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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 ))
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