Hey all, having a pretty tough time with this math problem, wondering if anyone here can push me in the right direction. Here is the problem:

You are given a 3D quad Q, where all 4 vertices are known. You are also given a point P where Px and Pz are known, and lie within the bounds of Q. It is also safe to assume that the x and z components of the vertices of Q form a square (as in, if you looked at Q down the Y-axis, you would see a square). Given this information, find Py.

It seems as if the cross product would be useful here, but I'm not sure which vectors to cross to get my answer. Anyone have any ideas?

In short: calculate the plane of the quad. Solve the plane equation for Y.

An attempt at a few more details...

Solving for y: a plane equation is in the form Ax + By + Cz + D = 0, so y = -(Ax + Cz + D) / B.

Plane for the quad: the cross product of two adjacent edges in the quad gives the normal of the plane (A, B, C) and then any point in the quad can be used to solve for D.

[indent=1]N = (Q2 - Q1) x (Q1 - Q0)

[indent=1]A = Nx
[indent=1]B = Ny
[indent=1]C = Nz
[indent=1]D = -(N dot Q0)

(where Q0 through Q3 are the points of the quad. D could be rewritten as D = -(A*x + B*y + C*z) where x,y,z are the components of Q0, if that makes more sense to you.)

so expanding it out a bit:

[indent=1]Py = -(Nx * Px + Nz * Pz - (N dot Q0)) / Ny


I'm sure a google search or two could explain the details better than I have, but hopefully this is enough info to get you going.

Hey all,

Thanks for the help! I was able to figure out the solution using Ray/Plane intersection, which looks similar to what you guys did. Here is the code snipit that performs the primary calculations, for future people:


vec4 S = vec4(Px, 0.0f, Pz, 1.0f); // Ray start position
vec4 V = vec4(0.0f, 1.0f, 0.0f, 0.0f); // Ray directional

vec3 N = glm::normalize(glm::cross(P1 - P0, P2 - P0)); // Normal Vector
vec4 L = vec4(N, glm::dot(-N, P0)); // Plane Vector
float Py = -(glm::dot(L, S) / glm::dot(L, V)); // Calculate final Y value
return Py;


Thanks again for the response.