Vedanta Opt. Maths Book 10 Final (2078)

vedanta Excel In Opt. Mathematics - Book 10

9. Prove that
((a)
tan Sc + T(((( = secT + tanT
((b) 4 2( (((
Sc 1 - sinT
tan 4 – T (=1 + sinT
( ((c) 2 (

sec Sc + T(( . sec Sc – T = 2secT
((d) 4 2(( 4 2

tan Sc – T( = cosT
( ((e) 4 2(( 1 +sinT
2cosT
cot T + Sc ( – tan T – Sc = 1 +sinT
( ((f) 2 4 ( 2 4

tan Sc + T + tan Sc - T = 2secT
4 2 4 2

10. Prove that
((a) (cosD - cosE)2 + (sinD - sinE)2 = 4sin2
((b) (sinD + sinE)2 + (cosD + cosE)2 = 4cos2 D-E
2
D-E
2

11. Prove that
( ( ( ((a)
cos 2Sc . cos 4Sc . cos 8Sc . cos 16Sc = 1
( ( ( ((b) 15 15 15 15 16

1+cos Sc 1+cos 3Sc 1+cos 5Sc 1+cos 7Sc = 1
8 8 8 9 8

Project Work

12. Discuss how to find the value of tan 7 1° . Then show that
2
1°
tan 7 2 = 6– 3+ 2 – 2.

3. (a) 3 (b) 1 (c) - 3 (d) - 44
2 2 125
3 -1 3 + 1
4. (a) 2 2 , 2 2 , 2 – 3

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10Vectors

10.0 Review

Discuss the following questions:
a. Classify the following quantities to vectors or scalars with your reasons.

(i) Temperature (ii) Pressure (iii) Force (iv) Length

(v) Area (vi) Acceleration (vii) Displacement (viii) Time

b. Give an example of each of the following vectors:

(i) Column vector (ii) Row vector (iii) unit vector

(iv) Equal vectors (v) Negative vectors (vi) position vectors

(vii) Unequal vectors (viii) Like and unlike vectors.

c. What role do vectors play in our daily life ? Give some examples.

o o oo oo
d. Let a = (2, -3) and b = (1, 2), find a + b and a - b . Express them in the
oo
form of x i + y j .

10.1 Scalar (or Dot) Product of Two Vectors

oo
Let OA = (2, 3), OB = (-3, 2) be two vectors. Plot them in a graph paper. Then,

(a) Find the product of x-components and y-components Y
of the vectors.

(b) Add the products.

(c) What is the result ? BA X
oo X' O

(d) Are OA and OB perpendicular to each other?
Here, 2 × (-3) + 3 × 2 = - 6 + 6 = 0

oo
OA is perpendicular to OB .

There are two types of product of two vectors: Y'
(a) Scalar product or dot product.
(b) Vector product or cross product.

We discuss only about scalar product or dot product of two vectors.

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Scalar Product of Two Vectors

The scalar product of two vectors oa and ob is defined as the
product of the magnitude of two vectors multiplied by the cosine
of the angle T between their directions.

Thus, oa · ob = |oa | |ob | cosT = ab cosT
where, |oa |= a and |ob |= b.

AO = oa and OB = ob

Now, draw perpendicular BM from B to OA.

Now, oa · ob = |oa | |ob |cosT = a b cos T

= (OA) (OB) cos T

= OA (OB cos T)
= (OA) (OM) = (magnitude of oa ) (component of ob in the direction of oa )

So, it is clear that the scalar product of two vectors is equivalent to the product of the
magnitude of one vector with the component of the other vector in the direction of this
vector.

If we write oa · ob , the rotation of oa towards ob is anti-clockwise and the angle T is taken to
be positive.

? oa · ob = ab cosT

If we write ob ;oa , the rotation of ob towards oa is clockwise and the angle T is taken to be
negative.

? ob · oa = b a cos(–T) = ba cosT
Hence, oa · ob = ob · oa

Note :
Let oa = x1oi + y1oj and ob = x2oi + y2oj be two vectors in terms of the components. Then
we know that oi and oj are unit vectors along X-axis and Y-axis respectively.

oi . oj = |oi ||oj |cos 90° = 0, oi . oi = |oi | |oi | cos0° = 1

Now, oa . ob = (x1 oi + y1) . (x2oi + y2oj ) = x1oi .(x2oi + y2oj ) + y1 oj . (x2oi + y2 oj )
= x1x2 oi . oi + x1y2 oi . oj + y1 x2 oj . oi + y1y2 oj . oj = x1x2 + y1y2

? oa . ob = x1x2 + y1y2

Hence, the scalar product of two vectors is equal to the sum of the product of their
corresponding components.

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Two Important Results of Scalar Product of Two Vectors.

(a) If oa and ob are perpendicular to each other, i.e., T = 90° B A
Then, oa . ob = |oa | |ob | cosT
= |oa | |ob | cos90° = 0 o
Hence, oa . ob = 0 b

o
a

Conversely, of oa . ob = 0, then T = 90°

oo
(b) If a and b are parallel to each other,

i.e., T = 0° or T = 180°

o o |oa | |ob | cos0° = 1
a. b=

and o o = |oa | |ob | cos180° = ab.(-1) = - ab
a. b

oo oo
Conversely, if a . b = ± ab, T = 0° or 180°, a and b are parallel to each other.

Some Important Results

a. If |oa | = a and |ob | = b, oo
a . b = abcosT.

oo (i) |oa |= 0 (ii) |ob | = 0 or o o
b. If a . b = 0, a A b

oo
c. If T = 0°, the value of a . b = ab is maximum.

oo
d. If T = 180°, the value of a . b = - ab is minimum.

oo oo
e. a . b = b . a .

oo o o oo
f. a . (k b ) = (k a ) . b = k( a . b ), where k z 0. k  R.

g. o o = |oa | |oa | cos0 = a.a. =a2
a. a

Also,

(i) o o = |oi | |oi | cos0° = 1.1.1 = 1
i. i

(ii) o o = |oj | |oj | cos0° = 1.1.1 = 1
j. j

(iii) o o = |oi | |oj | cos90° = 1.1.0 = 0
i. j

o . o = |oj | |oi | cos90° = 1.1.0 = 0
(iv) j i

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Angle between Two Vectors

Let us consider two points A(a1, a2) and B (b1, b2) in the plane. Then,

position vector of A = OA = oa = a1
a2
= ob = b1
Position vector of B = OB b2

Magnitudes of oa and ob are

| OA | = OA = a = |oa |

| OB | = OB = b = |ob |

Let ‘XOA = E, ‘XOB = D and ‘AOB = T. Then, ‘AOB = D – E = T.

