Rotating Vectors - Vector Analysis for Computer Graphics - page 176

But from Fig. 7.1(b), we have

r = p sin (7.5)

Therefore Eq. (7.4) equals Eq. (7.5) and

w = r

From Fig. 7.1(c), we know that

QP
NP =

QP QP

= =

sin

r w

Therefore,

− QP =

w sin = n ˆ × p sin

and then

p =

n ˆ · p n ˆ + p − n ˆ · p n cos ˆ + n ˆ × p sin

and

p =

p cos + n ˆ · p 1 − cos n ˆ + n ˆ × p sin

We make the following substitution:

K = 1 − cos

then

p = p cos + n ˆ · p K n ˆ + n ˆ × p sin

and

p = x p i z p k cos + ax p + cz p K a i

+ y p j + by p + + b j + c k

+ bz p − cy p i + cx p − az p j + ay p − bx p k sin

p = x p cos a ax p + cz p K + bz p − cy p sin i

+ by p +

+ y p cos b ax p + cz p K + cx p − az p sin j + by p +

+ z p cos c ax p + cz p K + ay p − bx p sin k

+ by p +

p = x p a 2 K b sin i

+ + − + +

cos y p abK c sin z p acK

+ x p abK

a sin j

+ +

y p b 2 K

+ + −

c sin cos z p bcK

+ x p acK cos k z p c 2 K

− b sin + y p bcK + a sin + +