VECTORS 1. Define coplanar vectors.

Definition : Three or more vectors lie on the same plane are called as coplanar vectors.

2. Define collinear vectors.

Definition : Two or more vectors are said to be collinear vectors, when they are along the same lines or parallel lines.

If a and b are parallel vectors then a = kb. For any scalar ‘k’.

3. If a = 2 i + 3j – 5k,

( i ) Find the unit vector in the direction of the vector a

( ii ) Find the direction cosines of the vector a and hence prove that,

Cos2 α + Cos2 β + Cos 2γ= 1

Solution : consider, a = 2 i + 3j – 5k = ( 2 , 3 , - 5 )

| a | = √ ( 4 + 9 + 25 ) = √ 38 units

a 2 i + 3j – 5k ( i ) the unit vector in the direction of the vector a = = | a | √ 38

( ii ) direction cosines of a are cosα = 2/√ 38 , Cosβ = 3/√ 38 & Cosγ = - 5 / √ 38

Consider,

Cos2 α + Cos2 β + Cos 2γ = 4/38 + 9/38 + 25/38 = 1

4. If the direction cosines of the vector are 1/5 , 3/5 & n , Find ‘n’.

Solution : It is given that,

cosα = 1/5 , Cosβ = 3/5 & Cosγ = n

since,

Cos2 α + Cos2 β + Cos 2γ = 1

1/25 + 9/25 + n² = 1

n² = 1 – 1/25 – 9 /25

n² = 15/25 = 3/5

n = ±√( 3/5)

5. Define dot product or scalar product of two vectors.

Solution; If a and b are any two vectors and θ be the angle between them, then dot product of a and b is defined by

a • b = | a | | b |Cosθ

b θ a

6. Define cross product or vector product of any two vectors.

Solution; If a and b are any two vectors and θ be the angle between

Them & η unit vector perpendicular to both a & b , then,

a X b = | a | | b | η Sinθ

a θ

η b

7. If a = i – 2j + 2k & b = 2i + j – 3k , Find

( i ) a • b

( ii ) cosine of the angle between the vectors a & b,

( iii ) projection of a in the direction of b

Solution ; consider,

a = i – 2j + 2k = ( 1 , -2 , 2 )

b = 2i + j – 3k = ( 2 , 1 - 3 )

( i ) a • b = 2 – 2 – 6 = - 6

( ii ) | a | = √ ( 1 + 4 + 4 ) = 3 units

| b | = √ ( 4 + 1 + 9 ) = √14 units

If θ be the angle between a & b then,

a • b - 6 - 2 Cos θ = = = | a | | b | 3√14 √14 ( iii ) a • b - 6 projection of a in the direction of b = = | b | √14

8. If | a | = 3 , | b | = 5 & | c | = 7 and a + b + c = 0 , find the angle between the vectors a & b.

Solution:

since, a + b + c = 0 , let θ be the angle between the vectors a & b

a + b = - c

| a + b |² = | -c |²

| a |² + | b |² + 2 a • b = | c |²

9 + 25 + 2 | a || b | Cosθ = 49

9 + 25 + 2 ( 3 ) ( 5 ) Cosθ = 49

30 Cosθ = 49 - 34

30 Cosθ = 15

Cosθ = ½

θ = 60°

9. If a & b are unit vectors inclined at an angle of 60° to each other , find | a + b |.

Solution:

It is given that | a | = | b |.= 1 and θ = 60°

Consider, | a + b |² = | a |² + | b |² + 2 a • b

= | a |² + | b |² + 2 | a | | b | Cosθ

= 1 + 1 + 2 (1) (1) cos60°

= 2 + 2( ½)

= 2 + 1

= 3

| a + b | = √ 3 units

10.If a & b are unit vectors inclined at an angle θ to each other , show that

│a – b │ = 2 Sin(θ/2)

Solution : Given that , │a │ = │ b │ = 1

consider , | a - b |² = | a |² + | b |² - 2 a • b

= | a |² + | b |² - 2 │a│ │ b │Cosθ

= 1 + 1 – 2 (1)(1) Cosθ

= 2 - 2 Cosθ

= 2( 1 – Cosθ )

= 2 { 2 Sin²(θ/2 )}

= 4 Sin²(θ/2)

Therefore │a – b │ = 2 Sin(θ/2)

11. If a = 3 i – 2j + k & b = i + 3j + k , Find

( i ) a x b ( ii ) | a x b | ( iii ) sinθ where θ is the angle between a & b

( iv ) unit vector perpendicular to both a & b.

