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
,θ
=
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
.