Graphics Reference
In-Depth Information
6.13 The Inverse of a Vector
In traditional vector analysis we accept that it is impossible to divide by a vector, but
that is not so in geometric algebra. In fact, we don't actually divide a multivector by
another vector but find a way of representing the inverse of a vector. For example,
we know that a unit vector
a
is defined as
ˆ
a
a
=
|
|
a
and using the geometric product
a
2
a
2
ˆ
=
2
=
1
|
a
|
therefore,
a
2
b
|
b
=
a
|
2
and exploiting the associative nature of the geometric product we have
a
(
ab
)
|
b
=
.
(6.14)
a
|
2
Equation (
6.14
) is effectively stating that, given the geometric product
ab
we can
recover the vector
b
by pre-multiplying by
a
−
1
:
a
a
−
1
=
2
.
|
a
|
Similarly, we can recover the vector
a
as follows by post-multiplying by
b
−
1
:
(
ab
)
b
|
a
=
.
2
b
|
For example:
=
e
1
+
a
2
e
2
=
3
e
1
+
b
2
e
2
their geometric product is
=
−
4
e
12
.
Therefore, given
ab
and
a
, we can recover
b
as follows:
ab
7
e
1
+
(
7
2
e
2
b
=
−
4
e
12
)
5
1
5
(
7
e
1
−
=
4
e
112
+
14
e
2
−
8
e
212
)
1
5
(
7
e
1
−
=
4
e
2
+
14
e
2
+
8
e
1
)
=
3
e
1
+
2
e
2
.
