Computer Aided Engineering Design

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