UNIT- II
VECTOR ALGEBRA – II
2.1. Vector product of two vectors:
Definition of vector product of two vectors-Geometrical meaningproperties-angle between two vectors–unit vector perpendicular
to two vectors-simple problems.
2.2. Application of vector product of two vectors and Scalar
Triple Product:
Definition of moment of a force, definition of scalar product of
three vectors - geometrical meaning – coplanar vectors - simple
problems.
2.3. Product of more vectors:
Vector Triple Product - Scalar and Vector product of four vectors.
Simple problems
2.1.VECTOR PRODUCT OF TWO VECTORS OR CROSS
PRODUCT
The vector product of two vectors a and b , whose directions are
inclined at an angle θ, is the vector whose modulus is a b sin θ, and
whose direction is perpendicular to both a and b , being positive
relative to a rotation from a to b .
/ / / /
/
i.e. a × b = a b sin θ n
23
where n is an unit vector perpendicular to the plane of a and b ,
having the same direction as the translation of a right handed screw
due to the rotation from a to b . From this it follows that b × a ,has the
same length, so that
a × b =-b × a .
Properties of vector product:
∧
1.
b × a = b a sin θ (- n )
∧
= - a b sin θ n
=- a × b
2.
If a and b are parallel, the angle θ = 0
∧
∴ a × b = a b sin θ n
∧
= a b sin 0 n
= 0 [∴sin 0=0]
/
/
∴The condition for two vectors a and b to be parallel is a × b =0
In particular, a × a = 0
/
/
/
If a × b = 0 , either a = 0 or b = 0 or a and b are parallel vectors
3.
If a, b, c are three vectors,
/ /
/ /
/ /
a × b + c = a × b + (a × c )
4.
If i, j, k are the unit vectors along x, y & z axes respectively,
/ / / /
i × i = j × j = k×k = 0
/ /
/ /
i × j = −j× i = k
Also, /
/ /
j × k = −k × j = i
/
/
/
k × i = −i ×k = j
(
) (
)
24
Hence
5.
×
/
i
/
j
/
i
0
k
/
j
-k
k
j
0
/
-i
k
/
j
/
i
0
/
/
/ /
/
/
/
If a = a1 i + a 2 j + a3k, b = b1 i + b2 j + b3k ,
/
/
i
j
/
Then a × b = a 1 a 2
b1 b 2
/
k
a3
b3
Proof:
G
G
G
G
Let a =a1 i +a 2 j +a 3 k, b =b1 i +b2 j +b3 k
G
G
G
ax b = a1 i +a 2 j +a 3 k x b1 i +b2 j +b3 k
G G
G K
G G
=a1b1 i × i +a1b2 i × j +a1b3 i × k
G G
K G
K K
+a 2 b1 j × i +a 2 b2 j × j +a 2 b3 j × k
G G
G K
G G
+ a 3b1 k × i +a 3 b2 k × j + a 3b3 k × k
G
G
= o +a1b2 k +a1b3 − j +a 2 b1 − k + o
G
G
G
+ a 2b3 i +a 3 b1 j +a 3 b2 − i + a 3 + b3
G a 2 a 3 G a1 a 3
a a
=i
−j
+k 1 2
b1 b2
b1 b3
b 2 b3
G G
i
j k
(
( )
( )
( )
) (
( )
(
)
( )
( )
( )
)
( )
( )
( )
( )
= a1 a 2 a 3
b1 b2 b3
25
6.
7.
/
a x b = a b sin θn
/ /
∴ axb =|a ||b |sin θ …(1)
K G
a xb
n= G G
|a ||b |sin θ
K G
ax b
∴ n = K G by using (1)
|axb |
If ‘θ’ is the angle between the vectors a and b then
Sin θ =
8.
[|n |= 1]
axb
/
a b
Geometrical meaning of vector product
Let a = OA and b = OB
Complete the parallelogram OACB with the sides OA and OB
(See figure) Draw BL perpendicular to OA.
Let the angle between triangle a and b i.e., AÔB=θ
From right angled triangle OBL,
BL
Sin θ =
OB
∴BL = OB Sinθ = b Sinθ
Now, axb = a b Sinθ
= OA .OB Sinθ
= OA .BL
= (base )x (height )
=Area of the Parallelogram OACB
∴If a and b are the adjacent sides of a parallelogram , the area
of the parallelogram= axb .
26
9.
If a and b are two sides of a triangle,
1 / /
area of the triangle = axb
2
10.
If d1 and d2 are the diagonal vectors of a parallelogram,
1
d1 x d2
2
11. If three points A,B and C are collinear
area of the parallelogram =
ABxBC = BCx AC = ACx AB = 0.