Consider,

a = 3 i – 2j + k = ( 3 , - 2 , 1 )

b = i + 3 j + k = ( 1 , 3 , 1 )

i j k

( I ) a x b = 3 - 2 1

1 3 1

= i [ - 2 – 3 ] – j [ 3 – 1 ] + k [ 9 + 2 ]

= - 5 i – 2 j + 11 k

( ii ) | a x b | = √ ( 25 + 4 + 121 ) = √150 = 5√6 units.

( iii ) | a x b | = 5√6 units.,

| a | = √ ( 9 + 4 + 1 ) = √14 units

| b | = √ (1 + 9 + 1 ) = √11 units

| a x b | 5√6 Sinθ = = | a | | b | √14 √11

( I v ) a x b - 5 i – 2 j + 11 k Unit vector perpendicular to both a & b = = | a x b | 5√6

11. Find the area of the parallelogram whose adjecent sides are given by the vectors,

i + 2j + 3k & - 3 i – 2 j + k.

solution : consider,

a = i + 2j + 3k = ( 1 , 2 , 3 )

b = - 3 i – 2 j + k = ( - 3 , - 2 , 1 )

i j k

( I ) a x b = 1 2 3

- 3 - 2 1

= i [ 2 + 6 ] – j [ 1 + 9 ] + k [ - 2 + 6 ]

= 8 i - 10 j + 4 k

= 2 { 4 I – 5 j + 2 k }

| a x b | = 2 √ ( 16 + 25 + 4 ) = 2 √45 units.

Area of the parallelogram = | a x b |= 2 √45 square units.

12. Find the area of the parallelogram, whose diagonals are represented by the vectors

3 i + j + 2k & i – 3j + 4k

Solution : Let

a = 3 i + j + 2k = ( 3 , 1 , 2 )

b = i – 3j + 4k = ( 1 , - 3 , 4 ) represents diagonals of the parallelogram,

i j k

( I ) a x b = 3 1 2

1 - 3 4

= i [ 4 + 6 ] – j [ 12 - 2 ] + k [ - 9 - 1 ]

= 10 i - 10 j - 10 k

= 10 { i – j - k }

| a x b | = 10 √ ( 1 + 1 + 1 ) = 10 √3 units.

| a x b | 10 √3 Area of the parallelogram = = = 5√3 square units. 2 2

13. Find the area of the triangle , whose two adjecent sides are given by the vectors,

i + 4 j – k & i + j + 2k.

solution : Let

a = i + 4 j – k = ( 1 , 4 , - 1 )

b = i + j + 2k = ( 1 , 1 , 2 )

i j k

a x b = 1 4 - 1

1 1 2

= i [ 8 + 1 ] – j [ 2 + 1 ] + k [ 1 – 4 ]

= 9 i – 3 j – 3 k

= 3 { 3 i - j – k }

| a x b | = 3√( 9 + 1 + 1 ) = 3√11 units.

| a x b | 3√11 Area of the triangle = = square units 2 2

14. Find the area of the triangle whose vertices are ( 1 , - 1 , 2 ) , ( 2 , 1 , - 1 ) and

( 3 , -1 , 2 )

Solution : Let O be the fixed point,

OA = position vector of A = ( 1 , - 1 , 2 )

OB = position vector of B = ( 2 , 1 , - 2 )

OC = position vector of C = ( 3 , - 1 , 2 )

AB = OB – OA = ( 1 , 2 , - 4 )

AC = OC – OA = ( 2 , 0 , 0 )

i j k

AB x AC = 1 2 - 4

2 0 0

= i [ 0 – 0 ] – j [ 0 + 8 ] + k [ 0 – 4 ]

= 0 i – 8 j – 4 k

= 4 { 0 i – 2 j – k }

| AB x AC | = 4√( 0 + 4 + 1 ) = 4√5 units

| AB x AC | 4√5 Area of the triangle ABC = = = 2√5 square units 2 2

15. Prove that , ( 2 a + 3 b ) x ( a + 4 b ) = 5 ( a x b )

Solution : consider,

( 2 a + 3 b ) x ( a + 4 b ) = 2 ( a x a ) + 8 ( a x b ) + 3 ( b x a ) + 12 ( b x b )

