NAME OF THE CHAPTER

VECTOR

PART-A ONE

MARKS QUESTIONS

PART-B TWO

MARKS QUESTIONS

PART-C FIVE

MARKS QUESTIONS

PART-D SIX OR FOUR

MARKS QUESTIONS

PART-E TWO OR

FOUR QUESTIONS

TOTAL MARKS ALLOTED

APPROXIMATELY

1 1 1 1 1 16

VECTORS ONE MARK QUESTIONS :

1.Define a vector and give an example.

Solution; A vector is a physical quantity, which has both magnitude and direction.

Example; velocity , acceleration , force, etc.

2. Define a scalar and give an example.

Solution ; Scalar is a physical quantity which has only the magnitude but not the direction.

Example; mass ,volume , density , speed , etc.

3.Define a null vector ( or Zero vector )

Solution ; A vector of magnitude zero is called a null vector .

4. Define a unit vector.

Solution : A vector of magnitude unity is called as unit vector.

Note ; If a is a vector then a â = unit vector in the direction of a = | a |

5.Define co-initial vectors.

Solution; Vectors with same initial point are called as co-initial vectors.

6.Define collinear vectors.

Solutions; 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’.

7. Define coplanar vector.

Solution ; Three or more vectors lie on the same plane are called as coplanar vectors.

8. Define dierection cosines of a vector

Solution; If a vector a makes angles α, β & γ with the positive direction of x y & z axes respectively, then Cos α , Cos β & Cos γ are called as direction cosines of the vector a.

NOTE; If a = a1 i + a2 j + a3 k then , a1 Cos α = | a| a2 Cos β = | a| a3 Cos γ = | a| and Cos2 α + Cos2 β + Cos 2γ= 1 9. 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

10.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

b

η

11. If the position vectors of P & Q are 3i + 2j – 7k and 4i + 7j – 11k

Then, Find PQ & |PQ|.

Solution ; Let ‘O’ be the fixed point , then

Position vector of P = OP= 3i + 2j – 7k= ( 3 , 2 , -7 )

Position vector of P = OQ= 4i + 7j – 11k = ( 4 , 7 , -11 )

Therefore, PQ = OQ-OP = ( 1 , 5 , -4 ) =i + 5j – 4k

PQ= |PQ| = √ ( 1 + 25 + 16 )= √42 units.

12. If a = 2i – 3j + k , b = i + 2j – k & c = 3i + 2j + 6k , then find

|2a + b – 3c |.

Solution ; 2a = 4i – 6j + 2k = ( 4 , -6 , 2 )

b = i + 2j – k = ( 1 , 2 , -1 )

3c= 9i + 6j + 18k = ( 9 , 6 , 18 )

2a + b – 3c = ( -4 , -10 . -17 )

|2a + b – 3c | = √( 16 + 100 + 289 ) = √ 405 units.

13. Find the direction cosines of a vector 2i – 3j + k

Solution ; Let a = 2i – 3j + k = ( 2 , -3 , 1 )

| a | = √ ( 4 + 9 + 1 ) = √14

Direction cosines of a are

Cos α = 2/√14 , Cos β = -3/√14 & Cos γ = 1/√14.

14.If the direction cosines of a vector are 1/4 , 3/4 & n , then find ‘n’.

Solution : Let Cos α = 1/4 , Cos β =3/4 & Cos γ= n

Since, Cos2 α + Cos2 β + Cos 2γ= 1

1/16 + 9/16 + n2= 1

n2 = 1 – 1/16 – 9/16 = 6/16=3/8

n = ± (√3) / 2√2

15. find the scalar product of the vectors 2i + 3j – k & i - 2j – 5k

Solution : Let a = 2i + 3j – k = ( 2 , 3 , -1 )

b = i - 2j – 5k = ( 1 , -2 , -5 )

a • b = 2 – 6 + 5 = 1

16. If a = 2i – j + 3k & b = i + 2j + k & c = 2i + j + k , find

(a + b ) • (b – c ) .

Solution : a = 2i – j + 3k =( 2 , -1 , 3 )

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

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

a + b = ( 3 , 1 , 4 ) & b – c = ( -1 , 1 . 0 )

( a + b ) •( b – c ) = -3 +1 + 0 = -2

17.Prove that the vectors 3i –j – 2k & 2i -2j + 4k are orthogonal vectors.

Solution : Let a = ( 3 , -1 , -2 ) , b = ( 2 , -2 , 4 )

Consider a • b = 6 + 2 – 8 = 0

Therefore a and b are orthogonal vectors.

18. Find ‘m’ , if the vectors i + 3j – 2k & 2i – 4j + mk are orthogonal vectors.

Solution : Let a = ( 1 , 3 , -2 ) & b = ( 2 , -4 , m ) are orthogonal vectors,

Therefore, a • b = 0

2 – 12 – 2m = 0

-2m = 10

m = -5

19.Find the cosine of the angle between the vectors 3i + j – 2k &

3i – 5j – 2k.

