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Here's my problem. Say I have three orthogonal planes (one with a normal vector along the $x$-direction, one along the $y$-direction, and one along the $z$-direction). Initially they intersect at $(x_0,y_0,z_0) = (0,0,0)$. First I rotate these planes by some known (non-orthogonal) rotation matrix, to come up with 3 new planes (all of this is easy, i'm getting to my tricky part). Lets assume that none of the planes are parallel, so we get a single intersection point. I can then shift these planes along their normal vector (which can easily be done by adding some fraction of the normal vector to the plane's point). I can now easily find the new intersection point by using the equation given here: http://mathworld.wolfram.com/Plane-PlaneIntersection.html .

Here's my question, instead of finding the intersection point, I want to instead find the values $a_1, a_2, a_3$ (the coefficients of shifting that translate my three rotated matrices) so that my end point (after i have rotated the matrices), is a given point $(\bar{x},\bar{y},\bar{z})$. Basically it's an inverse problem. Instead of having 3 planes with known shifts and translates and finding the intersection point, i know the rotation, and the intersection point, but i don't know the shift values. Any help would be greatly appreciated to get me going on the right path. Thanks very much!

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  • $\begingroup$ non-orthogonal rotation matrix - I'm not positive what this means. What kind of rotation isn't orthogonal? Maybe you are saying rotations can occur about any point (not just the origin?) $\endgroup$ – rschwieb Sep 19 '13 at 19:23
  • $\begingroup$ ah ok, maybe I didn't word this properly. What I mean is each of the 3 planes are being rotated by their own rotation matrix (which is, itself orthogonal, but all 3 are different). You're right to say non-orthogonal didn't make sense, there's a reason why i wrote that but ignore it cause it's misleading and wrong in my context. $\endgroup$ – Karl Landheer Sep 19 '13 at 20:26
  • $\begingroup$ But the three rotations all take place leaving the origin fixed, right? $\endgroup$ – rschwieb Sep 19 '13 at 23:23
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The rotations are a bit of a red herring. What you’re basically looking for is the distance of the target intersection point from a plane through the origin. This is simply the length of the orthogonal projection of the point onto the plane’s normal, positive if it’s on the same side of the plane as the normal, negative if it’s on the opposite side.

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