Graphics Reference
In-Depth Information
Y
P
t
λ
v
p
T
X
Z
Figure 6.11.
We use the same analysis as was used for the line-circle intersection.
The line equation is
p
=
t
+
v
where is a scalar and
1.
Using Eq. (6.16), we can state that
v
=
x
P
+
y
P
+
z
P
=
r
2
which is identical to
r
2
p
·
p
=
or
r
2
t
+
v
·
t
+
v
=
Expanding and simplifying gives
2
v
r
2
t
·
t
+
2
t
·
v
+
·
v
=
But as
v
is a unit vector,
v
·
v
=
1. Therefore,
2
2
r
2
+
2
t
·
v
+
t
−
=
0
which is a quadratic in and solved using
±
√
B
2
=
−
B
−
4AC
2A
where
A
=
1
B
=
2
t
·
v
C
=
t
·
t
−
r
2