Gooday, lads! Recently I've been trying to find out how to find a point of intersection between two n-dimensional planes ('a' instead of 'the', as 3D intersections use lines and thus infinite points). I've devised a way from scratch (like a big boy) and want ya'll to critique it and help me improve upon it and iterate. I've also drawn up a rough diagram, who's variables and client maths I'll explain shortly. The constants here are p, n, p', and n'. p is an arbitrary point on the first plane, and n is it's surface normal, p' and n' are similarly that but of the second plane. The goal is to find I, which is an intersection between the first and second plane, or proof that it doesn't exist, or could be expressed by any point on either plane (making it worthlessly broad). The derived values here are j, which is a version of p, projected onto the second plane; d, which is the magnitude of j's projection; t, which is a further projected version of j back onto the first plane; i, which is the magnitude of t's projection; and finally t', which is a normal perpendicular to n and who's creation is the sole reason for variables like t. Here's the equation (sorry if my notation's a tad odd), as always, vector multiplication is the dot-product, absolute values of vectors are their lengths (at least to me they've always been): If |nn'| = 1 —› No meaningful I exists. If (p' - p)n < 0 —› Invert n to -n. If (p - p')n' < 0 —› Invert n' to -n'. I = (t' ÷ |t'(-n')|)d + p t' = (t - p) ÷ |t' - p| d = |j - p| t = -ni + j j = -n'(p' - p) + p i = |(j - p)n'| * In 3D you can use "t' × n" to find a normal that when paired with I can be used to properly describe the line that is their intersection. So ya, hoping for critique and for ya'll to help explain to me how I've almost certainly screwed this up, or at least deeply overcomplicated it. PS. If there's a better place to put this, plz get a mod to move it there!