B"(6, 3)
and C"(7, 2).

Here, R1 = reflection about Y-axis (x = 0) and R2 = reflection about x = 4. Since the axes of reflectin

R1 and R2 are parallel, the combined reflection is a single translation by twice the distance between the axes.

To show it, we have the combined translation vector is 2M→N = 2 →OR = 2 4 = 8
0 0
8
In translating ∆ABC by vector 0 , we have

A(–3, 1) → A1(–3 + 8, 1 + 0) i.e. A1(5, 1)

B(–2, 3) → B1(– 2 + 8, 3 + 0) i.e. B1(6, 3)

C(–1, 2) → C1(–1 + 8, 2 + 0) i.e. C1(7, 2)

From above result ∆A"B"C" and ∆A1B1C1 are same.

∴ R2R1 is a translation.

246 / Optional Mathematics Class 10