Graphics Reference
In-Depth Information
Ta b l e 6 . 1
Hamilton's
quaternion product rules
i
j
k
i
−
1
k
−
j
j
−
k
−
1
i
k
j
−
i
−
1
6.16 The Relationship Between the Vector Product and the
Outer Product
We have already discovered that there is a very close relationship between the vector
product and the outer product, and we will see what happens when we form the cross
and wedge product of two 3D vectors:
a
=
a
1
e
1
+
a
2
e
2
+
a
3
e
3
b
=
b
1
e
1
+
b
2
e
2
+
b
3
e
3
a
×
b
=
(a
2
b
3
−
a
3
b
2
)
e
1
+
(a
3
b
1
−
a
1
b
3
)
e
2
+
(a
1
b
2
−
a
2
b
1
)
e
3
(6.16)
∧
=
(a
2
b
3
−
a
3
b
2
)
e
2
∧
e
3
+
(a
3
b
1
−
a
1
b
3
)
e
3
∧
e
1
+
(a
1
b
2
−
a
2
b
1
)
e
1
∧
e
2
a
b
=
(a
2
b
3
−
a
3
b
2
)
e
23
+
(a
3
b
1
−
a
1
b
3
)
e
31
+
(a
1
b
2
−
a
2
b
1
)
e
12
.
(6.17)
Multiplying (
6.17
)by
I
123
we obtain
I
123
(
a
∧
b
)
=
(a
2
b
3
−
a
3
b
2
)
e
123
e
23
+
(a
3
b
1
−
a
1
b
3
)
e
123
e
31
+
(a
1
b
2
−
a
2
b
1
)
e
123
e
12
a
2
b
1
)
e
3
)
which is identical to the cross product (
6.9
) apart from its sign. Therefore, we can
state:
=−
((a
2
b
3
−
a
3
b
2
)
e
1
+
(a
3
b
1
−
a
1
b
3
)
e
2
+
(a
1
b
2
−
a
×
b
=−
I
123
(
a
∧
b
).
6.17 The Relationship Between Quaternions and Bivectors
Hamilton's rules for the imaginaries
i
,
j
and
k
are shown in Table
6.1
, whilst Ta-
ble
6.2
shows the rules for 3D bivector products. Although there is some agreement
between the table entries, there is a sign reversal in some of them. However, if we
switch to a left-handed axial system the bivectors become
e
32
,
e
13
,
e
21
and their
products are as shown in Table
6.3
. If we now create a one-to-one correspondence
(isomorphism) between the two systems:
↔
↔
↔
e
21
there is a true correspondence between quaternions and a left-handed set of bivec-
tors.
i
e
32
,j
e
13
,k
