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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
( i ) the unit vector in the direction of the vector a =
2 i + 3j – 5k
=
|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
η
θ
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
Cos θ =
-6
=
|a| |b|
-2
=
3√14
√14
( iii )
a • b
projection of a in the direction of b =
-6
=
|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 &
(i)axb
b = i + 3j + k , Find
( 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+3j+k =(1 ,3 ,1)
(I)axb =
i
j
k
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
|axb|
Sinθ
=
5√6
=
|a| |b|
√14 √11
(Iv)
Unit vector perpendicular to both a & b
axb
=
=
|a x b|
- 5 i – 2 j + 11 k
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= -3i–2j+k=(-3 ,-2 ,1)
(I)axb =
i
j
k
1
2
3
-3
-2
1
=i[2+6]–j[1+9]+k[-2+6]
= 8 i - 10 j + 4 k
=2{4I–5j+2k}
| 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)axb =
i
j
k
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.
|axb|
Area of the parallelogram =
10 √3
=
2
= 5√3
2
square units.
13. Find the area of the triangle , whose two adjecent sides are given by the vectors,
i+4j–k
&
i + j + 2k.
solution : Let
a= i+4j–k =(1 ,4 ,-1)
b = i + j + 2k
axb=
=(1 ,1 ,2 )
i
j
k
1
4
-1
1
1
2
=i[8+1]–j[2+1]+k[1–4]
=9i–3j–3k
=3{3i-j–k}
| a x b | = 3√( 9 + 1 + 1 ) = 3√11 units.
|axb|
Area of the triangle =
3√11
=
2
square units
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
AB x AC =
j
k
1
2 -4
2
0
0
=i[0–0]–j[0+8]+k[0–4]
= 0i–8j–4k
=4{0i–2j–k}
| AB x AC | = 4√( 0 + 4 + 1 ) = 4√5 units
| AB x AC |
Area of the triangle ABC =
4√5
=
2
= 2√5 square units
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+2j- k
= ( 1 , 2 , -1 )
c= 3i- j +2k
=(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=3i+4j+2k
c=3i+j+5k
consider,
[ a
b
c ] =
1
1
1
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,
[ AB AC AD ] =
= -
=0
1
2
3
2
0
1
-2 0
-1
1
2
3
2
0
1
2
0
1
{ 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,
[ AB AC AD ] =
9
-5
2
11
4
1
14
-3
-7
= 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 =3i+j–2k,
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
3
m
5
= 0
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
-1
5
-3
=0
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)●{(bxc)+(bxa)+(cxc)+(cxa)}
= ( 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 ●(bxc)+b ●(bxa)+b ●(cxa)}
=[a
b
c]+[a
=[a
b
c]+0+0+0+0+[a
=2[a
b
b
a]+[a
c
a ]+[b
b
b
c]
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
axi
=
j
k
a1 a2 a3
1 0 0
= i [ 0 – 0 ] – j [ 0 - a3 ] + k [ 0 - a2 ]
= 0i + a3 j - a2 k
Next, ,
i
i x (a x i ) =
j
k
1
0 0
0
a3 - a2
= i [ 0 – 0 ] – j [- a2 - 0 ] + k [ a3 - 0 ]
= 0i + a2 j + a3 k ,
Next,
i
axj
=
j
k
a1 a2 a3
0 1 0
= i [ 0 – a3 ] – j [ 0 - 0 ] + k [a1 - 0 ]
= - a3 i + 0 j + a1 k
c]+[b
b
a]+[b
c
a]
Therefore,
,
j x (a x j ) =
i
j
k
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 )
=(axb)+(axc)+(bxc)+(bxa)+(cxa)+(cxb)
=(axb)-(cxa)+(bxc)-(axb)+(cxa)-(bxc)
=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+2j+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}
=
|( a x b ) x c|
- i + j + 0k
=
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,
[axb ,bxc ,cxa]= ( axb)•{( b xc) x( c xa)}
------ ( 1 )
Let ( b x c ) = p, then,
( b xc) x( c xa)= p x ( c x a )
= (a•p)c- (c•p)a
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 xc) x( c xa)= (λ)c–(0)a=λc
Therefore ( 1 ) becomes,
[axb ,bxc ,cxa]= ( axb)•λc
=λ {( axb)• c}
=λ { c•( axb)}
=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,
[axb ,bxc ,cxa]= 0
[ a , b , c ]² = 0
[ a , b, c ] =0
Therefore , a
b & c are coplanar vectors.
------- ( 2 )
26. Prove by vector method that, The angle in a semi circle is a right angle.
Solution:
P
O 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
=0
Therefore PA perpendicular to PB
Therefore, APB = 90º
since , │PO│ = │OA│= radius of the circle.
27. In any triangle ABC, prove by vector method
a
b
(a)
=
SinA
c
=
SinB
SinC
( b ) a² = b² + c² - 2bcCosA
( c ) a = bCosC + c CosB
π–A
A
π-C
B
π–B
Solution : Let BC = a , CA = b & AB = c
Then, a + b + c = 0
Solution for ( a ): consider, a + b + c = 0
ax( a+b+c)=ax0
(axa)+(axb)+(axc)=0
0 + (axb)–(cxa)=0
(axb)=(cxa)
------ ( 1 )
C
Next, consider,
a + b + c =0
Similarly , as above,
bx( a+b+c)=bx0
(bxa)+(bxb)+(bxc)=0
-(axb)+ (bxb)+(bxc)=0
-(axb)+0+(bxc)=0
(axb)=(bxc)
------ ( 2 )
from ( 1 ) & ( 2 ) , we have,
(axb)=(bxc)=(cxa)
│ 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
=
SinA
c
=
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
=
(c•b)
= λ ( say )
(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 =λi
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
A ‐ B |OQ|= √ (Cos²B + Sin²B) = 1
X
O 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
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
A ‐ B |OQ|= √ (Cos²B + Sin²B) = 1
X
O |OP x OQ| = |OP||OQ|Sin( A – B )
= ( 1 ) ( 1 ) Sin( A – B )
= Sin( A – B )
------- ( 1 )
But,
i
OP x OQ =
j
k
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
Therefore , from (1) & (2) , we have,
Sin ( A – B ) = SinACosB – CosASinB
------ ( 2 )
Next, since, Sin ( - θ ) = - Sinθ
& Cos( - θ ) = Cos θ
Consider,
Sin ( A + B ) = SinACosB + CosASinB
= SinACos( - B ) + CosASin( - B )
= SinACosB - CosASinB