12. If OA , OB, OC are the position vectors of the vertices of a triangle
ABC,
1
1
1
area of the triangle ABC = ABXBC = BCX AC = ACX AB
2
2
2
2.1 WORKED EXAMPLES
PART A
1.
/
/
/
If a = 2 i + 3 j − k,b = j − 2k, find a x b
Solution:
i j k
/
/
axb = 2 3 − 1 = i (− 6 + 1) − j (− 4 + 0) + k (2 − 0)
0 1 −2
/
/
= −5 i + 4 j + 2k
2.
(
) (
) (
Prove that a x b x a − b = 2 b − a
Solution:
( )( )
= (a × a ) − (a × b) + (b × a ) − (b × b)
= o + (b × a ) + (b × a ) − o
= 2(b × a )
L .H .S = a x b x a − b
= R.H. S
27
)
3.
( ) ( ) ( )
/
Prove that a x b + c + b x c + a + c x a + b = o
Solution:
( ) ( ) ( )
L .H .S = ax b + c + bx c + a + cx a + b
= a ×b + a ×c + b×c + b×a + c×a +
= a ×b + a × c + b× c − a ×b − a × c − b× c
=0
/
/
/
/
Prove that i − 2 j + 4k and 3 i − 6 j + 12k are parallel vectors.
/
/
Solution: Let a = i − 2 j + 4k
/
/
b = 3 i − 6 j + 12k
4.
Now
/
i
/
j
k
axb = 1
−2
4
3
− 6 12
/
/
i
j k
=31 −2 4
1 −2 4
()
=30 =0
R3
3
R 2 ≡R 3
∴ The given vectors are parallel.
28
PART B
1.
( ) + (a.b)
Prove that a × b
2
2
= a
2
2
b .
Solution
(axb) + (a.b)
2
2
= ª a b Sinθ nº 2 + ª a b Cosθ º 2
«¬
»¼ «¬
¼»
= a 2 b 2 Sin2 θ(n)2 + a b Cos 2 θ
2
2
2
2
2
[
= a b Sin2 θ + Cos 2 θ
=a b
2
=a b
2
]
.1
2
= R.H.S
2
/ /
/
/
Find the unit vector perpendicular to 2 i − j + k and 3 i + 4 j − k
find also the sine of the angle between these vectors.
Solution
/ /
Let a = 2 i − j + k
/
/
b = 3i + 4 j − k
/
/
i
j
k
a×b = 2 −1
1
3
4 −1
/
/
= i (1 − 4) − j (− 2 − 3) + k (8 + 3)
/
/
/
= −3 i + 5 j + 11 k
29
a×b =
(− 3)2 +5 2 +112
= 9 + 25 + 121 = 155
a = 2 (− 1)2 +12 = 4 + 1 + 1 = 6
2
b = 3 2 +42 +(− 1)2 = 9 + 16 + 1 = 26
/
/
/ a × b − 3 i + 5 j + 11 k
The unit vector ⊥ to both a and b =
=
155
a×b
r
Sinθ =
a×b
ab
3.
=
155
6 . 26
=
155
156
Find the area of the parallelogram whose adjacent sides are
/ /
/
i + j + k and 3 i − k
Solution
Let the adjacent sides of the parallelogram be
/ /
/
a = i + j + k and b = 3 i − k
/
i
/
j
k
axb = 1
1
1
3 0 1
/
/
= i (− 1 − 0) − j (− 1 − 3) + k (0 − 3)
/
/
= − i + 4 j − 3k
Area of the Parallelogram= a × b
=
(− 1)2 +42 +(− 3)2
= 1 + 16 + 9 = 26 sq.Units
30
4.
Find the area of the triangle whose vertices are having the
/
/
/
/
/
/
position vectors 2 i + 3 j + 4k , 3 i + 4 j + 2k and 4 i + 2 j + 3k .
Solution:
The position vectors of the vertices of the
/
/
triangle be OA = 2 i + 3 j + 4k
/
/
OB = 3 i + 4 j + 2k
/
/
OC = 4 i + 2 j + 3k
/
/
/
/
AB = OB − OA = 3 i + 4 j + 2k − 2 i + 3 j + 4k
/ /
/
= i + j − 2k
/
/
/
/
BC = OC − OB = 4 i + 2 j + 3k − 3 i + 4 j + 2k
/
/ /
= i −2j +k
/ /
i j k
/
ABxBC = 1 1 − 2 = i (1 − 4) − J(1 + 2) + k (− 2 − 1)
(
)(
)
(
)(
)
1 −2 1
/
= −3 i − 3J − 3k
1
ABX BC
2
1
(− 3)2 +(− 3)2 +(− 3)2
=
2
Area of the triangle ABC =
1
27
9+9+9 =
sq.Units
2
2
/
/
/
/
5.Prove that the position vectors of the points i − 2 j + 3k , 2 i + 3 j − 4k
/
and − 7 j + 10k form collinear points.