Draw perpendiculars AM and BN from A and B to the x-axis. Then,

OM = a1, MA = a2, ON = b1 and NB = b2.

From the right-angled triangle OMA,

cosE = OM = aa1 ? a1 = a cos E
OA ? a2 = a sinE
aa2
sinE = MA =
OA

Similarly from the right-angled triangle ONB,

b1 = b cos D and b2 = b sin D
Now, a1b1 + a2b2 = a cos E b cosD + a sin E b sin D

= ab cos(D – E) = |oa | |ob | cos T ................(i)
But, by the definition of scalar product of two vectors, oa and ob

oa · ob = | oa | |ob | cos T .............(ii)

Now, from (i) and (ii) oa · ob = a1b1 + a2b2.

This result leads us to define scalar product of two vectors in another way.

Let oa = a1 and oa = b1 be two vectors. Then the scalar product of oa and ob is denoted
a2 b2
by oa · ob and is defined by oa · ob = a1 b1
a2 · b2 = a1b1 + a2b2.

Again from (i), |oa | |ob | cos T = a1b1 +a2b2

or, cos T = a1b1 + a2b2 or oa . ob
|oa | |ob | |oa | |ob |
oa ob .
This result gives us angle between two vectors and

cosT = a1b1 + a2b2 or oa . ob
|oa | |ob | |oa | |ob |

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Worked out Examples

oo oo
Example 1. (a) If a = (4,5) and b = (2, 3), find the scalar product of a and b .
Solution:
(b) If |oa |= 4 and o = 3, T = 45°. find o o .
Example 2. b a. b
Solution:
oo
Example 3. (a) Here, a = (4, 5), b = (2, 3)
Solution:
Since (a1, a2). (b1,b2) = a1b1 + a2b2
oo
a . b = (4, 5).(2,3) = 4.2 + 5.3 = 8+15= 23

oo
(b) Here, | a | = 4, | b | = 3, T = 45°

oo o o 1 = 6 2
a . b = | a | | b | cosT = 4.3 cos45° = 12. 2

o oo ooo
If a = 2 i + j and b = i - 2 j , then, find the following:

oo oo
(i) a . b (ii) angle between a and b

o o oo o o
Here, a = 2 i + j , b = i - 2 j

oo o o o o
(i) a . b = (2 i + j ). ( i - 2 j ) = 2.1 + 1.(-2) = 2 - 2 = 0

oo oo
Since a . b = 0, a and b are perpendicular to each other.

o
(ii) Here,| a | = 22 + 12 = 4 + 1 = 5

o
| b | = 12+(-2)2 = 1 + 4 = 5

oo
Let T be the angle between a and b , we get,

cosT = oo = 0 =0
a. b 55
| oa || ob |

? T = 90°

ooo o oo o
(a) If a = 3 i + 4 j and b = -8 i + 6 j , show that a is perpendicular
oo o oo
to b , (b) If a = (2, 4) and b = (4, 8), show that a and b are parallel to

each other.

o o oo oo
(a) Here, a = 3 i + 4 j , b = - 8 i + 6 j

oo oo oo
a . b = (3 i + 4 j ) . (-8 i + 6 j )

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= 3.(-8) + 4.6 = - 24 + 24 = 0

oo o o
Since a . b = 0, a and b are perpendicular to each other.

( ) ( )(b)
o 2 o 4
Here, a = 4 , b= 8

( ) ( )o o 2 . 4 = 2.4 + 4.8 = 8 + 32 = 40
4 8
a. b =
o
| a | = 22 + 42 = 4 + 16 = 20 = 2 5

o
| b | = 42 +82 = 16+64 = 80 = 4 5

oo
Let T be the angle between a and b

Now, cosT oo =2 40 5 = 1 = cos0°
= a. b 5 ×4

oo
| a || b |

? T = 0°

oo
Hence, a and b are parallel to each other.

oo oo
Example 4. (a) Find the value of m if two vectors m i + 7 j and 4 i + 8 j are
Solution:
perpendicular to each other.

( ) ( )o

(b) If OA =
2 o -3 and ‘AOB = 90°, find the value of m.
m and OB = 4

(a) Here, let o = o + o o oo
a mi 7 j, b =4 i+8 j

oo oo
Since a and b are at right angle, a . b = 0

oo oo oo
i.e. a . b = (m i + 7 j ) . (4 i + 8 j )

or, 0 = 4m + 56

or, 4m = - 56 ? m = -14

( ) ( )(b) o o
oo 2 , OB = b= -3 , ‘AOB = T = 90°
Let OA = a = m 4

Then, by definition of dot product,

( ) ( )o o 2 . -3 = 2(-3) + m.4 = - 6 + 4m
m 4
a. b= oo

Since, T = 90° or, a . b = 0

or, - 6 + 4m = 0

or, 4m = 6 ? m= 3
2

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Example 5. If o = oo and o o o oo
Solution: OA 2 i +6 j OB = 2 i+ j find the angle between OA and OB.

Example 6. o o oo o o
Solution: Here, OA = 2 i + 6 j , OB = 2 i + j .

Example 7. oo
Solution: Let T be the angle between OA and OB . Then,

oo oo
a. b OA . OB
cosT = =
| oa || o OoA || o
b | | OB |

oo o o oo
But, OA . OB = (2 i + 6 j ) . (2 i + j )

= 2 × 2 + 6 × 1 = 10

o
| OA | = 22 + 62 = 4 + 36 = 40 = 2 10

o
| OB | = 22 +12 = 4+1 = 5

? cosT = 2 10 5 = 10 2 = 1 = cos45°
10. 10 2

? T = 45°

Hence, the angle between them is 45°.

