Intersections - Vector Analysis for Computer Graphics - page 160

Therefore,

k = t − p 1 = 4 j − 2 j = 2 j

p 21 = p 2 − p 1 = 4 k − 2 j =− 2 j + 4 k

p 31 = p 3 − p 1 = 3 i + j + 2 k − 2 j = 3 i − j + 2 k

and using Eq. (6.30), we obtain

003
1 DET = − 2 − 1 = 6

− 142

003
2

1
6

= 1

042

− 2 − = 4

00 3
12

1
6

r = − 1 = 1

− 10 2

000
1

1
6

s = − 22 = 0

− 140

r = 1 confirms that the ray intersects the triangle at P 2 . Similarly, with = 4, the ray intersects

the point p 4 k , which is P 2 .

Using Eq. (6.31), we find that

= 4 j + 4 − j + k =

i j k
020
0

m = = 8 i

− 24

i j k

n = 0 − 11 =− i + 3 j + 3 k

3 − 12

n · p 21 = − i + 3 j + 3 k · − 2 j + 4 k = 6

1
6 8 i

= · 3 i − j + 2 k = 4

1
6

r = − i + 3 j + 3 k · 2 j = 1

1
6 8 i

s = · − j + k = 0

which are identical to the above results.