Solution : Let a = ( 3 , 1 , -2 ) b = ( 3 , -5 , -2 )

a • b = 9 – 5 + 4 = 8

|a| = √ ( 9 + 1 + 4 ) = √ 14

|b| = √ ( 9 + 25 + 4 ) = √38

a • b 8 4

Cosθ = = =

|a| |b| √ 14 √38 √133

20. Find the angle between the vectors 2i + j + 2k & i – 2j + 2k

Solution : Let a = ( 2 , 1 , 2 ) b = ( 1 , -2 , 2 )

a • b = 2 - 2 + 4 = 4

|a| = √ ( 4 + 1 + 4 ) = 3

|b| = √ ( 1 +4 + 4 ) = 3

a • b 4 4

Cosθ = = =

|a| |b| 3 x 3 9

θ = Cos -1 (4/9)

21. Find the projection of the vector 2i + 3j – 2k in the dierection of the vector i - 2j + 3k.

Solution : Let a = ( 2 , 3 , -2 ) b = ( 1 , -2 , 3 )

a • b = 2 - 6 - 6 = -10

|b| = √ ( 1 + 4 + 9 ) = √ 14

a • b -10

projection of a along b = =

|b| √ 14

22. Find the angle between the vectors a + b & a - b if |a| =|b|.

Solution; Let |a| =|b| = λ ( say )

Consider, ( a + b ) • ( a - b ) = |a|2 -|b|2 = λ2 - λ2 = 0.

Therefore angle between ( a + b ) & ( a - b ) is 90°.

23.If |a + b|= 5 and a is perpendicular to b , Find |a - b|.

Solution;

Since, |a + b |2 - |a - b|2 = 4 a • b.

|a - b|2 = |a + b |2 - 4 a • b.

= (5)2 - 4(0) since a ⊥lar to b , a • b = 0

= 25

|a - b| = 5

24. If a & b are unit vectors and , |a + b | = 1 , find |a - b|.

Solution;

Since, |a + b |2 + |a - b|2 = 2 { |a|2 + |b|2 }

|a - b|2 = 2 { |a|2 + |b|2 } - |a + b |2

= 2{ 12 + 12 } - 12. Since, |a| =|b|= 1.

= 4 - 1

= 3

|a - b| = √3

25. If a , b , c are 3 vectors , such that a + b + c = 0 and |a| = 1 ,|b| = 2 & |c|=3 , find the value of a • b + b • c+ c • a.

Solution; a + b + c = 0

| a + b + c |2= 0

|a |2 + |b|2 + |c|2 + 2 { a • b + b • c + c • a } = 0

2 { a • b + b • c + c • a } = - (|a |2 + |b|2 + |c|2 )

= - ( 1 + 4 + 9 )

= - 14

a • b + b • c + c • a = -7

26.Find the cross product of the vectors j - 3k & i - j + 2k.

Solution; Let a = j - 3k = ( 0 , 1 , -3 )

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

i j k

a x b = 0 1 -3

1 -1 2

= i [ 2 - 3 ] - j [ 0 + 3 ] + k [ 0 - 1 ]

= - i - 3j - k

27. If b = 3a + c , prove that a x b = a x c.

Solution; Given that,

b = 3a + c

a x b = a x (3a + c )

= 3 ( a x a ) + ( a x c )

= 3(0) + ( a x c )

= ( a x c )

28. Prove that (2a + b ) x ( a + 2b ) = 3 ( a x b )

Solution; consider,

(2a + b ) x ( a + 2b ) = 2( a x a ) + 4 ( a x b ) + ( b x a ) + 2 ( b x b )

= 2 ( 0 ) + 4 ( a x b ) - ( a x b ) + 2 ( 0 )

= 3 ( a x b )

29.Show that the vectors 5i + 6j + 7k, 3i + 20j + 5k & 7i - 8j + 9k

are coplanar.

Solution; Let a = 5i + 6j + 7k = ( 5 , 6 , 7 )

b = 3i + 20j + 5k = ( 3 , 20 , 5 )

c= 7i - 8j + 9k = ( 7 , -8 , 9 )

consider,

[ a b c ] = a • ( b x c ) = 5 6 7

3 20 5

7 -8 9

= 5 [ 180 + 40 ] - 6 [ 27 - 35 ] + 7 [ -24 - 140 ]

= 0

Therefore , vectors a , b & c are coplanar.

30. If the vectors 2i - 3j + mk , 2i + j - k & 6i - j + 2k are coplanar,then, Find ‘m’.

Solution; Let a = 2i - 3j + mk = ( 2 , -3 , m )

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

c = 6i - j + 2k = ( 6 , -1 , 2 ) are coplanar vectors.