If oo = 12, T = 45°, | oa | = 4, find | ob |.
a. b

Here, o o = 12, o = 4
a. b | a|

By definition of dot product of two vectors,

o . o = | oa | | ob | cosT
a b

o
or, 12 = 4 | b | cos45°

or, o 1 o
3=| b| 2 ? b =3 2

oo o o oo
If | a + b | = | a - b | , then, prove that a and b are perpendicular to

each other.

o o oo
Here, | a + b | = | a - b |

Squaring on both side, we get

oo oo
| a + b |2 = | a - b |2

or, (oa + ob )2 = (oa - ob )2 (?|oa |2 = a2)

oo oo
or, a2 + 2 a . b + b2 = a2 - 2 a . b + b2

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oo oo
or, 4 a . b = 0 ? a.b =0

oo
Hence, a and b are perpendicular to each other.

Example 8. If P(0,2), Q(2,5) and R(8,1) are the vertices of a triangle ∆PQR, prove that
Solution: ∆PQR is a right angled triangle

oo o
Here, we have OP = (0,2), QR = (2, 5), OR = (8, 1)

o oo
Now, PQ = OQ - OP = (2, 5) - (0,2) = (2, 3)

o oo
QR = OR - OQ = (8, 1) - (2, 5) = (6, -4)

o oo
RP = OP - OR = (0, 2) - (8,1) = (-8, 1)

oo
PQ . QR = (2, 3) . (6,- 4) = 2.6 + 3.(-4) = 12 - 12 = 0

oo
? PQ . QR = 0, angle between PQ and QR is 90°.

Hence, ∆ PQR is a right angled triangle.

o oo
Example 9. Find the angle made by a = 6 i - 8 j with X - axis.

Solution: oo
We know that unit vectors along X-axis and Y-axis are i and j respectively.

oo
Let b = unit vector along X-axis = (1, 0) = i

o oo
Given a = 6 i - 8 j

oo
Let T be the angle between a and b .

oo o o .oi
a. b (6 i - 8 j )
cosT oo = = 6 = 3
62 + (-8)2 12 + 02 100 50
| a || b |
( )?
T = cos-1 3 = 86.57°
50
o o o
Example 10. If p+ q+ r = (0, 0), | o o o 127 , find the angle
oo p| = 6, | q | =7,| r|=
between p and q .
oo
Solution: Let T be the angle between p and q .

oo o
| p | = 6, | q | = 7, | r | = 127
o oo
Here, p + q + r = 0

oo o
or, p + q = - r

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Squaring on both sides,

oo
( p + q )2 = r2

oo
or, p2 + 2 p . q + q2 = r2

oo
or, 62 + 2| p | | q | cosT + 72 = 127

oo
or, 2 × 6 × 7 × cosT = 127 - 85, where T is angle between a and b .

or, cosT = 42
84
1
or, cosT = 2 = cos60° ? T = 60°

Hence, the angle between two vectors is 60°.

Example 11. Given that o + o and o - o are orthogonal vectors and o o are
(a 2 b) 5a 4b oo a, b

unit vectors. Find the angle between a and b .

oo oo
Solution: Here, a and b are unit vectors. | a | = 1, | b | = 1 .

Given vectors are orthogonal to each other. Hence, their dot product is zero.

o ooo
? ( a + 2 b ).(5 a - 4 b ) = 0

o o o oo o
or, a . (5 a - 4 b ) + 2 b .(5 a - 4 b ) = 0

oo oo
or, 5a2 - 4 a . b + 10 a . b - 8b2 = 0

o o oo
or, 5.1 - 4 a . b + 10 a . b – 8.1 = 0

oo
or, 6 a . b = 3

or, | o | o . cosT = 1 or, 1.1. cosT = 1
a| b| 2 2

or, cosT = 1 or, cosT = cos60°
2
? T = 60°

Hence, angle between them is 60°.

Exercise 10.1

Very Short Questions

1. (a) Define scalar product of two vectors.

(b) If T is the angle between oa and ob , what is the formula of dot product of o and
o a

b?

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(c) If o o = 0, what is the relation between o and ob ?
a. b a

(d) If o = o y1 o and o o y2 o find o o
a x1 i + j b= x2 i + j, a. b.

( ) ( )(e) If oa =a1 and o = b1 , find the value of oa . ob .
a2 b b2

oo
(f) Under which condition two vectors a and b are perpendicular to each other ?

oo
2. Find the dot product of a and b in following cases:

oo
(a) | a | = 2, | b | = 3, angle between them (T) = 60°

oo
(b) | a | = 2, | b | = 5, angle between them = 90°

oo
(c) | a | = 15, | b | = 25, angle between them = 45°

oo
3. Find the dot product of the vectors a and b in the following cases:

( ) ( )(a) a = 4 , b = 1
o 3o 2 oo oo o o
(b) a = i + 2 j , b = 3 i - j

o o oo o o o oo o
(c) a = 4 i + 2 j , b = - i + 2 j (d) a = 10 i , b = 5 j

oo o o oo oo oo
4. If a = (2, 1), b = (2, –3), find (i) a . b (ii) b . a (iii) show that a . b = b . a

o 2o 4o 2

( ) ( ) ( )5. If a = 3 , b = 2 , c = -3 , find:

(a) a2 (b) b2 (c) c2

oo oo oo
(d) a . b (e) b . c (f) a . c

Short Questions

6. Find the angles between given pair of vectors:

o o oo o o oo
(a) | a | = 4, | b | = 5, a . b = 10 (b) | a | = 3, | b | = 5, a . b = 7.5

( ) ( )(c) a = -3 , b = -2
o 2o 1 oo
(d) p = (3, 4), q = (-4, 3)

o o oo o o
(e) p = 4 i + 2 j , q = - i + 2 j

7. Show that following pair of vectors are parallel:

o 2o 4 ( ) ( )o 1 o -3
(b) p -3 , q = 9
( ) ( )(a) p = 3 , q = 6
o o oo o o
o o oo o o (d) a = i + 5 j , b = -2 i - 10 j
(c) a = 2 i - 3 j , b = 4 i - 6 j

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8. Show that pair of vectors are perpendicular to each other in the following cases :

oo oo
(a) a = (3, 6), b = (6, -3) (b) a = (1, 2), b = (-2, 1)

( ) ( )(c) p = 2 i - 3 j , q = 2 j + 3 i (d) p = 4 , q = 3 -4
o o oo o o o 3o

9. Find the value of k if each pair of the following vectors are orthogonal to each other:

oo
(a) a = (2, -3), b = (k, 4)

o o oo oo
(b) a = (4 i + k j ) , b = (3 i - 6 j )

o o oo o o
(c) a = 3 i - 2k j , b = 10 i - 6 j

o o oo o o
(d) p = 2k i +4 j , b = 3 i +2 j

Long Questions

oo oo o
10. Express PQ and RS in the form of x i + y j and show that PQ is perpendicular to
o
RS in the following cases.

