Graphics Reference
In-Depth Information
then
×
N
=
×
+
=
×
+
×
=
×
N
N
N
D
N
N
N
D
N
D
(10.11)
If we now substitute Eq. (10.10) in Eq. (10.11), we obtain
N
N
⎛
⎝
⎞
⎠
F
u
P
v
F
v
P
u
×
−
×
×
N
=
×
=
×
N
N
D
N
N
N
N
N
N
F
u
P
v
F
v
P
u
×
×
−
×
×
N
=
N
×
(10.12)
N
Equation (10.12) now contains a triple product, which, as we discovered in Section 2.9.2, has
the following identity:
a
×
b
×
c
=
a
·
c
b
−
a
·
b
c
Expanding Eq. (10.12) produces
N
N
N
N
F
u
P
v
N
P
v
F
v
P
u
N
P
u
·
−
·
−
·
−
·
N
N
N
=
N
×
N
But
2
N
·
N
=
N
and
P
v
=
P
u
=
N
·
0 and
N
·
0
Therefore,
F
u
2
P
v
F
v
2
P
u
F
v
−
N
−
−
N
P
u
−
F
u
P
v
N
=
N
×
=
N
N
which can be expressed as
×
N
=
N
N
A
(10.13)
where
P
v
Because of the cross product in Eq. (10.13),
A
must be perpendicular to the plane containing
N
and
N
. Therefore, one can imagine that
N
has been rotated about
A
to
N
.
Now since
F
v
P
u
−
F
u
A
=
N
N
×
=
N
×
D
