Graphics Reference
In-Depth Information
Fig. 12.7
Rotating a vector
by 120°
1
2 ( e 1 +
=
e 3 +
e 2 +
e 231 +
e 3
e 1 +
e 312
e 31231 )
p =
e 3
the rotation is clockwise about e 2 .
e 2 +
Example 2 Figure 12.7 shows a scenario where vector p is to be rotated 120° about
the bivector B , where
m
=
e 1
e 3 ,
n
=
e 2
e 3 =
120° ,
p
=
e 2 +
e 3 .
First, we compute the bivector:
B
=
m
n
=
( e 1
e 3 )
( e 2
e 3 )
=
e 12 +
e 23 +
e 31 .
B :
Next, we normalise B to
1
3 ( e 12 +
B
=
e 23 +
e 31 )
and
p = cos 60°
sin 60° B p cos 60°
sin 60° B
+
3
2
3
2
1
2
e 31 ( e 2 +
e 3 ) 1
e 31 )
3 e 12 +
1
1
3 ( e 12 +
=
e 23 +
2 +
e 23 +
1
2
( e 2 +
e 3 ) 1
e 12
2
e 23
2
e 31
2
e 12
2 +
e 23
2 +
e 31
2
=
2 +
1
4 ( e 2 +
=
e 3
e 1
e 123 +
e 3
e 2
e 312 +
e 1 )( 1
+
e 12 +
e 23 +
e 31 )
1
2 ( e 3
=
e 123 )( 1
+
e 12 +
e 23 +
e 31 )
1
2 ( e 3
=
e 2 +
e 1 +
e 3 +
e 1 +
e 2 )
=
e 1 +
e 3 .
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