(a) P(3, 5), Q(4, 6), R(7, 6), S(8,5) (b) P(3, -2), Q(5, 1), R(-1, 4), S(2, 2)

11. (a) In ∆PQR if o = o - o , o = o o prove that ∆PQR is a right
PQ 5i 9j QR 4i + 14 j

angled triangle.

o o o o oo
(b) If PQ = – 3 i + 4 j and PR = – 7 i + j prove that ∆PQR is an isosceles right

angled triangle.

oo
12. (a) Find the angle between 4 i – 3 j with x-axis.

oo
(b) Find the angles between 2 i + j and y - axis.

oo ooo
13. (a) Show that the angle between two vectors a and c is 90° if a + b + c = (0, 0),
ooo
| a | = 3, | b | = 5, | c | = 4.

o o ooo o o
(b) Find the angle between a and b if a + b + c = 0, | a | = 6, | b | = 7,
o
| c|= 127 o o o
ooo
(c) If a + b + c = 0, | a | = 3, | b | = 5, | c | = 7, show that the angle between
oo
| a | and | b | is 60°.

oo oo oo
14. If ( a + b ) and (2 a - b ) are orthogonal vectors, a and b are unit vectors, find the
oo
angle between a and b .

oo o o oo
15. If | a - 3 b | = | a + 3 b | , prove that a and b are orthogonal vectors.

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2. (a) 3 (b) 0 (c) 375 2
3. (a) 10 (b) 1 2
4. (i) 1 (ii) 1
(d) 14 (c) 0 (d) 0
(c) 13 (b) 60°
6. (a) 60° 5.(a) 13 (b) 20
(b) 2
9. (a) 6 (b) 63.43° (e) 2 (f) -5
12. (a) 71.56°
(c) 7.13° (d) 90° (e) 90°

(c) - 5 (d) - 4
2 3

13.(b) 60° 14. 180°

10.2 Vector Geometry

Review : A

From the given figures, answer the following questions :

oo
(a) Find the sum of AB and BC in the triangle ABC.

oo o B (i) C
(b) In ∆ABC, can we write BC + CA + AB = 0 ? S
R
oo
(c) In the given rectangle PQRS, write a single vector representing PQ + QR .

oo P
(d) Is PQ . QR = 0?

oo Q 10cm
(e) What is difference between PQ . QR and PQ.QR ? (ii)
(f) State the triangle law and parallelogram laws in vector addition.

Triangle Law of Vector Addition

"If the magnitude and direction of two vectors are represented by two sides of a triangle
taken in order, then the magnitude and direction of their sum is given by the third side taken
in reverse order."

Parallelogram Law of Vector Addition

"If two adjacent sides of a parallelogram through a point represent two vectors in magnitude
and direction, their sum is given by the diagonal of the parallelogram through the same
point in magnitude and direction."

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10.3 Theorems on Vector Geometry
THEOREM 1

Mid Point Theorem

oo
If a and b are the position vectors of two points A and B and M is the mid-point of the
o o o
line segment AB, the position vector of M is OM = 1 (a + b)
2
Proof : Let O be the origin and AB be a line segment.

Position Vector of A = o = o oB
OA a b

Position Vector of B = o = o
OB b

M is the mid-point of AB. O M
A
Then by using triangle law of vector addition, we have, o
a
o oo o 1 o
OM = OA + AM = OA + 2 AB

= o 1 ( o - o
OA + 2 OB OA )

= o 1 o - o
a+ 2 (b a)

= oo o oo
2 a+ b- a a +b

2 = 2

? o 1 o o Proved.
OM = 2 (a + b)

Alternative Method

Since, M is the mid point of AB

oo
AM = MB

oo oo
or, OM – OA = OB – OM

ooo
or, 2 OM = OA + OB

? o 1 o + o Proved.
OM = 2 (a b)

Note :

If A(x1, y1) and B(x2, y2) are any two points and M is the mid point of line segment AB, then
the position vector of M is given

( )o

OM =
x1 + x2 , y1 + y2
2 2

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THEOREM 2

Section formula for internal division.

oo
If a and b are the position vectors of two points A and B respectively and P divides the
line segment AB internally in the ratio m:n, the position vector of vector of P is

oo oo
OP = p = mb + na
m+n
Proof : Let O be the origin and AB be a line segment. P divides AB internally in the ratio

m:n. B

oo o n
Here, position vector of A = OA = a bo P

position vector of B = o = o p
OB b

position vector of P = o = o m
OP p
O oA
Here, AP = m a
PB n
o oo
Now, AP = m ( AP and PB are in the same direction)?
o n
PB Proved.

oo
or, n AP = m PB

oo oo
or, n( OP - OA ) = m ( OB - OP )

oo oo
or, n( p - a ) = m b - m p

oo oo
or, m p + n p = m b + n a

oo
o mb + na
? p = m + n

Note :

o 1 ( o o where o = o o = o o = o
If P is the mid point of AB, then p = 2 a + b ), OP p, OA a, OB b

THEOREM 3

Section formula for external division

oo
If a and b are position vectors of two points A and B respectively and the point
P divides the line segment externally in the ratio m:n, the position vector of P is

oo
o mb – na
p = m–n

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Proof : Let O be the origin and AB be a line segment. The point P divides line segment AB

in ratio m:n externally. P

Here, position vector of A = o o o
OA = a po

position vector of B = o o bB
OB = b

position vector of P = o o O
OP = p
o
Now, AP = m a A
BP n
o
or, AP = m oo?
o n ( AP and BP have same direction)
BP
oo
or, m BP = n AP

oo oo
or, m( OP - OB ) = n ( OP - OA )

oo oo
or, m p - m b = n p - n a

o oo
or, (m - n) p = m b - n a

oo
o mb – ma
? p = m–n Proved.

