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1. 1
Classical Mechanics P. No.1
Given the two vectors
& ,
find the following;
(a) &
(b) 3 2
(c) .
(d) &
i j j k
= + = +
+ +
−
 
A B
A B A B
A B
A B
A B A B
Solution
( )
2
1 4 1
6
( )
3 2 3( ) 2( )
3 3 2 2
3 2
a
i j k
b
i j j k
i j j k
i j k
+ = + +
+ = + +
=
− = + − +
= + − −
= + −
A B
A B
A B

2. 2
2
3 4
( )
. 1(0) 1(1) 0(1) 1
( )
1 0
1 1 0 ( 1)
1 1
0 1 1
1 0 1 1
( 1) ( 1)
0 1 0 1
1 1 1 3
c
d
i j k
i
j k
i j k
= + + =
 = = −
+ − + −
= − +
 = + + =
A B
A B
A B
Classical Mechanics P. No.2
Given the three vectors
2 ,
,
4 ,
find the following
(a) .( ) & ( ).
(b) .( ) & ( ).
(c) ( ) & ( )
i j
i k
j
= +
= +
=
+ +
 
   
A
B
C
A B C A B C
A B C A B C
A B C A B C

3. 3
Solution
( )
4
.( ) (2 ).( 4 )
2X1 1X 4 0X1 2 4 6
3
( ). (3i j k).(4 j)
3X 0 1X 4 1X 0
4
a
i j k
i j i j k
i j k
+ = + +
 + = + + +
= + + = + =
+ = + +
 + = + +
= + +
=
B C
A B C
A B
A B C
2
3 4
(b)
0 1
1 0 1 ( 1)
4 0
0 4 0
1 1 1 0
( 1) ( 1)
0 0 0 4
4 0 4 4 4
.( ) (2 ).( 4i 4k)
2X( 4) 1X0 0X 4 8
i j k
i
j k
i j k i k
i j
 = = −
+ − + −
= − − + = − +
 = + − +
= − + + = −
B C
A B C

4. 4
2
3 4
1 0
2 1 0 ( 1)
0 1
1 0 1
2 0 2 1
( 1) ( 1)
1 1 1 0
2
( 2 ).(4 )
1X(0) ( 2)X 4 ( 1)X0
0 8 0 8
i j k
i
j k
i j k
i j k j
 = = −
+ − + −
= − −
 = − −
= + − + −
= − + = −
A B
(A B).C
2
3 4
(c)
0 1
1 0 1 ( 1)
4 0
0 4 0
1 1 1 0
( 1) ( 1)
0 0 0 4
4 0 4 4 4
( ) 2 1 0
4 0 4
(4 0) j(8 0) k(0 ( 4))
4i 8 j 4k
i j k
i
j k
i j k i k
i j k
i
 = = −
+ − + −
= − − + = − +
  =
−
= − − − + − −
= − +
B C
A B C

5. 5
2
3 4
1 0
2 1 0 ( 1)
0 1
1 0 1
2 0 2 1
( 1) ( 1)
1 1 1 0
2
1 2 1
0 4 0
(0 4) j(0) k(4 0)
4i 4k
i j k
i
j k
i j k
i j k
i
 = = −
+ − + −
= − −
  = − −
= + − + −
= +
A B
(A B) C
Analytical Mechanics, Grant R. Fowles, George L. Cassiday