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or
p
=
cos
(θ/
2
)
sin
(θ/
2
)
B
p
cos
(θ/
2
)
sin
(θ/
2
)
B
.
−
+
(12.11)
Let's demonstrate how (
12.11
) works with two examples.
Example 1
Figure
12.6
shows a scenario where vector
p
is to be rotated 90° about
e
2
which is perpendicular to
B
, where
B
θ
=
90°
,
a
=
e
2
,
p
=
e
1
+
e
2
,
=
e
31
.
Therefore,
p
=
cos 45°
sin 45°
e
31
(
e
1
+
e
2
)
cos 45°
sin 45°
e
31
−
+
√
2
2
−
√
2
2
e
2
)
√
2
√
2
2
e
31
(
e
1
+
e
31
=
2
+
√
2
2
√
2
2
√
2
2
√
2
2
e
312
√
2
√
2
2
e
31
=
e
1
+
e
2
−
e
3
−
2
+
1
2
(
e
1
−
=
e
3
+
e
2
+
e
231
−
e
3
−
e
1
−
e
312
−
e
31231
)
=
e
2
−
e
3
which is correct.
Observe what happens when the bivector's sign is reversed to
−
e
31
:
p
=
cos 45°
sin 45°
e
31
(
e
1
+
e
2
)
cos 45°
sin 45°
e
31
+
−
1
2
(
1
=
+
e
31
)(
e
1
+
−
e
2
)(
1
e
31
)
1
2
(
e
1
+
=
e
2
+
e
3
+
e
312
)(
1
−
e
31
)
Fig. 12.6
Rotating a vector
by 90°
