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Here's my problem. Say I have 3 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 (x0,y0,z0) = (0,0,0). I can shift these planes along their normal by some value. (lets call the values x1, y2 and z3). I then rotate these planes by some known 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 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, x1, y2, z3 (the amounts I can initially translate my three orthogonal matrices) so that my end point (after i have rotated the matrices), is a given point (xbar,ybar,zbar). 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 initial shift values. Any help would be greatly appreciated to get me going on the right path. Thanks very much!

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  • $\begingroup$ Is the rotation the same for all planes? If yes, then it's sufficient to applythe inverse rotation to the know intersection point, you will obtain your shifts. $\endgroup$ – TZakrevskiy Sep 19 '13 at 15:55
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I think it should just be $(x_1, y_2, z_3) = R^{-1}(\overline{x}, \overline{y}, \overline{z})$, right? Because after you translate the planes, you are left with the new intersection point $(x_1, y_2, z_3) $, then $R(x_1, y_2, z_3) = (\overline{x}, \overline{y}, \overline{z})$.

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  • $\begingroup$ Hi, I realized shortly after writing this that was indeed the case, and my problem is actually shifting after the rotation (but this is still just as easy). Thanks very much for your help. $\endgroup$ – Karl Landheer Sep 19 '13 at 16:10
  • $\begingroup$ You're welcome. $\endgroup$ – Eric Auld Sep 19 '13 at 17:15

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