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So, I have a plane P.

I've got the 3 coordinates of the plane P normal vector n(n1, n2, n3) and I've got the 3 coordinates of a given point A(a1, a2, a3) which belongs to P. So from that, I can easily get the plane equation.

Now, if the 3D projection of A onto the plane gives the coordinates (x, y), what I want is to find the 3D coordinates of a B point (belonging to the plane), assuming that the 3D projection of B onto the plane gives the coordinates (x + x', y + y'). Of course, x' and y' are known.

I'm sure this must not be hard, but I fail to find the solution..

zadamantine
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  • What is a basis for (x,y) coordinates? – MBo Sep 28 '14 at 04:42
  • exactly you need 2 basis vectors of the plane coordinate system your projection uses. This might help to see what you need: http://stackoverflow.com/a/25906665/2521214 (see the second solution) when you have the U,V basis vectors then `P(x,y,z)=Q.u*U+Q,v*V` where P is your 3D point you want and Q is your 2D point. Assuming your projection is linear and orthogonal of coarse – Spektre Oct 03 '14 at 07:54

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