Graphics Reference
In-Depth Information
p
20
p
22
p
21
p
12
p
10
p
11
u
n
p
02
p
00
v
p
01
Figure 8.5.
The surface at
p
00
is tangential to the vectors
u
and
v
. Therefore, the normal vector
n
is
given by
n
=
u
×
v
where
u
=
p
10
−
p
00
=
−
111
=−
i
+
j
+
k
and
v
=
p
01
−
p
00
=
101
=
i
Therefore,
i
j k
n
=
−
111
101
=
i
+
2
j
−
k
Although one would not expect the magnitude of this vector to equal that of Eq. (8.9), it must
equal a scaled version. In this case the scaling factor is 4:
4
i
+
2
j
−
k
=
4
i
+
8
j
−
4
k
Example 2
Consider next the parametric equations for a sphere:
x
=
sinu cos vy
=
sinu sinvz
=
cos u
or as a vector equation
r
=
x
i
+
y
j
+
z
k
(8.10)
r
=
sinu cos v
i
+
y sinu sinv
j
+
cos u
k