THEOREM 4

Centroid Formula

oo o
If a , b , and c are position vectors of the vertices A, B and C of ∆ABC, the position
o o o o o
vector of centroid of triangle is OG = g = 1 ( a+ b+ c)
3
Proof : Let O be the origin and A,B and C the vertices of ∆ABC. o
a A

Then, oo O o
position vector of A = OA = a g

position vector of B = o = o o Go
OB b b c

position vector of C = o = o DC
OC c

Consider a median AD and G the centroid of the triangle, B

G divides AD into 2:1 ratio. i.e. AG : GD = 2:1 = m:n

Now, o 1 ( o + o = 1 ( oo
OD = 2 OB OC ) 2 b +c)

o oo 2× 1 ( oo o 1 o o o
OG m OD + n OA = 2 b +c) + 2. a 3 a b + c)
2+1
? = m+n = ( + Proved.

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Worked out Examples

oo oo
Example 1. The position vectors of A and B are 6 i + 2 j and 2 i + 4 j respectively.
Solution:
If M is the mid-point of line segment AB, find the position vector of M.
Example 2.
Solution: Let O be the origin.

oo o o
Then, position vector of A = OA = a = 6 i + 2 j

position vector of O ob B
oo o o M
oa
B = OB = b = 2 i + 4 j

M is the mid-point if line segment AB.

? position vector of M o 1 ( o + o ) A
= OM = 2 a b

= 1 ( 6 o + 2 o + 2 o o
2 i j i +4 j )

= 1 ( o + o ) = 4 o o
2 8i 6j i +3 j

oo oo
The position vectors of A and B are respectively 4 i +7 j and 6 i - 2 j .

Find the position vector of C and D if

(a) C divides AB in 2:1 ratio internally. B

(b) D divides AB in 3:2 ratio externally. 1
C
(a) Let O be the origin . 2

Then, ooo A
positive vector of A = OA = 4 i + 7 j

positive vector of B = o = o o . O
OB 6i –2 j

C divides AB in ratio 4:2 internally. i.e. m:n = 4.2

oo
m OB + n OA
Now, position vector of C = m+n

2(6 o – 2 o )+ 1(4 o + 7 o )
i j i j
= 2+1
)o o o o
= 12 i - 4 j + 4 i +7 j
3
oo
16 i +3 j
= 3

= 16 o o
3 i +j

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(b) Here, D divides AB in 3:2 ratio externally. i.e. m:n = 3:2

oo
m OB – n OA
Now, position vector of D = m+n

3(6 o – 2 o )- 2(4 o + 7 o )
i j i j
= 3-2
)o o o o
= 18 i – 6 j - 8 i - 14 j
1
oo
= 10 i – 20 j

Example 3. If A(4, 4), B(3, 2), and C(3, 4) are the vertices of ∆ABC, find the position
Solution: vector of centroid the triangle.

Let O be the origin. Then we have,
o

position vector of A = OA = (4, 4)
o

position vector of B = OB = (3, 2)
o

position vector of C = OC = (3, 4)

Now, position vector of centroid G of ∆ABC is given by

o = 1 o oo
OG 3 { OA + OB + OC }

= 1 {(4, 4) + (3, 2) + (3, 4)}
3
1
= 3 (10, 10)

= 10 , 10
3 3

Exercise 10.2

Short Questions
oo

1. Let the position vectors of A and B be respectively a and b , AB is the line segment
joining them. Then, write the formula in the following conditions:
(a) position vectors of the mid point M of AB.
(b) position vector of the point P on AB which which divides AB in ratio of m:n
internally.
(c) position vector of the point P of AB which divides AB in ratio of m:n externally.

2. (a) The position vectors of A and B are respectively 4 and 6 , find the

( ) ( )5 3

position vector of the mid point of the line segment AB.

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oo oo
(b) The position vectors of P and Q are respectively. (4 i + 2 j ) and (3 i + 3 j ) ,

then find the position vector of the mid point of line segment PQ.

oo oo
(c) The position vector of M and N are respectively (3 i + 5 j ) and (- i + 3 j ), if
oo
MP = PN , find the position vector of P.

oo oo
(d) The position vectors of A and B are (2 a - 3 b ) and (5 a - 4 b ) respectively. Find

the position vector of mid point of AB.

oo
3. (a) If the position vector of the mid point of the line segment AB is (3 i - 2 j ), where

oo
the position vector of B is (5 i + 2 j ), find the position vector of A.

oo
(b) The position vector of the mid point of the line segment PQ is (4 i + 2 j ), where

oo
the position vector of P is (3 i - 7 j ). Find the position vector of Q.

o o o oo o
4. (a) The position vectors of A and B are respectively a =3 i +4 j and b = i - 2 j . If

C divides AB in the ratio of 3:2 internally, find the position vector of C.

oo oo
(b) The position vectors of A and B are respectively (4 i + 6 j ) and (2 i + 3 j ) .

Find the position vector of C which divides AB in the ratio 2:3 internally.

(c) The position vectors of A and B are 2 o - 3 o oo
a b and 5 a – 4 b respectively. Find

the position vector of the point which divides AB in ratio 3:2 internally.

oo oo
5. (a) The position vectors of A and B are respectively (3 i - 2 j ) and (3 i + 6 j ).

Find the position vector of P which divides AB in ratio 2:3 externally.

oo oo
(b) The position vector of A and B are respectively ( i + j ) and (3 i + 5 j ). Find

the position vector of P which divides AB in ratio of 2:1 externally.