Then, [ a b c ] = 0

2 -3 m

2 1 -1 = 0

6 -1 2

2 [ 2 - 1 ] + 3 [ 4 + 6 ] + m [ -2 - 6 ] = 0

2 + 30 - 8m = 0

8m = 32

m = 4

31.Prove that [ i - j , j - k , k - i ] = 0

Solution; consider,

[ i - j , j - k , k - i ] = 1 -1 0

0 1 -1

-1 0 1

= 1 [ 1 - 0 ] + 1 [ 0 - 1 ] + 0 [ 0 + 1 ]

= 0

 

 

 

 

VECTORS

TWO MARKS QUESTIONS

1.Prove that the position vector of a point dividing the points A & B internally in the ration m:n is given by mb + na r = m + n where a & b are the position vectors of A & B w.r.t some fixed point. Solution : m n

Let O be the fixed point . Let P divides the A P B

line joining the points A & B

internally in the ration m : n a r b

O

Let a = position vector of A = OA

b = position vector of B = OB

r = position vector of P = OP

since, P divides the line joining the points A & B internally in the ration m:n,

AP/PB = m/n

nAP = mPB

nAP = mPB

n ( OP - OA ) = m ( OB - OP )

n OP - n OA = m OB - m OP

m OP + n OP = m OB + n OA

m OB + n OA OP = m + n m b + n a r = m + n 2. ABCD is a parallelogram and E is the point of intersection of two diagonlas, if O is any fixed point , prove that, OA + OB + OC + OD = 4 OE Solution : B A

C D

Since , E is the midpoint of the diagonals AC & BD.

If O is any fixed point,

Then, position vector of E = ( OC + OA) / 2

OE = ( OC + OA) / 2

2 OE = ( OC + OA) ------ ( 1 )

Similarly,

position vector of E = ( OB + OD) / 2

OE = ( OB + OD) / 2

2 OE = ( OB + OD) ------ ( 2 )

By adding ( 1 ) & ( 2 ) , we have,

OA + OB + OC + OD = 4 OE

              E 

3. If A = ( 2 , 3 , -4 ) and B = ( 1 , -1 . -2 ) Find the co-ordinates of the point dividing AB internally in the ratio 2 : 3. Solution : 3 A 2 P B

Let P be a point which divides line joining the points AB internally in the ration 2:3,

If O be the fixed point ,

OP = position vector of P

OA = position vector of A = (2 , 3 , -4 )

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

2 OB + 3 OA ( 2 . -2 , -4 ) + ( 6 , 9 , -12 ) ( 8 , 7 , -16 ) OP = = = 2 + 3 5 5 = ( 8/5 , 7/5 , -16/5 )

4. Show that the vectors 2i - j + k , i - 3j - 5k & 3i - 4j - 4k form a right angled triangle.

Solution : Let a = 2i - j + k

b = i - 3j - 5k

c = 3i - 4j - 4k

since , a + b = c , a , b & c represents sides of the triangle,

a = | a | = √ ( 4 + 1 + 1 ) = √6 units

b = | b | = √( 1 + 9 + 25 = √ 35 units

c = | c | = √ ( 9 + 16 + 16 ) = √ 41 units

since, c² = a² + b²

therefore, a, b & c forms a right angled triangle.

5. Show that the points with position vectors i + 2j + 3k , - i -j + 8k & - 4i + 4j + 6k form an equilateral triangle.

Solution: Let O be the fixed point.

Let OA = position vector of A = i + 2j + 3k = ( 1 , 2 , 3 )

OB = position vector of B = - i - j + 8k = ( -1 , -1 , 8 )

OC = position vector of C = - 4i + 4j + 6k = ( -4 , 4 , 6 )

AB = OB - OA = ( -2 , -3 , 5 )

BC = OC - OB = ( -3 , 5 , -2 )

CA = OA - OC = ( 5 , -2 , -3 )

Since , AB + BC + CA = 0

ABC is a triangle.

AB = |AB| = √ ( 4 + 9 + 25 ) = √38 units,

BC = |BC| = √ (9 + 25 + 4 ) = √38 units,

CA = |CA| = √ (25 + 4 + 9 ) = √38 units,

Since, AB = BC = CA , ABC is an equilateral triangle.

6. If a = 5i - j -3k , b = i + 3j + 5k , show that , ( a + b ) & ( a - b ) are orthogonal vectors.

Solution : consider, a = 5i - j -3k = ( 5 , -1 , -3 )

b = I + 3j + 5k = ( 1 , 3 , 5 )

( a + b ) = ( 6 , 2 , 2 )

( a - b ) = ( 4 , -4 , -8 )

( a + b ) • ( a - b ) = 24 -8 - 16 = 0

therefore ,

( a + b ) & ( a - b ) are orthogonal vectors.