Solution:
Let the position vectors of the three points A,B,C be
/
/
OA = i − 2 j + 3k
/
/
OB = 2 i + 3 j − 4k
/
OC = −7 j + 10k
=
31
(
) (
/
/
/
/
AB = OB − OA = 2 i + 3 j − 4k − i − 2 j + 3k
/
/
= i + 5 j − 7K
/
/
/
BC = OC − OB = − 7 j + 10k − 2 i + 3 j − 4k
/
/
= 2 i − 10 j + 14K
/
/
i
j
k
(
ABxBC =
1
5
− 2 − 10
/ /
i
j
= (− 2)1 5
1 5
) (
)
)
−7
14
k
−7
−7
= (− 2)O
[R 2 ≡R 3 ]
=O
∴ The points A,B,C are collinear points
/ /
/ /
6. If i − j − 3k and 2 i − j − 3k are the diagonals of a
parallelogram, find the area of the parallelogram.
Solution:
/ /
/ /
The diagonals of the parallelogram are d1= i − j − 3k,d2 = 2 i − j − 3k
/
/
i
j
k
d1 x d2 = 1 − 1 − 3
2 −1 − 3
/
/
= i (3 − 3 ) − j (− 3 + 6 ) + k(− 1 + 2)
/
/
/
= i (0) − 3 j + k = −3 j + k
∴ Area of the parallelogram
1
= d1 x d2
2
=
1
2
(− 3)2 +12
=
1
10
9 +1 =
sq. units
2
2
32
2.2 APPLICATION OF VECTOR PRODUCT OF TWO VECTORS
Moment or Torque of a force about a point
/
Let O be any point and r be the position vector relative to the
/
point O of any point P on the line of action of the force F . The
moment of the force about the point O is defined as
/
M = r ×F
The magnitude of the moment
M = r×F
= r F Sinθ
Scalar Triple product of the three vectors:
(
)
Scalar triple product of three vectors is defined as a. b × c It is
/ / /
denoted as a • b × c = a,b, c . It is called box product of three vectors
( ) [
]
a, b, and c .
Let a = OA , b = OB, c = OC be the three vectors not lying in the
same plane and meeting at the point O.
33
Complete the parallelopiped with a, b, c as the adjacent edges
with a common intersection. Let AL be the height of the parallelepiped.
b× c is the perpendicular vector of the parallelogram OBDC ∴ b × c
is area of the parallelogram OBDC.
( )
Let θ be the angle between a and b × c .
AL
AL
Cosθ =
=
OA
a
∴ AL = a cos θ
( )
∴ a. b × c = a b × c Cosθ
From the right triangle AOL
= a Cosθ b × c
= AL b × c
=(Height of the parallelepiped) × (Area of the base parallelopiped)
(
)
∴ a. b × c =Volume of the parallelopiped.
Properties of scalar Triple Product
1.
If a, b, c are the three vectors of a parallelopiped (three edges
meeting at a common point),
34
( )
= b.(c × a )
= c.(a × b)
Volume of the parallelopiped, V = a. b × c
2.
…(1)
Scalar product is commutative,
( ) ( )
b • (c × a ) = (c × a ) • b
c • (a × b ) = (a × b ) • c
∴ From (1) and (2) a • ( b × c ) = ( a × b )• c
a • b × c = b × c •a
…(2)
∴ In scalar triple product , • and x may be interchanged.
3.
In a box product of three vectors, if any two of the vectors are
equal or parallel to each other, then the scalar triple product is
zero.
/ //
/ / /
/ / /
i.e, a,a,b = [a,c,c] = b,c,b = 0
/ / /
If a is parallel to b, then a, b, c = 0
[
4.
5.
6.
]
[
]
[
]
In the box product of three vectors for each interchange of two
vectors, the sign will change.
/// / /
/ /
/ /
i , j,k = i . j x k = j. kx i = k. i x j
// //
= i i = j. j = k.k
[
] ( ) ( ) ( )
= 1.
