# Linear algebra — the intersection of 3 rotated planes

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!

• 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. – TZakrevskiy Sep 19 '13 at 15:55

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 $R(x_1, y_2, z_3) = (\overline{x}, \overline{y}, \overline{z})$.