7. Prove that , (i) | a + b |² = | a |² + | b |² + 2 a • b

(ii) | a - b |² = | a |² + | b |² - 2 a • b

Solution : | a + b |² = ( a + b ) • ( a + b )

= a • a + a • b + b • a + b • b

= | a |² + a • b + a • b + | b |²

= | a |² + 2 a • b + | b |²

Next consider,

| a - b |² = ( a - b ) • ( a - b )

= a • a - a • b - b • a + b • b

= | a |² - a • b - a • b + | b |²

= | a |² - 2 a • b + | b |²

8. Prove that

( i ) | a + b |² + | a + b |² = 2 { | a |² + | b |² }

( i i ) | a + b |² - | a + b |² = 4 a • b

Solution : consider,

| a + b |² + | a + b |² = { | a |² + 2 a • b + | b |² } + { | a |² - 2 a • b + | b |² }

= 2 { | a |² + | b |² }

Next consider,

| a + b |² - | a + b |² = { | a |² + 2 a • b + | b |² } - { | a |² - 2 a • b + | b |² }

= 4 a • b

9. 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°

10. 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.

11.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)

12. ABC is an equilateral triangle of side ‘a’ then prove that,

AB • BC + BC • CA + CA • AB = - 3/2 a²

Solution: Since ABC is a triangle,

AB + BC + CA = 0

│ AB + BC + CA │² = 0

│AB│² +│BC│² + │CA│² + 2 { AB • BC + BC • CA + CA • AB } = 0

2 { AB • BC + BC • CA + CA • AB } = - {│AB│² +│BC│² + │CA│² }

= - { a² + a² + a² } since ,│AB│=│BC│=│CA│=a

= -3 a²

{ AB • BC + BC • CA + CA • AB } = -3/2 a²

13.Find a unit vector perpendicular to both the vectors 2i – 2j + k & 4i + j – k.

Solution : Let a = 2i – 2j + k

b = 4i + j – k

i j k

a x b = 2 -2 1 = i [ 2 – 1 ] –j [ -2 -4 ] + k [ 2 + 8 ] = i + 6j + 10k

4 1 -1

│a x b │ = √ ( 1 + 36 + 100 ) = √137 units

a x b 1 { i + 6j + 10k } η = = │a x b │ √137 14. Find a unit vector perpendicular to the plane determined by the points

( 1 , -1 , 2 ), ( 2 , 0 , -1 ) & ( 0 , 2 , 1 ).

Solution : Let O be the fixed point.

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

OB = position vector of B = ( 2 , 0 , -1 )

OC = position vector of C = ( 0 , 2 , 1 ).

AB = OB - OA = ( 1 , 1 , -3 )

AC = OC – OA = ( -1 , 3 , -1 )

i j k

AB x AC = 1 1 -3 = i [ -1 + 9 ] – j [ -1 -3 ] + k [ 3 + 1 ] = 8 i + 4j + 4k

-1 3 -1

│ AB x AC │ = √ (64 + 16 + 16 ) = √96 = 4√6 units

AB x AC 8 i + 4j + 4k 4 { 2i + j + k } 2i + j + k η = = = = │ AB x AC │ 4√6 4√6 √6

15. Find the Sine of the angle between the vectors 4i + 3j + 2k & i – j + 3k.

Solution: Let a = 4i + 3j + 2k

b = i – j + 3k.

i j k

a x b = 4 3 2 = i [ 9 + 2 ] – j [ 12 – 2 ] + k [ -4 -3 ] = 11i -10j -7k

1 -1 3

│ a x b │ = √ ( 121 + 100 + 49 ) = √270

│ a │ = √ ( 16 + 9 + 4 ) = √29

│ b │ = √ ( 1 + 1 + 9 ) = √11

│ a x b │ √270 Sinθ = = │ a ││ b │ √29√11

16. Find the area of the parallelogram whose adjecent sides are represented by the vectors i + j + k & i – j + k.

Solutions: Let a = i + j + k

b = i – j + k

i j k

a x b = 1 1 1 = i [ 1 + 1 ] – j [ 1 – 1 ] + k [ -1 – 1 ] = 2i + oj – 2k

1 -1 1

│ a x b │ = √ ( 4 + 0 + 4) = √ 8 = 2 √ 2 units

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

17. Find the area of the parallelogram whose diagonals are represented by the vectors - 4 i +2 j + k & 3 i – 2 j - k.

Solutions: Let d1 = - 4 i +2 j + k

d2 = 3 i – 2 j - k

i j k

d1 x d2= - 4 2 1 = i [ -2 + 2 ] – j [ 4 - 3 ] + k [ 8 – 6 ] = 0i - j + 2k

3 -2 -1

│d1 x d2│= √ ( 0 + 1 + 4 ) = √ 5 units

Area of the parallelogram = │d1 x d2│ = (√ 5)/2 square units.

2

18. Find the area of the triangle ,two of whose sides are represented by the vectors 3i + 4j & 5i + 7j + k.

Solutions: Let a = 3i + 4j

b = 5i + 7j + k

i j k

a x b = 3 4 0 = i [ 4 - 0 ] – j [ 3 - 0 ] + k [ 21 – 20] = 4i -3 j + k

5 7 1

│a x b│= √ ( 16+ 9 + 1 ) = √ 26 units

Area of the triangle = │a x b│ = (√ 26) /2 square units.

2

19. Find the area of the triangle whose vertices are represented by the position vectors i+ 3j + 2k , 2i – j + k & - i + 2j + 3k.