/
/
If a =a1 i +a 2 j +a 3 k
/
/
b =b1 i +b2 j +b3 k
/
/
c =c1 i +c 2 j +c3 k
[a, b, c ] = a • (b x c ) = b
a1 a 2
1 b2
c1 c 2
35
a3
b3
c3
/ /
i j k
/ b2 b3 / b1b3
b1b2
Proof : bx c = b1b2 b3 = i
−j
+k
c2c3
c1c 3
c1c 2
c1c 2 c 3
( ) (
)
/
/
ª / b2 b3 / b1b3
b1b2 º
∴ a. b × c = a1 i +a 2 j +a3 k . « i
−j
+k
»
c1c3
c1c2 »¼
«¬ c1c3
=a
b2 b3
c 2 c3
− a2
b1b3
c1c 3
+ a3
b1 b2
c1c 2
a1 a 2 a 3
= b1 b2 b3
c1 c 2 c 3
7.
8.
If a, b, c are Coplanar vectors (those vectors in the same plane),
[a,b, c] = 0
If [a, b, c] = 0 then
i. any one of the three vectors is zero or
ii. any two of the three vectors are parallel or
iii. a, b, c are coplanar vectors.
1.
2.2 WORKED EXAMPLES
PART A
// /
Find the value of i , j, k
[
]
Solution:
// / / /
i , j, k = i . j × k
[
] ( )
1 0 0
= 0 1 0
0 0 1
=1
1 0
0 1
−0+0
= 1 (−0) = 1
36
[
/ // / / /
2. Find the value of i + j, j + k, k + i
Solution:
1 1 0
/ / /
/
i + j , j + k, k + i = 0 1 1
[
]
]
1 0 1
3.
= 1(1 − 0 ) − 1(0 − 1) + 0
= 1+ 1 = 2
Find the scalar triple product of the vectors
/
/
/ /
/
i − 3 j + 3k, 2 i + j − k and j + k
Solution:
/
/
Let a = i − 3 j + 3k
/ /
b = 2i + j − k
/
c = j +k
(
)
1 −3 3
a. b × c = 2 1 − 1
0 1
1
= 1 ( 1 + 1) + 3(2 − 0) + 3(2 − 0)
= 2 + 6 + 6 = 14
Find the volume of the parallelepiped whose three edges
/
/
/
/ / / /
meeting at a point are 2 i − 3 j + 4k, i + 2 j − k, 3 i − j + 2k
Solution:
/
/
Let a = 2 i − 3 j + 4k
/
/
b = i + 2j −k
/ /
c = 3 i − j + 2k
4.
2 −3 4
Volume of the parallelepiped V = a,b,c = 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
∴ Volume of the parallelepiped =7 cubic Units
37
PART-B
/
Find the moment of the force 3 i + k acting along the point
/
/
/ /
i + 2 j − k about the point 2 i + j + 2k .
Solution:
/ /
Let F = 3 i + k
/
/
OA = i + 2 j − k
/ /
OB = 2 i − j + 2k
/
r = BA = OA − OB
/
/ /
/ /
= i + 2 j − k − 2 i − j + 2k
/
/
/
= − i + 3 j − 3k
/
Moment M = r × F
/ /
i j k
= −1 3 − 3
3 0 1
/
/
= i (3 + 0) − j (− 1 + 9) + k (0 − 9 )
/
/
= 3 i + − 8 j − 9k
/
/
/
= 3 i − 8 j − 9k
/
/
∴ M = 3 i − 8 j − 9k
1.
(
) (
)
( )
M = 32 +(− 8)2 +(− 9)2
= 9 + 64 + 81 = 154 units
O,A,B,C are points(0,0,0),(1,-2,3),(2,3,4) and (-1,0,2). Find the
volume of the Parallelepiped whose edges are OA, OB, and OC.
Solution:
O is the origin
/
/
∴OA = i − 2 j + 3k
/
/
OB = 2 i + 3 j + 4k
/
OC = − i + 2k
2.
38
OA,OB,OC are the edges.
∴ Volume of the parallelepiped V = [OA ,OB,OC ]
1− 2 3
= 2 34
−1 0 2
3.
= 1(6 − 0) + 2(4 + 4) + 3(0 + 3)
= 6 + 16 + 9 = 31 Cubic Units.
/
/
/
/
/ /
Prove that the vectors 3 i + 2 j − 2k, 5 i − 3 j + 3k, and 5 i − j + k
are coplanar vectors.
Solution:
/
/
Let a = 3 i + 2 j − 2k
/
/
b = 5 i − 3 j + 3k
/ /
c = 5i − j + k
3 2 −2
3
5 −1 1
[a,b, c] = 5 − 3
= 3(− 3 + 3) − 2(5 − 15) − 2(−5 + 15)
= 0 − 2(−10) − 2(10)
= 20 − 20 = 0
[a,b, c] = 0 Ÿ a,b and c are coplanar vectors.