6. (a) If the position vectors of the vertices A, B, and C of ∆ABC are respectively
o o oo oo
(3 i + 5 j ), (5 i - j ), and ( i + 8 j ), find the position vector of centroid

of the triangle.
oo oo
(b) The position vectors of P, Q and R of ∆PQR are respectively (3 i +4 j ), (4 i +5 j ),
oo
and (5 i + 6 j ), find the position vector of centroid of the triangle.

(c) In ∆LMN, o o o o = o + o and the position vector of
OL = 4i -5 j , OM 6i 4j
o oo o
centroid G, OG = 2 i + j . Find ON.

o o oo oo
(d) In ∆ABC, OA = 3 i - 5 j , OB = - 7 i + 4 j and centroid G,

o oo o
OG = 2 i + 6 j , find OC .

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(e) Find the position vector of centroid G of ∆ABC, with vertices A(-1, -1), B(-1, 5),
and C(5, 2).

Long Questions

oo
7. (a) If a and b are the position vectors of points A and B respectively. Find the

oo
position vector of C in AB produced such that AC = 3 BC

(b) OABC is a parallelogram. P and Q divide OC and BC in the ratio of
o oo o o
CP:PO = CQ:QB=1:3. If OA = a , OC = c , find the vector PQ and show that

PQ//OB.

8. (a) In the given figure if A

o 3 o oo
CB 2 AB AC = 3 AD
o = , then show that : 2 o +
o PQ ,
CD 1 o PB D C
(b) In the given figure PA = 6 p A

then show that o
Oa

( )o o
qQ
a=
1 oo
6 5 p+ q

9. (a) In ∆ABC, the medians AD, BE and CF are drawn from the vertices A, B and C
ooo
respectively. Then prove that , AD + BE + CF = (0, 0). A

(b) If G is centroid of ∆ABC, FE
then prove that G
o oo
GA + GB + GC = O B DC

Project Work

10. State the triangle law of vector addition. Is this theorem necessary to prove mid-point
theorem and section formulas in vector geometry? Illustrate with examples.

11. State differences between theorems proved in vector geometry and plane geometry.
Illustrate with examples.

2. (a) 5 (b) 7 i + 5 j (c) i + 4 j (d) 7 a – 7 b
4 2 2 2 2
9 2
3. (a) i – 6 j (b) 5 i + 11 j 4.(a) 5 i + 5 j

(b) 16 i + 24 j (c) 19 a – 18 b 5.(a) 3 i – 18 j (b) 5 i + 9 j
5 5 5 5 (d) 10 i + 19 j

6. (a) 3 i + 4 j (b) 4 i + 5 j (c) -4 i + 4 j

(e) (1, 2) 7.(a) 3 b – a
2

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10.4 Theorems Related to Triangles

THEOREM 5

The line segment joining the mid-point of two sides of a triangle is parallel to the third side
and is half of it. (prove vectorically).

Solution:

Let ABC be a triangle in which D and E are the mid-points of AB and AC respectively, EF is

the line segment joining E and F. A

To prove : o = 1 o o // o
DE 2 BC , DE BC

Proof : DE

o oo B C
Here, DE = DA + AE (by triangle law of vector additional)

or, o 1 o 1 o (E and F are mid-points of AB and AC)
DE = 2 BA + 2 AC

or, o 1 oo
DE = 2 ( BA + AC )

o 1 o
DE = 2 BC

oo o o 1
? DE // BC ( DE =k BC , k = 2 )
o o
? DE = 1 BC Proved.
2

THEOREM 6

The median to the base of an isosceles triangle is perpendicular to the base.

Solution:

Let ∆ABC be an isosceles triangle with AB = AC, D be the mid point of base BC. AD is
median drawn from vertex A to BC.

To prove : AD A BC A
Proof :

oo o
1. BC = BA + AC (by triangle law of vector addition)

o oo
AD = AB + AC (by mid-point theorem)
( )2. 1 BD C
2

oo
AC + AB
( )=1
2
oo o o oo
BC . AD = AC AB ) . AB + AC
( ( )3. - 1
2

oo oo oo
AC - AB AC + AB AC 2 - AB 2
( ) ( ) [( ) ( ) ]=1 . = 1
2 2
1 1
= 2 [AB2 - AB2] = 2 .0 =0 ( ? AB = AC)

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oo
Since, BC . AD = 0.

oo
? AB is perpendicular to BC . Proved.

THEOREM 7

The mid-point of the hypotenuse of a right angled triangle is equidistant from its vertices

Solution: A

Let ∆ABC be a right angled triangle with ‘ABC = 90°,

D the mid point of hypotenuse AC. i.e. AD = DC D
To prove : AD = BD = CD

Proof :

oo o B C

1. In ∆ABD, AB = AD + DB (by triangle law of vector addition)

o o o oo o o o o
2. In ∆BDC, BC = BD + DC = BD + AD = AD + BD ( AD = DC )

3. Since ‘B = 90°, so we can write,
oo
AB . BC = 0

oo o o
or, ( AD + DB ) . ( AD + BD ) = 0

or, ( o - o ) . ( o + o ) = 0
AD BD AD BD

or, ( o )2 = ( o )2
AD BD

or, AD2 = BD2

or, AD = BD

4. Again, D is mid-point of AC i.e. AD = DC

5. From (3) and (4), we get

AD = BD = CD Proved.

Worked out Examples

Example 1. Prove by vector method that the perpendicular drawn from the vertex to the
Solution: base of an isosceles triangle bisects the base.

Let ∆ABC be an isosceles triangle with AB = AC. AD is drawn perpendicular

to BC. A

To prove : BD = DC

Proof :

1. Since AD A BC B DC

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oo oo
then, AD . BD = 0 or, AD . DC = 0

ooo
2. AB = AD + DB (by using triangle law of vector addition)

ooo
AC = AD + DC

oo ( ABC is an isoceles triangle)
3. | AB | = | AC |

oo oo
or, | AD + DB | = | AD + DC |

squaring on both sides, we get

( o o )2 = ( o + o )2
AD + DB AD DC

oo oo
or, AD2 + 2 AD . DB + DB2 = AD2 + 2 AD . DC + DC2

or, 2.0 + DB2 = 2.0 + DC2 (by using 1)

or, DB2 = DC2 Ÿ BD = DC Proved.