Solution : Let O be the fixed point,

Let OA = position vector of A = i+ 3j + 2k = ( 1 , 3 , 2 )

OB = position vector of B = 2i – j + k = ( 2 , -1 , 1 )

OC = position vector of C= - i + 2j + 3k = ( -1 , 2 , 3 )

AB = OB – OA = ( 1 , -4 , -1 )

AC = OC – OA = ( -2 , -1 , 1 )

i j k

AB x AC = 1 - 4 -1 = i [ - 4 - 1 ] – j [ 1 - 2 ] + k [ -1 – 8] = -5 i + j -9 k

-2 -1 1

│ AB x AC │= √ ( 25+ 1 + 81 ) = √ 107units

Area of the triangle = │ AB x AC │ = (√107) /2 square units.

2

20.Find the perpendicular distance of A ( 1 , 4 , -2 ) from the line segment BC, where B ( 2 , 1 , -2 ) & C = ( 0 , -5 , 1 ).

Solution : A

B P C

Here , AP is the perpendicular distance from the line segment BC.

: Let O be the fixed point,

Let OA = position vector of A = ( 1 , 4 , -2 )

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

OC = position vector of C = ( 0 , -5 , 1 ).

AB = OB – OA = ( 1 , -3 , 0 )

AC = OC – OA = ( -1 , -9 , 3 )

i j k

AB x AC = 1 - 3 0 = i [ - 9 - 0 ] – j [ 3 - 0 ] + k [ -9 – 3] = -9 i - 3 j -12 k

-1 -9 3

│ AB x AC │= √ ( 81+ 9 + 144 ) = √ 234units = 3√ 26units

= Area of the triangle = │ AB x AC │ = (3√26) /2 square units.

2

BC = OC – OB = ( -2 , -6 , 3 )

BC = │ BC │ = √ 4 + 36 + 9 ) = √49 = 7 units

Since , = ½ (BC) (AP)

(3√26) /2 = 1/2 ( 7 ) AP

AP = (3√26)/7 units.

21. 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

22. prove that , │a x b │² + │ a • b │² = 2 { │ a │²│ b │² }

Solution : consider,

│a x b │² + │ a • b │² = {│ a ││ b │Sinθ }² + {│ a ││ b │Cosθ }²

=│ a │²│ b │² sin²θ + │ a │²│ b │² Cos²θ

= │ a │²│ b │² { sin²θ + Cos²θ }

= │ a │²│ b │²

23. If │a x b │ = 4 & │ a • b │= 2 , Find │ a │²│ b │² .

Solution : since , we know that,

│a x b │² + │ a • b │² = 2 { │ a │²│ b │² }

Therefore, 2 { │ a │²│ b │² } = 16 + 4 = 20

│ a │²│ b │² = 10

24. If θ be the angle between the vectors a & b , find the value of

│a x b │

│ a • b │

Solution : consider,

│a x b │ │ a ││ b │Sinθ = = tanθ │ a • b │ │ a ││ b │Cosθ

25. If a + b + c = 0 , prove that a x b = b x c = c x a

Solution: 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 )

Again consider,

a + b + c = 0

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

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

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

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

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

a x b = b x c = c x a

26. Find the volume of the parallelepiped whose co-terminal edges are represented by the vectors 2i + j – k , 3i – 2j + 2k & i - 3j – 3k.

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

b = 3i – 2j + 2k = ( 3 , -2 , 2 )

c = i - 3j – 3k. = ( 1 , -3 , -3 ) are represent the co-terminal edges of the parallelepiped object.

2 1 -1

Volume of the parallelepiped object = 3 -2 2

1 -3 -3

= 2 [ 6 + 6 ] – 1 [ -9 -2 ] -1 [ -9 + 2 ]

= 24 + 11 + 7

= 42 cubic units.

27. Find the vector triple product a x ( b x c ) , when

a = 2i + 3j – k , b = i + 2j – 5k & c = 3i + 5j - k

Solution : Given that a = 2i + 3j – k = ( 2 , 3 -1 )

b = i + 2j – 5k = ( 1 , 2 , -5 )

c = 3i + 5j – k = ( 3 , 5 -1 )

we know that,

a x ( b x c ) = ( c • a ) b – ( b • a ) c ------- ( 1 )

now, c • a = 6 + 15 + 1 = 22

b • a = 2 + 6 + 5 = 13

( 1 ) becomes,

a x ( b x c ) = 22 b – 13 c

= ( 22 , 44 , -110 ) – ( 39 , 65 , -13 )

= ( -17 , -21 -97 )

= - 17i – 21 j – 97 k

28. Find the value of │( a x b ) x c │, when a = ( 1 , 2 , 3 ) , b = ( 2 , 1 , 2 ) &

c = ( 3 , 3 , 2 )

Solution : it is given that, a = ( 1 , 2 , 3 )

b = ( 2 , 1 , 2 )

c = ( 3 , 3 , 2 )

we know that,

( a x b ) x c = ( c • a ) b – ( c • b ) a ----------- ( 1 )

Consider, c • a = 3 + 6 + 6 = 15

c • b = 6 + 3 + 4 = 13

( 1 ) becomes,

we know that,

( a x b ) x c = 15 b – 13 a

= ( 30 , 15 , 30 ) – ( 13 , 26 , 39 )

= ( 17 , - 11 , - 9 )

we know that,

│ ( a x b ) x c │ = √ ( 289 + 121 + 81 ) = √491 units.