4.
/ /
/
/
/
If the three vectors 2 i − j + k, i + 2 j − 3k, and 3 i + m j + 5k, are
coplanar, find the value of m.
Solution:
/ /
Let a = 2 i − j + k
/
/
b = i + 2 j − 3k
/
/
c = 3 i + m j + 5k
39
[ ]
since a, b and c are coplanar vectors then a, b, c = 0
2 −1 1
i.e, 1 2 − 3 = 0
3 m 5
i.e, 2(10 + 3m) + 1(5 + 9) + 1(m − 6) = 0
i.e, 20 + 6m + 14 + m − 6 = 0
i.e, 7m + 28 = 0
i.e,7M = −28
28
i.e, m = −
= −4.
7
5.
Show that the points whose position vectors are
/
/
/
/
/
/
/
4 i + 5 j + k,− j − k, 3 i + 9 j + 4k and − 4 i + 4 j + 4k lie on the same
plane.
Solution:
Let the position vectors of the four points be
/
/
OA = 4 i + 5 j + k
/
OB = − j − k
/
/
OC = 3 i + 9 j + 4k
/
/
OD = −4 i + 4 j + 4K
AB = OB − OA
/
/
/
= − j −k − 4i + 5 j + k
/
/
= −4 i − 6 j − 2k
(
)(
)
AC = OC − OA
/
/
/
/
= 3 i + 9 j + 4k − 4 i + 5 j + k
/
/
= − i + 4 j + 3k
(
)(
)
40
AD = OD − OA
/
/
/
/
= − 4 i + 4 j + 4k − 4 i + 5 j + k
/ /
= −8 i − j + 3k
)(
(
[
]
)
−4 −6−2
Now, AB, AC, AD = − 1
4
3
− 8 −1
3
= −4(12 + 3) + 6(− 3 + 24) − 2(1 + 32)
= −4(15 ) + 6(21) − 2(33 )
= −60 + 126 − 66
=0
∴ AB, AC, AD are coplanar vectors.
∴ The given four points A,B,C and D lie on a same plane.
6.
[
] [ ]
Prove that a + b, b + c, c + a = 2 a, b, c
Solutions:
[
]
= (a + b)• [( b + c )× ( c + a )]
= (a + b)• [(b × c ) + (b × a ) + (cx c ) + (c × a )]
/
= (a + b)• [(b × c ) + (b × a ) + 0 + (c × a )]
= a • (b × c ) + a • (bx a ) + a • (cx a ) + b • (bx c ) + b • (bx a ) + b • (cx 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[b, c,a ]
[∴[b,c,a] = [a,b,c]
= [a,b,c] + [a,b, c]
= 2[a,b, c].
LHS = a + b, b + c, c + a
41
2.3 PRODUCT OF MORE VECTORS
Vector Triple Product: Vectors Triple product of three vectors a, b
(
and c is a x b x c
) or ( a x b) x c
Results
1.
2.
3.
(a x b)x c = (a .c)b − (b.c) a
a x ( b x c )= ( a . c )b − (a . b) c
a x (b x c ) ≠ ( a x b) x c
Product of four vectors
Scalar Product of four vectors:
If a, b, c,d are four vectors, the scalar product of four vectors is
(a x b)• (c x d)
( )( )
Here, axb is a vector, cx d is a vector and ax b • cx d . is the dot
product of two vectors and hence it is a scalar quantity.
( )( )
Result: ax b • cx d . =
a.c a.d
b.c b.d
( )( ) ( )
[x = cxd]
= a • (b x x )
= a • [bx (cx d)]
= a • [(b.d)c − (b.c )d]
= (a.c )(b.d) − (a.d)(b.c )
Proof: ax b • cx d = ax b x
=
a.c
a.d
b.c
b.d
.
42
Vector product of four vectors:
If a, b, c, and d are four vector, axb is a vector, cx d is a vector
( )( )
and axb x cx d is a vector and it is called vector product of four
vectors.
Results:
1.
( )( ) [ ] [ ]
If a, b, c,d are four vectors, a × b × c × d = a, b, d c − a, b, c d
Proof:
Let axb = x
( )( )
= x x (c × d)
= (x • d)c − (x • c )d
= { (a × b)• d} c − { (a × b)• c} d
= [a,b, d] c − [a,b,c] d
L.H.S = ax b × c × d
R.H.S
2.
If a, b, c and d are Coplanar vectors, then axb and cx d are
parallel vector perpendicular to the plane of these four vectors.
