Graphics Reference
In-Depth Information
01
10
10
00
00
01
k
×
j
=
i
+
j
+
k
=−
i
which confirms that
×
=−
×
j
k
k
j
The areas of the parallelograms associated with the above examples all equal unity, which is the
length of the resulting vector.
Perhaps now you can understand how Sir William Rowan Hamilton must have felt when he
discovered the above!
Y
n
=
a
×
b
1
θ
b
a
1
X
1
Z
Figure 2.23.
As a further example, consider the scenario shown in Fig. 2.23. Vectors
a
and
b
are given by
a
=−
j
+
k
and
j
The task is to compute the vectors normal to both
a
and
b
. Therefore,
b
=
i
−
i
j k
a
×
b
=
0
−
11
1
−
10
−
11
10
01
0
−
1
a
×
b
=
i
+
j
+
k
−
10
1
−
1
a
×
b
=
i
+
j
+
k
which are correct. But what about the angle ? Well, this can be computed by Eq. (2.5) or by
rearranging Eq. (2.11). Using Eq. (2.5), we get
cos
−
1
x
a
x
b
+
y
a
y
b
+
z
a
z
b
=
a
b