29. prove that, ∑ a x ( b x c ) = 0

Solution : consider,

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

= { ( c • a ) b – ( b • a ) c } + { ( b • a ) c – ( b • c ) a } + { ( c • b ) a – ( c • a ) b }

= 0

VECTORS

THREE MARKS QUESTIONS

1. In a regular hexagon ABCDEF , Show that AB + AC + AD + AE + AF = 3 AD

Solution: E D

F C

A B

Let AB = a , BC = b

AC = AB + BC = a + b

AD = 2 BC = 2 b

AE = AD + DE = AD – ED = 2 b – a

AF = CD = CA + AD = - AC + AD = - ( a + b ) + 2 b = - a – b + 2b = b – a

Consider,

AB + AC + AD + AE + AF = a + ( a + b ) + 2 b + ( 2 b – a ) + ( b – a )

= 6 b

= 3 ( 2 b )

= 3 AD

2. Prove that position vector of the centroid of a triangle ABC is 1/3( a + b + c ), where, a , b & c are the position vectors of the vertices A , B & C w.r.t. some fixed point O.

Solution : Let O be the fixed point.

a = position vector of A = OA

b = position vector of B = OB

c = position vector of C = OC

let AD be the median of the triangle ABC , A

Since D is the mip point of BC

Position vector of D = OD = (OB + OC ) / 2

Since , G divides internally AD in the ratio 2 : 1

B D C

2 ( OD ) + 1 ( OA ) Position vector of G = 3 = 2 {(OB + OC ) / 2 } + 1 ( OA ) 3 = OB + OC + OA 3 = OA + OB + OC 3 = a + b + c 3

           •    G 

3. If the position vectors of the points P and Q are 2 i + 3j + 4k and 3 i – 2 j – 3 k , find the direction cosines of the vector PQ and hence prove that,

Cos²α + Cos²β + Cos²γ = 1

Solution: Let O be the fixed point.

OP = position vector of P = 2 i + 3j + 4k = ( 2 , 3 , 4 )

OQ = position vector of Q = 3 i – 2 j – 3 k = ( 3 , - 2 , -3 )

PQ = OQ – OP = ( 1 , -5 , -7 )

│PQ│= √ ( 1 + 25 + 49 ) = √ 75

Direction cosines of PQ are Cosα = 1/√ 75 , Cosβ = -5/√ 75 & cosγ = - 7 /√ 75.

Consider, Cos²α + Cos²β + Cos²γ = 1/75 + 25 /75 + 49 /75 = 1

4.Show that the points ( 1 , 2 , 1 ) , ( 2 , 4 , 2 ) ( 4 , 3 , -2 ) & ( 3 , 1 , -3 ) are the vertices of a parallelogram.

Solution: D C

A B

Let O be the fixed point.

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

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

OC = position vector of C = ( 4 , 3 , -2 )

OD = position vector of D = ( 3 , 1 , -3 )

AB = OB – OA = ( 1 , 2 , 1 )

BC = OC – OB = ( 2 , - 1 , - 4 )

CD = OD – OC = ( - 1 , -2 , -1 )

DA = OA – OD = ( -2 , 1 , 4 )

AC = OC – OA = ( 3 , 1 , - 3 )

 

BD = OD – OB = ( 1 , -3 , -5 )

AB = │AB│= √ ( 1 + 4 + 1 ) = √6 units

BC = │BC│= √ ( 4 + 1 + 16 = √21 units

CD = │CD│=√ ( 1 + 4 + 1 ) = √6 units

DA = │DA│=√ ( 4 + 1 + 16) = √21 units

AC = │AC│=√ (9 + 1 + 9) = √19 units

BD = │BD│=√(1 + 9 + 25) = √35 units

Since, AB = CD , BC = DA , but, AC ≠ BD

Therefore, ABCD form a parallelogram.

5. Find a unit vector perpendicular to both the vectors a & b , Also, find the

Sine of the angle between the vectors a & b , where, a = 6 i – 2j + k &

b = 3 i + j – 2 k .