Example 2. In the given ∆PQR, RM and QN are medians of ∆PQR such that RM = QN,
Solution: then prove vectorically that ∆PQR is an isosceles triangle.

In ∆PQR, medians RM = QN

To prove : ∆PQR is an isosceles P
Proof :

We have, QN = RM M
i.e
| o | = | o | N
QN RM R

squaring on both sides, we get, Q
oo

| QN|2 = | RM|2

oo oo
or, ( QP + PN )2 = ( RP + PM )2

oo oo
or, QP2 + 2 QP . PN + PN2 = RP2 + 2 RP . PM +PM2

QP2 +( ) ( )or,142=QPo43QP 2.P-R122PoPoQR.+ 1 PR 2 = RP2 +2 o 1 o + 1 PQ 2
o 2 PR2 - 1 PR2 o RP . 2 PQ 2
or, PR = 4
PQ2 -or, - PR o
. PQ

3 PQ2
4
or, PQ2 = PR2

oo
or, | PQ |2 = | PQ |2

? PQ = PR.

Hence, ∆PQR is an isosceles. Proved.

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Exercise 10.3 S T
Q
Short Questions R
1. (a) In the given figure S and T are the mid points of sides P

PQ and PR. Write the relation between ST and QR. Q MR
(b) In the given ∆PQR, PQ = PR, PM is a median. Write the P

relation between PM and QR. M
(c) In the given ∆PQR, M is the mid-point of PR and
QR
‘PQR = 90°. Write the relations among PM, MR and
QM.

(d) In the given ∆PQR if PM = MR = QM. Find the relation
between PQ and QR.

oo
(e) If O is origin and OA = a ,

o oo o oo A
OB = b , OC = c . Write the relation of a , b ,
oa
oo og
c and g , where G is the centroid of the triangle. O
G
ob C
oc
B
D

Long Questions P
2. (a) In ∆PQR, ‘PQR = 90°.

Prove that PR2 = PQ2 + QR2

QR

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(b) ∆ABC is a right angled triangle, ‘B = 90°. A
Let ‘C = T, AC = h, BC = b, AB = p, then prove B
that : h2 = p2 + b2

CT
A

(c) In the figure ∆ABC is an isosceles triangle. AD is a
oo
median, show that AD . BC = 0
B DC

A

3. In triangle ABC, M and N are the mid points of AB

and AC respectively. Then, prove vertically that MN
(a) MN//BC

(b) o = 1 o PB C
MN 2 BC

4. In the given triangle, PM is a median of an isosceles,

triangle PQR with PQ = PR. Then show that PM is

perpendicular to QR, vertically. Q MR

5. Prove vectorically that the mid point of hypotenuse of a right angled triangle is

equidistant from its vertices.

6. Prove vectorically that the line joining the vertex and the mid point of P

the base of an isoscles triangle is perpendicular to the base.

7. ∆PQR is an isosceles triangle with PQ = PR and PM A QR. Then

vectorically show that QM = RM. Q MR

oo o P
8. If a , b , and c are the position vectors of the vertices
G
A, B, C of ∆ABC respectively. Prove vectorically that the position MR

ooo
a+b+ c
o 1 Q
3
( )vector of centroid G of ∆ABC is g =

1. (a) ST // QR, o = 1 o (b) PM A QR (c) PQ A QR, ‘PQR = 90°
ST 2 QR

(d) PM = RM = QM (e) o = 1 o + o + o
OG 3 (a b c)

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10.5 Theorem on Quadrilateral and Semi - circle

THEOREM 8

The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is
a parallelogram.

Solution: AS D

In the figure, P,Q,R and S are the mid-points of sides AB, BC, CD,

and AD respectively of a quadrilateral ABCD. Joining P,Q,R, and S P R
successively a quadrilateral PQRS is formed.

To prove : PQRS is a parallelogram. B QC
Construction : Join BD.

Proof :

1. In ∆ABD, o = 1 o
PS 2 BD (In ∆ABD, P and S are the mid-points of AB and AD)

2. In ∆BCD, o = 1 o (In ∆BCD, Q and R are the mid-points of BC and CD)
QR 2 BD

oo
3. PS = QR = [From statement (1) and (2)]

4. | PoS| = | QoR|, PQ // SR (Magnitudes and directions of equal vectors o and o equal)
PS QR

o o oo
5. Similarly, | SR | = | PQ | , SR // PQ . (as in statement (4))

6. ? PQRS is a parallelogram. (from (4) and (5) opposite sides are parallel).

THEOREM 9

The diagonals of a parallelogram bisect each other.

Solution: O
M
In the figure ABCD is a parallelogram. Diagonals D C
AC and BD are drawn. B

Let the mid point of diagonal BD be M.

To prove : M is the mid-point of AC. A

Construction : Let the position vectors of A,B,C,D
oooo o
and M be OA , OB , OC , OD and OM
respectively.

Proof : o o o
OM = ( OD + OB )
1. In ∆ OBD, 1 (by mid-point theorem)
2

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or, o = 1 o + o + o (In ∆ OAD, o = o o
OM 2 ( OA AD OB ) OD OA + AD )

or, o 1 o o o
OM = 2 ( OA + OB + BC )

? o = 1 o o ooo
OM 2 ( OA + OC ) (In ∆ OBC, OB + BC = OC )

2. ? M is the mid-point of AC. (By mid-point theorem o = 1 o o
OM 2 ( OA + OC ))

3. Hence, the diagonals the parallelogram bisect each other C from statement (1) and
(2). Proved.

Alternative Method

Let O be the origin. O

o ooo D C
Then OA , OB , OC , OD are the position vectors of A B
o oo o
A, B C and D respectively. Let OA = a , OB = b ,
oo oo
OC = c and OD = d

Proof

1. By using mid-point theorem, position vector of mid point of diagonal
o o o o
BD = 1 ( b + d ). Position vector of mid-point of diagonal AC = 1 ( a + c ).
2 2

oo
2. Now, AB = DC (opposite sides of a parallelogram)

o o oo
or, OB - OA = OC - OD

oo oo
or, b - a = c - d

oo oo
b+ d a+ c
or, 2 = 2 (dividing both sides by 2)

3. Position vector of mid-point of BD = position vector of mid point of AC, [from (2)].

4. Hence, the diagonals of a parallelogram bisect each other. Proved.

THEOREM 10

The diagonals of a rhombus bisect each other at right angle. Prove it vectorically.