= 2 ( 0 ) + 8 ( a x b ) - 3 ( a x b ) + 12 ( 0 )

= 5 ( a x b )

16. Find the volume of the parallelipiped whose co-terminal edges are represented by the

vectors 2 i - 3 j + 4 k , i + 2 j - k & 3 i - j + 2 k.

solution : Let a = 2 i - 3 j + 4 k = ( 2 , -3 , 4 )

b = i + 2 j - k = ( 1 , 2 , -1 )

c = 3 i - j + 2 k = ( 3 , - 1 , 2 )

then, volume of the parallelopiped object with co-terminal edges a , b & c

= [ a , b , c ]

2 - 3 4

= 1 2 -1

3 - 1 2

= 2 [ 4 – 1 ] + 3 [ 2 + 3 ] + 4 [ - 1 – 6 ]

= 2 ( 3 ) + 3 ( 5 ) + 4 ( -7 )

= 6 + 15 – 28

= - 7

= 7 cubic units.

17. Show that the vectors i + j + k , 3 i + 4 j + 2 k & 3 i + j + 5 k are coplanar.

Solution : let a = i + j + k

b = 3 i + 4 j + 2 k

c = 3 i + j + 5 k

1 1 1

consider, [ a b c ] = 3 4 2

3 1 5

= 1 [ 20 – 2 ] - 1 [ 15 – 6 ] + 1 [ 3 – 12 ]

= 0

Therefore , the vectors a , b & c are coplanar.

18. Show that the points with position vectors,

( i ) i + j + k , 2i + 3 j + 4 k , 3 i + j + 2 k & - i + j

( ii ) - 6a + 3 b + 2 c , 3 a – 2 b + 4 c , 5 a + 7 b + 3 c & - 13 a + 17 b – c

are coplanar.

Solution : ( i ) Let O be the fixed point.

OA = position vector of A = ( 1 , 1 , 1 )

OB = position vector of B = ( 2 , 3 , 4 )

OC = position vector of C = ( 3 , 1 , 2 )

OD = position vector of D = ( - 1 , 1 , 0 )

AB = OB – OA = ( 1 , 2 , 3 )

AC = OC – OA = ( 2 , 0 , 1 )

AD = OD – OA = ( - 2 , 0 , -1 )

Consider, 1 2 3

[ AB AC AD ] = 2 0 1

- 2 0 -1

1 2 3

= - 2 0 1

2 0 1

= 0 { since, second and third rows are identical }

Therefore the points A , B , C & D are coplanar.

Solution : ( ii ) Let O be the fixed point.

Let O be the fixed point.

OA = position vector of A = - 6a + 3 b + 2 c

OB = position vector of B = 3 a – 2 b + 4 c

OC = position vector of C = 5 a + 7 b + 3 c

OD = position vector of D = - 13 a + 17 b – c

AB = OB – OA = 9a – 5b + 2c

AC = OC – OA = 11a + 4b + c

AD = OD – OA = - 7a + 14b – 3c

Consider, 9 - 5 2

[ AB AC AD ] = 11 4 1

- 7 14 -3

= 9 [ - 12 – 14 ] + 5 [ - 33 + 7 ] + 2 [ 154 + 28 ]

= 0

Therefore the points A , B , C & D are coplanar.

19. If the vectors 3 i + j – 2 k , I + 2 j – 3k & 3 i + m j + 5 k are coplanar , find ‘m’.

Solutions: Let a = 3 i + j – 2 k , b = i + 2 j – 3k & c = 3 i + m j + 5 k

are coplanar vectors,

therefore, [ a b c ] = 0

3 1 - 2

1 2 - 3 = 0

3 m 5

3 [ 10 + 3m ] – 1 [ 5 + 9 ] – 2 [ m – 6 ] = 0

30 + 9m – 14 – 2m + 12 = 0

7m + 28 = 0

m = - 4

20. A , B , C & D are the points with position vectors 3 i – 2 j – k , 2 i + 3 j – 4 k ,

- i + 2 j + 2 k & 4 i + 5 j + λk respectively . If the points A , B , C & D lie on a plane, Find the

value of ‘λ’.