Solution, Consider,

a = 6 i – 2j + k = ( 6 , -2 , 1 )

b = 3 i + j – 2 k = ( 3 , 1 , -2 )

i j k

a x b = 6 -2 1 = i [ 4 – 1 ] – j [ - 12 - 3 ] + k [ 6 + 6 ] = 3 i + 15 j + 12 k

3 1 -2

a x b = 3 { i + 5 j + 4 k }

│ a x b │ = 3√ ( 1 + 25 + 16) = 3√42 units

│ a │ = √ ( 36 + 4 + 1) =√ 41 units

│ b │ = √ ( 9 + 1 + 4) = √ 14 units

η = unit vector perpendicular to both a & b

a x b 3 { i + 5 j + 4 k } i + 5 j + 4 k

= = =

│ a x b │ 3√42 √42

│ a x b │ 3 √42

Sinθ = =

│ a │ │b│ √ 41 √ 14

6. If a , b & c are the position vector of the vertices of a triangle ABC , Prove

that ,Vector area of the triangle ABC = ½│ ( a x b ) + ( b x c ) + ( c x a )│square

units.

Solution : Let O be the fixed point . A

a = OA = position vector of A

b = OB = position vector of B

c = OC = position vector of C

AB = OB – OA = b – a B C

AC = OC – OA = c – a

AB x AC = ( b – a ) x ( c – a )

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

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

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

│AB x AC│ = │( b x c ) + ( a x b ) + ( c x a ) │

Area of the triangle ABC = 1/2│AB x AC│

= │( b x c ) + ( a x b ) + ( c x a ) │square units.

7. 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 ]

8.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

9.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

Again, consider,

, 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

10.Show that the points ( - 6 , 3 , 2 ) , ( 3 , -2 , 4 ) , ( 5 , 7 , 3 ) & ( -13 , 17 , -1 ) are coplanar.

Solution : Let O be the fixed point.

OA = position vector of A = ( - 6 , 3 , 2 )

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

OC = position vector of C = ( 5 , 7 , 3 )

OD = position vector of D = ( -13 , 17 , -1 )

Therefore,

AB = OB – OA = ( 9 , -5 , 2 )

AC = OC – OA = ( 11 , 4 , 1 )

AD = OD – OA = ( -7 , 14 , -3 )

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.

VECTORS 4 OR 5 MARKS QUESTIONS

1. 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

2.Find a unit vector which is coplanar with a & b and perpendicular to a, where,

a = 2i + j + k & b = i + 2j – k

Solution : Given that,

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

b = i + 2j – k = ( 1 , 2 , - 1 )

consider,

( a x b ) x a = ( a • a ) b – ( a • b ) a -------- ( 1 )

Now, a • a = 4 + 1 + 1 = 6

a • b = 2 + 2 – 1 = 3

from ( 1 ) , we have,

( a x b ) x a = 6 b – 3 a

= 3 { 2b – a }

= 3 { ( 2 , 4 , - 2 ) – ( 2 , 1 , 1 ) }

= 3 ( 0 , 3 , - 2 )

= 3 { 0 i + 3 j – 3 k }

= 9 { 0 i + j – k }

|( a x b ) x a | = 9√ ( 0 + 1 + 1 ) = 9 √2 units

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

( a x b ) x a 9 { 0 i + j – k } 0 i + j – k

= = =

|( a x b ) x a | 9 √2 √2

3. If ( a x b ) x c = a x ( b x c ) , then prove that either a is parallel to c or b is

perpendicular to both a & c

Solution : Given that,

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

( a • c ) b – ( b • c ) a = ( c • a ) b – ( b • a ) c

Since , ( a • c ) = ( c • a )

( b • c ) a =( b • a ) c

p a = q c by taking, ( b • c ) = p & ( b • a )= q

where , p & q are scalars

therefore , a is parallel to c

Next consider, ( b • c ) a =( b • a ) c

( b • c ) a - ( b • a ) c = 0

b is perpendicular to both a & c.

5. Prove by vector method , that the medians of a triangle are concurrent.

Solution :

A

R Q

B P C

Let us consider a triangle ABC. Let O be the fixed point,

Solution : Let O be the fixed point.

OA = position vector of A = a

OB = position vector of B = b

OC = position vector of C= c

Position vector of the mid point P is OP = ½(OB + OC) = ½( b + c )

If G divides AP in the ratio 2 : 1

Then, the position vector of G = OG = 2{½( b + c )} + 1 (a)

2 + 1

= b + c + a

3

The symmetry of this result show that , the point which divides the other two medians in the ratio 2 : 1 will also have the same position vector b + c + a

3

Hence, the medians of a triangle are concurrent at G.

               G 

5. Prove that diagonals of a parallelogram bisect each each other.

Solution: D C

A B

Let the diagonals AC and BD of the parallelogram ABCD intersect at P. Opposite sides of a parallelogram are equal and parallel.

Therefore, AB = CD

If O be the Origin of reference,

OA = position vector of A

OB = position vector of B

OC = position vector of C

OD = position vector of D

Since , AB = DC

OB – OA = OC – OD

OB + OD = OA + OC

OB + OD OA + OC = 2 2 That is mid points of BD & AC are coincide.

Therefore , the diagonals AC & BD are bisect each other.

            P 

6. 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 

    

7. 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

Solution for ( c ): consider,

a + b + c = 0

a = - b – c

a ● a = - (a ● b ) – ( a ● c )

│a│² = - │a││b│Cos(π – C) - │a││c│Cos(π – B)

a² = - ab { - CosC } – ac { - CosB }

a² = abCosC +ac CosB

a = bCosC + c CosB

8. 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.