( )( )
∴ a×b × c×d = o
2.3 WORKED EXAMPLES
PART – A
1.
( )
/ /
Find the value of i × j × k
Solution:
/ /
/ /
ix j ×k = i × i
( )
=o
43
2.
( )
/
/
. If a = i + j, b = j + k, c = k + i find ax bx c .
/
i j k
bx c = 0 1 1
1 0 1
/
/
= i (1 − 0) − j (0 − 1) + k(0 − 1)
/ /
= i + j −k
(
)
/ /
i j
/
k
a x bxc = 1 1 0
1 1 −1
/
/
= i (− 1 − 0 ) − j (− 1 − 0) + k(1 − 1)
/ /
= −i + j
PART B
/ /
/
/ /
If a = i − j + k, b = i − 2 j, c = 2 i − j + k, prove that
/ / /
a × b × c = a .c b − a .b c
Solution:
/ /
a = i − j +k
/
/
b = i − 2j
/ /
c = 2i − j + k
/
i j k
1.
(
) ( ) ( )
bx c = 1 − 2 0
2 −1 1
/
= i (− 2 − 0) − j(1 − 0) + k (− 1 + 4)
/
= −2 i − j + 3k
44
/
i
ax(bx c) =
j k
1 −1
1
− 2−1 3
/
= i (− 3 + 1) − j(3 + 2) + k (− 1 − 2)
(i.e ) a x (bx c) = −2 i − 5 j − 3k
/
(
)(
(
)(
…(1)
)
/ /
/
a.c = i − j + k . 2 i − j + k
= 1(2) + (−1)(−1) + 1(1)
= 2 + 1+ 1 = 4
/
/
a.b = i − j + k . i − 2 j
= 1(1) + (− 1)(− 2) + 1(0)
= 1+ 2 = 3
)
( ) ( )
( ) (
∴ a.c b − a.b c
/
/
= 4 i − 2j − 3 2 i − j + k
)
= 4i − 8 j − 6i + 3 j − 3k
/
= −2 i − 5 j − 3k
…(2)
( ) ( ) ( )
From (1) & (2) , ax bx c = a.c a − a.c c
/
/
/
/
2. If a = i − j + k, b = − i + 2 j − k,c = i + 2 j, d = i − j − 3k
/
Find a × b • c × d
Solution:
/
a = i − j+k
/
b = − i + 2j − k
/
c = i + 2j
/ /
d = i − j − 3k
( )( )
45
/
i j k
axb = 1 − 1 1
−1 2 −1
/
= i (1 − 2) − j(− 1 + 1) + k(2 − 1)
/
= −i + k
/
i j k
cx d = 1
2 0
1 −1 − 3
/
= i (− 6 − 0) − j(− 3 − 0) + k(− 1 − 2)
/
= −6 i + 3 j − 3k
/
/
axb • cx d = − i + k • − 6 i + 3 j − 3k
= (− 1)(− 6) + 0(3) + 1(− 3 )
= 6−3 = 3
Alternate Method
/
/
a.c = i − j + k • i + 2 j
= 1(1) + (− 1)2 + 1(0)
= 1 − 2 = −1
/
/
a.d = i − j + k • i − j − 3k
= 1(1) + (− 1)(− 1) + 1(− 3)
= 1 + 1 − 3 = −1
/
/
b.c = − i + 2 j − k • i + 2 j
= (− 1)1 + 2 (2) + (− 1)0
= −1 + 4 = 3
/
·
§/
b.d = − i + 2 j − k • ¨ i − j − 3k ¸
©
¹
= (− 1)1 + 2(− 1) + (− 3)
= −1 − 2 + 3 = 0
( )( ) (
)(
(
)(
(
)(
(
)(
(
)
)
)
)
)
46
…(1)
−1
=0+3=3
0
b.c b.d
/
/
/
3.
If a = i + j + k, b = i − j − k, c = − i + j + 2k,
/
d = 2 i + j, find axb x cx d .