Solution : Le O be the fixed point.

OA = position vector of A = 3 i – 2 j – k = ( 3 , - 2 , - 1 )

OB = position vector of B = 2 i + 3 j – 4 k = ( 2 , 3 , - 4 )

OC = position vector of C = - i + 2 j + 2 k = ( -1 , 2 , 2 )

OD = position vector of D = 4 i + 5 j + λk = ( 4 , 5 , λ )

AB = OB – OA = ( - 1 , 5 , -3 )

AC = OC – OA = ( - 4 , 4 , 3 )

AD = OD – OA = ( 1 , 7 , λ + 1 )

Since , A , B , C & D are coplanar ,

[ AB AC AD ] = 0

[ AD AC AB ] = 0

1 7 λ + 1

- 4 4 3 = 0

- 1 5 - 3

1 [ - 12 – 15 ] – 7 [ 12 + 3 ] + (λ + 1) [ - 20 + 4 ] = 0

- 27 - 105 - 16λ – 16 = 0

- 148 – 16 λ = 0

16 λ = - 148

λ = - 148/8 = - 37/4

20..Prove that , [ a + b b + c c + a ] = 2 [ a b c ]

Solution : Consider,

[ a + b b + c c + a ] = ( a + b ) ● { ( b + c ) x ( c + a ) }

= ( a + b ) ● { ( b x c ) + ( b x a ) + ( c x c ) + ( c x a ) }

= ( a + b ) ● {( b x c ) + ( b x a ) + 0 + ( c x a ) }

= { a ● ( b x c ) + a ● (b x a ) + a ● (c x a) }

+ { b ● ( b x c ) + b ● ( b x a ) + b ● ( c x a ) }

= [ a b c ] + [ a b a ] + [ a c a ] + [ b b c ] + [ b b a ] + [ b c a ]

= [ a b c ] + 0 + 0 + 0 + 0 + [ a b c ]

= 2 [ a b c ]

21. Show that , ∑ i x ( a x i ) = 2 a

Solution : Let a = a1 i + a2 j + a3 k = ( a1 , a2 , a3 )

Consider, i j k

a x i = a1 a2 a3

1 0 0

= i [ 0 – 0 ] – j [ 0 - a3 ] + k [ 0 - a2 ]

= 0i + a3 j - a2 k

Next, , i j k

i x (a x i ) = 1 0 0

0 a3 - a2

= i [ 0 – 0 ] – j [- a2 - 0 ] + k [ a3 - 0 ]

= 0i + a2 j + a3 k ,

Next, i j k

a x j = a1 a2 a3

0 1 0

= i [ 0 – a3 ] – j [ 0 - 0 ] + k [a1 - 0 ]

= - a3 i + 0 j + a1 k

Therefore,

, i j k

j x (a x j ) = 0 1 0

- a3 0 a1

= i [a1 – 0 ] – j [0 + 0 ] + k [ 0 + a3 ]

= a1 i + 0 j + a3 k

Similarly we may show that,

k x (a x k ) = a1 i + a2 j + 0 k

hence,

, ∑ i x ( a x i ) = { i x (a x i ) } + { j x (a x j ) } + { k x (a x k ) }

= 2 a1 i + 2 a2 j +2 a3 k

= 2 { a1 i + a2 j + a3 k }

= 2a

22. Prove that ∑ a x ( b + c ) = 0

Solution : consider,

.∑ a x ( b + c ) = a x ( b + c ) + b x ( c + a ) + c x ( a + b )

= ( a x b ) + ( a x c ) + ( b x c ) + ( b x a ) + ( c x a ) + ( c x b )

= ( a x b ) - ( c x a ) + ( b x c ) - ( a x b ) + ( c x a ) - ( b x c )

= 0

23.Find a unit vector which should lie on the plane determined by the vectors

2 i + j + k & i + 2 j + k and perpendicular to i + j + 2k.