VECTORS

6 MARKS QUESTIONS

1. Define Dot product and vector product of any two co-initial vectors.

If a = a 1 i + a 2 j + a 3 k and b = b 1 i + b 2 j + b 3 k , Prove that ,

a • b = a 1 b 1 + a 2 b 2 + a 3 b 3 and

i j k

a x b = a 1 a 2 a 3

b 1 b 2 b 3

Definition of dot product ( scalar product ) of any two vectors:

If a & b are any two co-initial vectors, & if θ is the angle between the vectors , then,

a • b = | a | | b | Cosθ B

b

θ

O A

, a

Definition of vector product ( cross product ) of any two vectors:

If a & b are any two co-initial vectors, if θ is the angle between the vectors , and η be the unit vector perpendicular to both a & b then,

a x b = | a | | b | η Sinθ B

b

θ

O A

, η a

Derivation of a • b Y

j

O i X

k

Z

i , j & k are mutually perpendicular unit vectors along X , Y & Z axes,

i • i = j • j = k • k = 1 & i • j = j • k = k • i = 0 = j • i = k • j = i • k

consider,

a • b = { a 1 i + a 2 j + a 3 k } • { b 1 i + b 2 j + b 3 k }

= a 1 b 1 ( i • i ) + a 1 b 2 ( i • j ) + a 1 b 3 ( i • k ) + a 2 b 1 ( j • i )

+ a 2 b 2 ( j • j ) + a 2 b 3 ( j • k ) + a 3 b 1 ( k • i ) + a 3 b 2 ( k • j )

+ a 3 b 3 ( k • k )

= a 1 b 1 ( 1 ) + a 1 b 2 ( 0 ) + a 1 b 3 ( 0 ) + a 2 b 1 ( 0 )

+ a 2 b 2 ( 1 ) + a 2 b 3 ( 0 ) + a 3 b 1 ( 0 ) + a 3 b 2 ( 0 )

+ a 3 b 3 ( 1 )

= a 1 b 1 + a 2 b 2 + a 3 b 3

Derivation of a x b Y

j

O i X

k

Z

i , j & k are mutually perpendicular unit vectors along X , Y & Z axes,

i x i = j x j = k x k = 0 & i x j = k , j x k = i , k x I = j , but ,

j x i = - k , k x j = - i & i x k = - j

consider,

a x b = { a 1 i + a 2 j + a 3 k } x { b 1 i + b 2 j + b 3 k }

= a 1 b 1 ( i x i ) + a 1 b 2 ( i x j ) + a 1 b 3 ( i x k ) + a 2 b 1 ( j x i )

+ a 2 b 2 ( j x j ) + a 2 b 3 ( j x k ) + a 3 b 1 ( k x i ) + a 3 b 2 ( k x j )

+ a 3 b 3 ( k x k )

= a 1 b 1 ( 0 ) + a 1 b 2 ( k ) + a 1 b 3 ( - j ) + a 2 b 1 ( - k )

+ a 2 b 2 ( 0 ) + a 2 b 3 ( i ) + a 3 b 1 ( j ) + a 3 b 2 ( - i )

+ a 3 b 3 ( 0 )

= a 1 b 2 ( k ) + a 1 b 3 ( - j ) + a 2 b 1 ( - k )

+ a 2 b 3 ( i ) + a 3 b 1 ( j ) + a 3 b 2 ( - i )

= i { a 2 b 3 - a 3 b 2 } – j { a 1 b 3 - a 3 b 1 } + k { a 1 b 2 - a 2 b 1 }

i j k

a x b = a 1 a 2 a 3

b 1 b 2 b 3

2. Prove that [ a b c ] = [ b c a ] = [ c a b ] & also show that

[ a b b ] = 0

Solution :Let a = m i + nj + l k

b = x i + y j + z k

c = p i + q j + rk

consider, m n l

[ a b c ] = x y z

p q r

perform C2 C1

l n m

[ a b c ] = - z y x

r q p

perform C2 C3

l m n

[ a b c ] = z x y

r p q

[ a b c ] = [ c a b ]

Simmillary , we may show that , [ a b c ] = [ b c a ]

Therefore, [ a b c ] = [ b c a ] = [ c a b ]

Nest consider, m n n

[ a b b ] = x y y

p q q

since, second and third columns are identical,

[ a b b ] = 0

3. 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.

3. 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

4. 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

 

                                   A ‐ B 

              O 

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

Consider,

Cos( A + B ) = Cos{ A – (- B) }

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

= CosACosB - SinA SinB

5. 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,

 

                                   A ‐ B 

              O 

i j k

OP x OQ = CosA SinA 0

CosB SinB 0

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

= 0i + oj – 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

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

Consider,

Sin ( A + B ) = SinACosB + CosASinB

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

= SinACosB - CosASinB