So A = (x1,y1,z1), B = (x2,y2,z2), and C = (x2,y1,z2), where <CAB = 45Â° (or <ABC = 45Â°)? And y1 < y2.

This actually makes it easy, because everything lies in a plane perpendicular to the x-z plane.

On the other hand, if you are trying to find point A, there are actually an infinite number of solutions (all on a circle).

Let R = (y2-y1) = |BC|. We have a 45-45-90 triangle, so |AC| = |BC| = R. It is also given that the y component of A is the same as the y component of C. However, we can choose x1 and z1 any way we want, so long as |AC| = R.

So x1 = x2 + R cos(t), z1 = z2 + R sin(t),

and we can choose t to be any angle we want.