Solution:
/
a = i + j + k,
/
b = i − j − k,
/
c = − i + j + 2k,
/
d = 2i + j
G K
i j k
axb = 1 1 1
1− 1 − 1
G
K
= i (− 1 + 1) − j (− 1 − 1) + k (− 1 − 1)
G
K
= i (0) − j (− 2) + k(− 2)
K
= 2 j − 2k
G
K
i
j k
cx d = − 1 1 2
2 1 0
G
K
= i (0 − 2) − j (0 − 4) + k(− 1 − 2)
G
K
= −2 i + 4 j − 3k
G
K
i
j
k
∴ axb x cx d = 0
2 −2
−2 4 −3
G
K
= i (− 6 + 8) − j (0 − 4) + k (0 + 4)
G
K
= 2 i + 4 j + 4k
∴
a.c
a.d
=
−1
3
( )( )
( )( )
47
Alternate Method
[a,b,d] =
1 1 1
1 − 1 −1
2 1 0
= 1(0 + 1) − 1(0 + 2) + 1(1 + 2)
= 1(1) − 2 + (3)
= 1− 2 + 3 = 2
[ ]
1 1 1
a,b,c = 1 − 1 − 1
−1 1 2
= 1(− 2 + 1) − 1(2 − 1) + 1(1 − 1)
= −1 − 1 = −2
[ ] [ ]
) (
(
)
[
][ ]
∴ a,b, d c − a,b,c d
G K
G K
= 2 − i + j + 2k − (− 2) 2 i + j
G
G
G
K
K
K
= −2 i + 2 j + 4k + 4 i + 2 j = 2 i + 4 j + 4k
3.
Prove that axb, bx c, cx a = a, b, c 2
Solution:
(bxc)x (cxa) = [b,c,a ]c − [b,c,c ]a
= [ b,c, a ]c − 0
[ [b,c,c] = 0
= [b,c,a ]c
[axb,bxc,cxa ] = (axb)• {(bxc)x(cxa)}
= (axb)• [b,c, a ]c
= [ b,c, a ](axb).c
= [b, c,a ][a,b,c ]
= [a,b, c ][a,b,c ]
[ [a,b,c] = [b,c,a ]
=[a,b,c ]
2
48
EXERCISE
PART - A
1.
2.
Find the vector product of two vectors.
G
G
K
K
(i) 2 i + 3 j − 4k and i − 2 j + 4k
G K
G K
(ii) i + j + k and 2 i − j + 3k
G
G
K
K
(iii) 2 i − 3 j + 5k and i − 2 j − 2k
G K
G
K
(iv) 3 i + j + k and 2 i − 3 j + 2k
G
G
K
K
Prove that the two vectors 5 i − 7 j + 3k and 15 i − 21 j + 9k are
3.
parallel vectors.
G
K
K
If a = 2i + j − k and b = i − 2 j + 2k Prove that axb. = 5 2
4.
Find the area of the Parallelogram whose adjacent sides are
G
G
K
K
(i) 2 i − 3 j and i − 2 j − 3k
G
G
K
K
(ii) 3 i + 2 j + 2k and i − 2 j + 3k
G K
G
K
(iii) 2 i + j − 2k and i + 2 j + 3k
5.
Find the Scalar triple product of
G
G K
K
K
(i) i − 2 j + 3k , 2 i + j − k, j + k
G
G
G
K
K
(ii) 2 i + 5 j + k , i − 2k, 5 i + 2 j − k
G G
G
G
K
K
(iii) i − j + k , 2 i + 3 j − 3k, 6 i − 2 j − k
6.
Find the Volume of parallelepiped whose edges are
G
G K
G
K
K
(i) 2 i − 4 j + 5k , i − j + k, 3 i − 5 j + 2k
G
G
G
K
K
K
(ii) 2 i + 3 j + 4k , 4 i + 3 j + k, i + 2 j + 4k
G
G
G
K
K
K
(iii) 3 i + 7 j + 2k , 2 i + 5 j − k, i + 6 j + k
49
7.
Prove the following three vectors are coplanar
G
G
G
K
K
K
(i) − i + 4 j − 3k , 3 i + 2 j + 5k, − 3 i + 8 j − 5k
G
G
G K
K
K
(ii) 3 i + 2 j − 2k , 5 i − 3 j + 3k, 5 i − j + k
G
G
K
K
K
(iii) i − 2 j + 3k ,− 2 i + 3 j − 4k, − j + 2k
8.
( )
G K
G K
G
K
If a = i − j, b = − + 2 i + j + k, c = i + 3 j + k, find ax bx c
PART B
1.
Find the Unit vector perpendicular to the following two vectors as
well as the angle between them.
G
G K
K
(i) 4 i + 3 j + k
and
2 i − j + 2k
G K
G
K
(ii) − i + j + 2k and − 4 i + 3 j − 2k
G K
G
K
and
i + 2j + k
(iii) 2 i + j + k
G K
G
K
i + 2j + k
(iv) 3 i + j + 2k and
2.
Show that the following points are Collinear
G K
G
G
K
K
(i) 2 i + j − k,4 i + 3 j − 5k ,2 i + 5 j − 9k
G
G
G
K
K
K
(ii) i + 2 j + 4k,4 i + 8 j + k ,3 i + 6 j + 2k
G K
G
G
K
K
(iii) 2 i − j + 3k,3 i − 5 j + k ,− i + 11 j + 9k
3.