Solution : D C
B
In the figure, let ABCD be a rhombus with diagonal AC and BD.

Then AB = BC = AD = DC.

To Prove : AC A BD, mid-point of AC = mid-point of BD. A

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Proof :

ooo (by using triangle law of vector addition)
1. In ∆ ABC, AC = AB + BC

2. In ∆ABD, (by triangle law of vector addition)
o oo oo
BD = BA + AD
oo ( AD = BC )
= BA + BC
oo
= BC – AB
oo

3. Taking dot product of BD and AC ,

oo oo o o
BD . AC = ( BC - AB ) . ( BC + AB )
oo
= BC2 – AB2 ( | BC | = | AB | )

= BC2 – BC2 = 0

4. BD A AC.
o oo oo oo o

5. Let OA = a , OB = b , OC = C , OD = d .

oo
a+ c
Position vector of the mid-point of AC = 2

oo
b+ d
Position vector of the mid-point of BD = 2

oo
6. AB = DC (Opposite sides of parallelogram)

oo oo
or, OB - OA = OC - OD

oo o o
or, b - a = c - d

oooo
? a+ c=b+d

oo oo
a+ c b+ d
? 2 = 2

i.e., Mid-point of diagonal AC = Mid-point of diagonal BD.

7. Hence, the diagonals of rhombus bisect each other at right angle. (From statement (4)
and (6).) Proved.

THEOREM 11

Prove vectorically that the diagonals of a rectangle are equal. D C
Solution: B

Let ABCD be a rectangle.

AC and BD are diagonals of the rectangle. A

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To prove : | o | = | o |
AC BD

Proof :

1. In ∆ABC, o = o + o (By triangle law of vector addition)
AC AB BC

o oo
or, ( AC )2 = ( AB + BC )2

oo oo o
or, ( AC )2 = ( AB )2 + 2. AB . BC + ( BC )2

oo
Since AB A BC, AB . BC = 0,

? AC2 = AB2 + BC2

2. In ∆ABD, o = o + o
BD BA AD

or, o = (- o ) + o . ( o = o )
BD AB BC AD BC

oo oo o
( BD )2 = (- AB )2 + 2( AB ). BC + ( BC ) 2

ooo
? ( DB )2 = ( AB )2 + ( BC )2

? DB2 = AB2 + BC2

3. From (1) and (2), we get.

AC2 = BD2

or, AC = BD

oo
? | AC | = | BD | Proved.

THEOREM 12

Prove vectorically that the angle at semi - circle is a right angle.
Solution : Let O be the centre of the circle and AB be a diameter.

‘ACB angle at semi-circle. OC is joined. C
To prove : ‘ACB = 90°

Proof : AO B
1. By triangle law of vector addition,

oo o
AC = AO + OC

o oo
CB = CO + OB

oo oo
= CO + AO ( AO = OB radii of the same circle)

oo
= AO - OC

oo
2. Taking dot product of AC and CB , we get,

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oo o o o o
AC . CB = ( AO + OC ). ( AO - OC )

oo
= ( AO )2 - ( OC )2

= AO2 - AO2 (OA = OC = OB, radii of same circle)

=0

3. Since o . o = 0, o A o
AC CB AC CB

Thus, shows that ‘ACB = 90° Proved.

Exercise 10.4

Short Questions

1. (a) In the adjoining figure P, Q, R, S are the mid-points of D SC
sides AD, AB, BC, and CD respectively. The points P, Q, R R

and S are joined. What type of quadrilateral is PQRS?

(b) What is the measure of circumference angle at semi- P
circle?

(c) Write relations of diagonals in each of the following A Q B
quadrilateral.

(i) Square (ii) Rhombus (iii) Parallelogram

Long Questions

2. In a quadrilateral ABCD, the mid-points of sides AB, BC, CD, and DA are respectively
P, Q, R and S. Prove vectorically that PQRS is a parallelogram.

3. Prove vectorically that the diagonals of parallelogrm PQRS bisect each other.

4. Prove vectorically that the diagonals of rhombus EFGH bisect each other at right angle.

5. Prove vectorically that the diagonals of rectangle PQRS are equal.

6. Prove vetcotically that the angle at semi-circle is right angle. R
7. In the figure O is the centre of the semi-circle. Prove that

‘PRQ is right angle.

8. If the diagonals of a quadrilateral bisect each other, prove by P OQ
vector method that is a parallelogram.
BLC

9. In the given figure, ABCD is a parallelogram. L and M are the M
mid-points of sides BC and CD respectively. D

Prove that o + o 3 o A R
AL AM = 2 AC S N
Q
10. In the given figure PQRS is a parallelogram M and N are the mid M
points of PS and QR, show that MQNS is a parallelogram. 297

P

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10

11. If a line is drawn from the centre of a circle to the mid-point of a chord, prove by vector

method the line is perpendicular to the chord.

12. In the given figure PQRS is a trapezium where PS//QR. M and N P S

are the mid-points of PQ and SR respectively. Prove vectorically N
that M

(i) o = 1 o + o Q R
MN 2 ( PS QR ) D C

oo
(ii) MN // QR

13. In the adjoining figure, PQRS is a parallelogram. G is the point O G
of intersection of the diagonals. If O is any point prove that

oooo A B
OA + OB + OC + OD SM R
o = 1
4 NQ
( )OG C

14. In the adjoining figure PQRS is a parallelogram. M and N are

two points on the diagonal SQ. If SM = NQ, then, prove by

vector method by PMRN is a parallelogram. P

oo D
15. ABCD is a parallelogram and O is the origin. If OA = a ,
o oo o o oo o
OB = b , OC = c find OD in terms of a , b and c .

Project Work A B
O
15. Write three statements of vector geometry. Show differences
between them in vector geometry and plane geometry.

1. (a) parallelogram (b) 90° 15. oa – ob + oc

298 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur