Graphics Reference
In-Depth Information
Similarly, given
ab
and
b
,
a
is recovered as follows:
4
e
12
)
3
e
1
+
2
e
2
=
−
a
(
7
13
1
13
(
21
e
1
+
=
14
e
2
−
12
e
121
−
8
e
122
)
1
13
(
21
e
1
+
=
14
e
2
+
12
e
2
−
8
e
1
)
=
e
1
+
2
e
2
.
Note that the inverse of a unit vector is the original vector:
a
|ˆ
ˆ
a
−
1
ˆ
=
2
= ˆ
a
.
a
|
6.14 The Imaginary Properties of the Outer Product
So far we know that the outer product of two vectors is represented by one or more
unit basis vectors, such as
λ
3
e
31
where, in this case, the
λ
terms represent areas projected onto their respective unit
basis bivectors. But what has not emerged is that the outer product is an imaginary
quantity, which is revealed by expanding
e
12
:
e
12
=
a
∧
b
=
λ
1
e
12
+
λ
2
e
23
+
e
1212
but as
e
21
=−
e
12
then
e
1
(
21
)
2
=−
e
1
(
12
)
2
=−
e
1
e
2
e
12
=−
1
.
Consequently, the geometric product effectively creates a complex number! Thus in
a 2D scenario, given:
a
=
a
1
e
1
+
a
2
e
2
b
=
b
1
e
1
+
b
2
e
2
their geometric product is
ab
=
(a
1
b
1
+
a
2
b
2
)
+
(a
1
b
2
−
a
2
b
1
)
e
12
and knowing that
e
12
=
i
, then we can write
ab
as
ab
=
(a
1
b
1
+
a
2
b
2
)
+
(a
1
b
2
−
a
2
b
1
)i.
(6.15)
