Ray and plane intersection

It's actually a very simple question. I forgot all about mathematics. Mark it.

Transferred from "computer graphics" Third Edition sun jiaguang p530

 

Assume that the ray is defined as X = DT + E (t> = 0) (9-6-1)

E = (E1, E2, E3) is the ray start point, corresponding to the viewpoint or visible point; D = (D1, D2, D3) is the ray direction, | d | = 1; X = (x, y, z) indicates the radiation point.

 

Assume that the plane equation is AX + by + cz + D = 0 (9-6-2)

(9-6-1) is substituted into (9-6-2) to obtain N * (DT + E) + d = 0, where N is the vector (A, B, C)

That is, (N * D) T + (n * E) + d = 0

Obtain T0 =-(N * E + d)/(n * D) (when N * D is not equal to 0)

When N * D = 0, the rays are parallel to the plane and can be treated as non-intersecting. When N * D is not equal to 0, the intersection is X0 = dt0 + E0