I wanted to write a blog post about an unusual usage of cross products. Namely, that you can take two points and cross them together to get a line, and take two lines and cross product them to get the intersection point  and this works in 3D too! Cross product 3 vectors and you get the plane that it represents, cross product 3 planes and get the point they intersect at!
I learned about this a great many years ago when learning about Computer Vision techniques for Epipolar Geometry and solving for the Essential Matrix. Where in epipolar geometry, a pixel in one view is a line in another view, and given a bunch of corresponding points you can calculate the Essential Matrix with the help of a cross product matrix. This blog post is not about that specifically, but bringing this idea over to computer graphics generally. First in 2D where you have points and lines. Lets say you have 2 points p0(x,y) and p1(x,y). You first make them into 3D points by adding a 1. Not going to explain why you do that right now as that would just confuse the point. Sufficed to say though, p0(x,y,1) and p1(x,y,1). Then you perform a 3D cross product of those points cross3D(p0,p1) and you get... the line equation they represent as the solution. So, the orientation of the line (x,y), and the offset from the origin (z). One thing to note though that it is not normalized. So most of the time you want to take an additional step where you take the length of (x,y) and normalize the line equation such that (x,y) == 1. So... (x,y,z) = (x,y,z)/(x,y). Then, you can find the intersection point of two line equations by doing a 3D cross product of those two lines! So cross3D(line0,line1) = point of intersection. Note: the intersection point is unnormalized, so you have to divide by Z. And of course this also works in 3D, where you have three 3D points. p0(x,y,z), p1(x,y,z), p2(x,y,z). Next up, you guessed it, you add a 1 as the W coordinate. so those points then become p0(x,y,z,1), p1(x,y,z,1), p2(x,y,z,1). Then you cross product those three 4D points together to get the plane equation they represent! planeEquation = cross4D(p0,p1,p2). Again th0ugh, this result is unnormalized, so you probably want to normalize the length via [x,y,z,w]/[x,y,z] And of course if you have 3 plane equations, you can cross product them and get the point they intersect at! point = cross4D(plane0,plane1,plane2). Note: the intersection point is unnormalized, so you have to divide by W. For easy reference, the calculation for a 4D cross product is specified as thus... void cross4D(const float v1[4], const float v2[4], const float v3[4], float result[4]) { result[0] = v1[1]*(v2[2]*v3[3]v3[2]*v2[3])v1[2]*(v2[1]*v3[3]v3[1]*v2[3] )+v1[3]*(v2[1]*v3[2]v3[1]*v2[2]); result[1] = v1[0]*(v3[2]*v2[3]v2[2]*v3[3])v1[2]*(v3[0]*v2[3]v2[0]*v3[3])+v1[3]*(v3[0]*v2[2]v2[0]*v3[2]); result[2] = v1[0]*(v2[1]*v3[3]v3[1]*v2[3])v1[1]*(v2[0]*v3[3]v3[0]*v2[3])+v1[3] *(v2[0]*v3[1]v3[0]*v2[1]); result[3] = v1[0]*(v3[1]*v2[2]v2[1]*v3[2])v1[1]*(v3[0]*v2[2]v2[0]*v3[2])+v1[2]*(v3[0]*v2[1]v2[0]*v3[1]); } and of course for 3D it is... void cross3D(const float v1[4], const float v2[4], float result[4]) { result[0] = v1[1]*v2[2]  v1[2]*v2[1]; result[1] = v1[2]*v2[0]  v1[0]*v2[2]; result[2] = v1[0]*v2[1]  v1[1]*v2[0]; } Its pretty useful or at least I find it useful to know this trick. I bet you will too!
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