Solution: Let a = 2 i + j + k = ( 2 , 1 , 1 )

b = i + 2 j + k = ( 1 , 2 , 1 )

c = i + j + 2k = ( 1 , 1 , 2 )

consider , ( a x b ) x c = (c ● a) b – ( c ● b ) a ---------- ( 1 )

c ● a = 2 + 1 + 2 = 5 c ● b = 1 + 2 + 2 = 5 ( a x b ) x c = 5 b – 5 a

= 5 ( b – a )

= 5 ( -1 , 1 , 0 )

|( a x b ) x c | = 5 √ ( ! + 1 + 0 ) = 5√2 units

η = unit vector coplanar with a & b and perpendicular to c

( a x b ) x c 5 { - I + j + 0k} - i + j + 0k = = = |( a x b ) x c| 5√2 √2

24. Prove that [ a x b , b x c , c x a ] = [ a , b , c ]² & Also, If

a x b , b x c & c x a are coplanar , then prove that , a , b & c are coplanar.

Solution : consider,

[ a x b , b x c , c x a ] = ( a x b ) • { ( b x c ) x ( c x a ) } ------ ( 1 )

Let ( b x c ) = p, then,

( b x c ) x ( c x a ) = p x ( c x a )

= ( a • p ) c - ( c • p ) a ------- ( 2 )

Now, a • p = a • ( b x c ) = [ a , b , c ] = λ ( say )

c • p = c • ( b x c ) = [ c , b , c ] = 0

therefore, ( 2 ) becomes,

( b x c ) x ( c x a ) = ( λ ) c – ( 0 ) a = λ c

Therefore ( 1 ) becomes,

[ a x b , b x c , c x a ] = ( a x b ) • λ c

= λ { ( a x b ) • c }

= λ { c • ( a x b ) }

= λ [ c , a , b ]

= λ [ a , b , c ]

= λ²

= [ a , b , c ]²

If a x b , b x c & c x a are coplanar , then,

[ a x b , b x c , c x a ] = 0

[ a , b , c ]² = 0

[ a , b , c ] = 0

Therefore , a b & c are coplanar vectors.

26. Prove by vector method that, The angle in a semi circle is a right angle.

Solution: P

A B

Let AB be a diameter and O be the centre of a circle.

Let P be a point on the semi-circle.

Join PA , PB & PO.

By the law of triangle of vectors

PA = PO + OA

PB = PO + OB = PO – OA since OB = - OA

Consider,

PA ● PB = ( PO + OA ) ● ( PO – OA )

= │PO│² - │OA│² since , │PO│ = │OA│= radius of the circle.

= 0

Therefore PA perpendicular to PB

Therefore, APB = 90º

  

       O 

    

27. In any triangle ABC, prove by vector method

a b c ( a ) = = SinA SinB SinC ( b ) a² = b² + c² - 2bcCosA ( c ) a = bCosC + c CosB

π – A A

π - C

B π – B C

Solution : Let BC = a , CA = b & AB = c

Then, a + b + c = 0

Solution for ( a ): consider, a + b + c = 0

a x ( a + b + c ) = a x 0

( a x a ) + ( a x b ) + ( a x c ) = 0

0 + ( a x b ) – ( c x a ) = 0 ( a x b ) = ( c x a ) ------ ( 1 )

Next, consider, a + b + c = 0

Similarly , as above,

b x ( a + b + c ) = b x 0

( b x a ) + ( b x b ) + ( b x c ) = 0

- ( a x b ) + ( b x b ) + ( b x c ) = 0

- ( a x b ) + 0 + ( b x c ) = 0

( a x b ) = ( b x c ) ------ ( 2 )

from ( 1 ) & ( 2 ) , we have,

( a x b ) = ( b x c ) = ( c x a )

│ a x b │= │ b x c │ = │ c x a │

│a││b│Sin(π – C) = │b││c│Sin(π – A) = │c││a│Sin(π – B)

a b SinC = bcSinA = caSinB

dividing through out by abc, we have,

a b c = = SinA SinB SinC

Solution for ( b ): consider,

a + b + c = 0

a = - b – c

│a │² = │- b – c │²

│a│² = │b│² + │c│² + 2 b ● c

│a│² = │b│² + │c│² + 2 │b││c│Cos(π – A )

a² = b² + c² - 2 bcCosA since , Cos(π – A ) = - CosA

27. prove that ( a x b ) x c = ( a • c ) b - ( b • c ) a

Solution : Since , ( a x b ) x c lies in the plane determined by a & b , their exits scalars x & y such that ,

