Graphics Reference
In-Depth Information
Now let's try another ray:
t
=
4
j
v
=
i
−
j
+
k
Using Eq. (6.30), we obtain
k
=
2
j
unchanged
−
103
1
DET
=
−
2
−
1
=
6
−
142
003
2
1
6
=
1
042
−
2
−
=
4
−
10 3
12
1
6
1
3
r
=
−
1
=
−
10 2
−
100
1
1
6
1
1
s
=
−
22
=
3
−
140
Because the value of s>1, the intersection point is outside the triangle.
Let's make sure that we obtain the same result using Eq. (6.31):
=
m
8
i
unchanged
i
j k
n
=
1
−
11
=−
i
+
j
+
2
k
3
−
12
n
·
p
21
=
−
i
+
j
+
2
k
·
−
2
j
+
4
k
=
6
1
6
8
i
=
·
3
i
−
j
+
2
k
=
4
1
6
1
3
r
=
−
i
+
j
+
2
k
·
2
j
=
1
6
8
i
1
1
3
s
=
·
i
+−
j
+
k
=
which confirm the previous values.
Finally, let's select another ray that hopefully intersects the triangle:
t
=
4
j
v
=
2
i
−
4
j
+
2
k
Using Eq. (6.30), we obtain
k
=
2
j
unchanged