Find the area of the triangle whose vertices are given as position
vectors:
G
G
G K
K
(i) i + 2 j − k,2 i + 3k ,3 i − j + 2k
G K
G
G
K
(ii) i + j + k,2 i + 3 j − k,3 i + k
(iii) (3,1,2),(1,-1,-3) and (4,-3,1)
(iv) (1,3,4),(-2,1,-1) and (0,-3,2)
50
4.
5.
6.
G K
Find the moment of the force 6 i + j − k acting along the point
(0,1,-1) about the point (4,3,-1).
G
G K
The force i + 2 + 3k is acting along the point i + j + k Find the
/ /
moment of the force about the point 2 i − j + k
G K
G K
A force 3 i + j + 2k is acting along i − j + 2k Find the moment of
the force about the point
G K
G
G K
G K
K
a = 3 i + j − k, b = − i + 2 j − 3k, c = − i − j + 3k, d = i + j − k,
7.
Prove that the following four points line in a same plane
G
G
G
G
G
G
K
K
(i) − 6 i + 3 i + 2k, 3 i − 2 j + 4 k, 5 i + 7 j + 3k, − 13 i + 17 i − k
(ii) (4,5,1),(0,-1,-1),(3,9,4),(-4,4,4)
(iii) (1,3,1),(1,1,-1),(-1,1,1),(2,2,-1)
(iv) (1,2,3),(3,-1,2),(-2,3,1),(6,-4,2)
G
G
G K
K
K
G
8. If a = 3 i + 2 j − 4k, b = 5 i − 3 j + 6k, c = 5 i − j + 2k, find (a × b) × c
/ /
G K
G
G
K
9. If a = i + j + k, b = 3 i − 2 j + k, c = 2 i − j − 4k find a × b × c
G
G
G
K
K
10. If a = i − 2 j + 3k, b = 2 i + 3 j, c = 3 i − k find (i) a.(b × c)
(
)
( )
(ii) a × b × c
G
G
G K
K
K
11. If a = 2 i + 3 j − k, b = j + k, c = i + k, d = i + j + k, find
(i) (a × b).(c × d)
G K
G
G
G K
G K
12. If a = i − j + k, b = 2 i + 3 i − 5k, c = 2 i + j − 2k, d = 3 i − j + 4k,
show that (a × b).(c × d) =
a.c a.d
b.c b.d
G K
G K
G
G K
K
13. If a = 2 i − j + k, b = − i − j − k, c = 2 i + 3 j − k, d = i + j − k, find
(a × b) × (c × d) .
51
14. If
G K
G
G K
G K
K
a = 3 i + j − k, b = − i + 2 j − 3k, c = − i − j + 3k, d = i + j − k, find
(a × b) × (c × d)
G
G
G K
K
K
K
15. If a = 3 i + 2 j − 4k, b = 5 i − 3 j + 6k, c = 5 i − j + 2k, d = 3 j − 4k,
find (a × b) × (c × d) .
16. Prove that a × (b × c) + b × (c × a ) + c × (a × b) = 0
1.
ANSWERS
PART - A
G
G K
K
(i) 4 i − 12 j − 7k (ii) 4 i − j − 3k
G
G
K
K
(iii) 16 i + 9 j − k (iv) 5 i − 4 j − 11k
4.
(i)
5.
(i) 12
(ii) -35
(iii) -15
6.
(i) 8
G
K
− 5 i − 5 j − 3k
(ii) 5
(iii) 26
8.
166
(ii)
69
(iii)
PART - B
1.
(i)
(ii)
G
K
7 i − 6 j − 10k
185
G
K
− 8 i − 10 j + k
, sin θ =
185
, sin θ =
165
234
3 165
174
G K
i − j +k
3
, sin θ =
(iii)
2
3
G K
i − j −k
3
, sin θ =
(iv)
3
84
52
90
4.
62
,
(ii)
2
G
K
2 i − 4 j + 8k, 84
5.
G
K
6 i + 3 j − 4k, 61
6.
G K
i − j − k, 3
8.
G
K
− 95 i − 95 j + 190k
9.
G
K
− 13 i + 8 j + 5k
10.
/
− 34,12 i − 4k
3.
(i)
62
2
(iii)
11. - 2
G
K
13. 2 i + 8 j + 4k
14.
G
K
− 14 i − 14 j + 18k
15.
G
K
− 190 i + 38 j − 76k
53
473
2
(iv)
933
2