( a x b ) x c = x a + y b ---- ( 1 )

Taking dot product with c on both sides , we have,

c • { ( a x b ) x c } = x ( c • a ) + y ( c • b )

[ c , ( a x b ) , c ] = x ( c • a ) + y ( c • b )

0 = x ( c • a ) + y ( c • b )

x ( c • a ) = - y ( c • b )

x - y

= = λ ( say )

( c • b ) ( c • a )

Therefore,

x = λ( c • b ) , y = - λ ( c • a )

therefore ( 1 ) becomes,

( a x b ) x c = { λ( c • b ) } a + { - λ ( c • a )} b ------- ( 3 )

To find λ , take a = i , b = j & c = j in ( 3 ), we have,

( i x j ) x j = { λ( j • j ) } i + { - λ ( j • i )} j

k x j = λ ( 1 ) i + λ ( 0 ) j

- i = λ i

λ = - 1

substitute value of λ in ( 3 ) , we have,

( a x b ) x c = { ( - 1)( c • b ) } a + { ( 1 ) ( c • a )} b ------- ( 3 )

( a x b ) x c = ( a • c ) b - ( b • c ) a

28.Prove that , Cos ( A - B ) = CosACosB + SinA SinB &

Cos ( A - B ) = CosACosB + SinA SinB

Solution : Consider a unit circle, x² + y² = 1

Let o be the point of reference, and it is fixed,

Let OP = position vector of P = ( CosA , SinA ) = CosA i + SinA j + 0 k

OQ = position vector of Q = (CosB , SinB ) = CosB i + SinB j + 0 k

are any two points on the circumference of the circle. Y

Let ∟XOP = A & ∟XOQ = B

Therefore, ∟QOP = A – B P

|OP|= √ (Cos²A + Sin²A) = 1 Q

|OQ|= √ (Cos²B + Sin²B) = 1 X

OP • OQ = |OP||OQ|Cos( A – B )

= ( 1 ) ( 1 ) Cos( A – B )

= Cos( A – B ) ------- ( 1 )

But,

OP • OQ = CosACosB + SinA SinB + 0

= CosACosB + SinA SinB ---------- ( 2 )

From ( 1 ) & ( 2 ) , we have,

Cos ( A - B ) = CosACosB + SinA SinB

Next, since, Sin ( - θ ) = - Sinθ & Cos( - θ ) = Cos θ

Consider,

Cos( A + B ) = Cos{ A – (- B) }

= CosACos( - B ) + SinA Sin( - B )

= CosACosB - SinA SinB

 

                                   A ‐ B 

              O 

29. prove that , Sin ( A – B ) = SinACosB – CosASinB and

Sin ( A + B ) = SinACosB + CosASinB

Solution : Consider a unit circle, x² + y² = 1

Let o be the point of reference, and it is fixed,

Let OP = position vector of P = ( CosA , SinA ) = CosA i + SinA j + 0 k

OQ = position vector of Q = (CosB , SinB ) = CosB i + SinB j + 0 k

are any two points on the circumference of the circle. Y

Let ∟XOP = A & ∟XOQ = B

Therefore, ∟QOP = A – B P

|OP|= √ (Cos²A + Sin²A) = 1 Q

|OQ|= √ (Cos²B + Sin²B) = 1 X

|OP x OQ| = |OP||OQ|Sin( A – B )

= ( 1 ) ( 1 ) Sin( A – B )

= Sin( A – B ) ------- ( 1 )

But,

i j k

OP x OQ = CosA SinA 0

CosB SinB 0

= i [ 0 – 0 ] – j [ 0 – 0 ] + k [ CosASinB – SinACosB]

= 0i + 0j – λk take , λ = SinACosB - CosASinB

|OP x OQ| = √( 0 + 0 + λ² ) = √λ² = λ = SinACosB - CosASinB ------ ( 2 )

Therefore , from (1) & (2) , we have,

Sin ( A – B ) = SinACosB – CosASinB

 

                                   A ‐ B 

              O 

Next, since, Sin ( - θ ) = - Sinθ & Cos( - θ ) = Cos θ

Consider,

Sin ( A + B ) = SinACosB + CosASinB

= SinACos( - B ) + CosASin( - B )

= SinACosB - CosASinB