TRANSFORMATIONS AND PROJECTIONS 33
v
v
v
v
vv
vv
1
*
2
*
1
2
11 12 13
21 22 23
31 32 33
11
22
= =
0
0
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
T T
xy
xy
aaa T
aaa
aaa
A
=
+ + +
+ + +
11 1 12 1 21 1 22 1 31 1 32 1
11 2 12 2 21 2 22 2 31 2 32 2
aa aa aa
aa aa aa
xy x y xy
xy x y xy
vvvvvvT
vvvvvv
⎡
⎣
⎢
⎤
⎦
⎥
Thus,
vv* 1 ⋅ = * 2 vv 1 * * 2 cos *θ
= (a 11 v 1 x + a 12 v 1 y)(a 11 v 2 x + a 12 v 2 y) + (a 21 v 1 x + a 22 v 1 y)(a 21 v 2 x + a 22 v 2 y)
+ (a 31 v 1 x + a 32 v 1 y)(a 31 v 2 x + a 32 v 2 y)
= (aaa 112 +^221 + 312 )vvvv 12 xx+ (aa a 122 + 222 + 322 ) 12 yy
+ (a 11 a 12 + a 21 a 22 + a 31 a 32 )(v 1 xv 2 y + v 1 yv 2 x) (2.13)
Also,
vv vvk 1 *× = 2 * 1 * * 2 sin *θ
= (a 11 v 1 x + a 12 v 1 y)(a 21 v 2 x + a 22 v 2 y)k – (a 11 v 1 x + a 12 v 1 y)(a 31 v 2 x + a 32 v 2 y)j
- (a 21 v 1 x + a 22 v 1 y)(a 11 v 2 x + a 12 v 2 y)k + (a 21 v 1 x + a 22 v 1 y)(a 31 v 2 x + a 32 v 2 y)i
+ (a 31 v 1 x + a 32 v 1 y)(a 11 v 2 x + a 12 v 2 y)j – (a 31 v 1 x + a 32 v 1 y)(a 21 v 2 x + a 22 v 2 y)i (2.14)
The angle between the original vectors v 1 and v 2 are given by
|v 1 ||v 2 | cos θ = (v 1 xi + v 1 yj) · (v 2 xi + v 2 yj) = (v 1 xv 2 x + v 1 yv 2 y)
|v 1 ||v 2 |k sin θ = (v 1 xi + v 1 yj)× (v 2 xi + v 2 yj) = (v 1 xv 2 y – v 1 yv 2 x)k (2.15)
For no change in magnitude or angle, vv 1 2 cos θ = | v 1 ||v 2 | cos θ and also vv 1 × * 2 = v 1 ×v 2.
On comparing results, we obtain
(v 1 xv 2 x + v 1 yv 2 y) = (aaa 112 + + ) 212 312 vvvv 12 xx+ (aa a 122 + + ) 222 232 12 yy
+ (aa11 12+ aa21 22+ aa31 32)(vvvv 1 xy 2 + 1 yx 2 ) (2.16)
and
(v 1 xv 2 y – v 1 yv 2 x)k = (a 11 v 1 x + a 12 v 1 y)(a 21 v 2 x + a 22 v 2 y)k – (a 11 v 1 x + a 12 v 1 y)(a 31 v 2 x + a 32 v 2 y)j
- (a 21 v 1 x + a 22 v 1 y)(a 11 v 2 x + a 12 v 2 y)k + (a 21 v 1 x + a 22 v 1 y)(a 31 v 2 x + a 32 v 2 y)i
+ (a 31 v 1 x + a 32 v 1 y)(a 11 v 2 x + a 12 v 2 y)j – (a 31 v 1 x + a 32 v 1 y)(a 21 v 2 x + a 22 v 2 y)i (2.17)
We can work out Eq. (2.17) and compare the coefficients of i,j and kand further, compare the terms
corresponding to v 1 yv 2 x and v 1 xv 2 y. Finally, after comparison from Eqs. (2.16) and (2.17), we would
get