Graphics Reference
In-Depth Information
R
90
°
=
−
cos 45°
sin 45°
e
12
1
−
e
12
√
2
=
R
†
90
°
=
cos 45°
+
sin 45°
e
12
1
+
e
12
√
2
=
.
So let's confirm these using (
12.14
) and (
12.15
):
1
+
e
2
e
1
R
90
°
=
√
2
(
1
+
e
1
·
e
2
)
1
−
e
12
√
2
=
1
+
e
1
e
2
R
†
90
°
=
√
2
(
1
+
e
1
·
e
2
)
1
+
e
12
√
2
=
which confirm our predictions.
In a previous example above, we used
sin 60°
B
R
120
°
=
cos 60°
−
R
†
sin 60°
B
120
°
=
cos 60°
+
1
√
3
(
e
12
+
B
=
e
23
+
e
31
)
to rotate
e
2
+
e
3
.
Let's use (
12.14
) and (
12.15
) to invert the process, but remember that we are
dealing with unit vectors, which means that we have to normalise
a
and
b
:
e
3
to
e
1
+
1
√
2
(
e
2
+
a
ˆ
=
e
3
)
1
√
2
(
e
1
+
b
=
e
3
).
Furthermore, although the bivector
B
formed the plane of rotation in the previous
example, this time, the plane of rotation is
∧
b
. Therefore,
a
ˆ
+
b
1
a
ˆ
R
120
°
=
2
(
1
·
b
)
+ ˆ
a
1
1
1
+
√
2
(
e
1
+
e
3
)
√
2
(
e
2
+
e
3
)
2
(
1
=
1
1
+
√
2
(
e
2
+
e
3
)
·
√
2
(
e
1
+
e
3
))