   3D segment intersection point
  Hello people !I have these two 3D line segments both defined by a start and end point. What I would like to do is find a point on segment A which if segment B was adjusted along the Y-axis, the two would then intersect. In other words I don't care if they actually have an intersection point in 3D space as long as there is one when they are projected onto the Y-plane. If this point was then somehow projected back onto one of the segments in 3D space, that's the point I'm interested in.I have a method to find an intersection point for a pair of 2D segments which I figured I could use with these 3D segments simply by ignoring their Y-components. This gives me a resulting 2D-vector if they do, and so I wonder if I can use this information to find out the remaining Y-component for one of these segments (or of course if there is a better way)?Many thanks in advance !- Dave   
  This doesn't really have anything to do with DirectX - did you mean to post it somewhere else? (Maybe Math & Physics?)If I understand your question correctly, given a point on the xz plane, you want to find the 'y' value of the corresponding point on a specified line (by 'corresponding', I mean 'the point that projects orthogonally onto the xz plane at the specified coordinates').We'll start by expressing the line in the parametric form, P(t) = O+t*D, where:O = line_start_pointD = line_end_point - line_start_pointGiven a coordinate, such as x, we can then compute the corresponding value of t:x = Ox+t*Dxt*Dx = x - Oxt = (x - Ox) / DxAnd similarly for y:t = (y - Oy) / DyFor numerical stability, I'd recommend using the former equation if |Dx| > |Dy|, and using the latter equation otherwise. (If the line is more or less parallel to the y axis, the problem either has no answer, or doesn't have a unique answer.)Once you've computed t, you can compute the desired y value as y = Oy + t * Dy.(I think all of the above is right, but no guarantee of correctness.)   ;  Yes, I realized after making the post that it turned out quite generically mathematical so I probably should have someone move it to that section.This solution seems to work nicely, thank you for taking the time to help this clueless fellow :)   