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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