Projecting 3D points to 2D plane [closed]

Question

Let A be a point for which I have the 3D coordinates x, y, z and I want to transform them into 2D coordinates: x, y. The projection shall be orthogonal on a plane defined by a given normal. The trivial case, where the normal is actually one of the axes, it's easy to solve, simply eliminating a coordinate, but how about the other cases, which are more likely to happen?

Answer 1

If you have your target point P with coordinates r_P = (x,y,z) and a plane with normal n=(nx,ny,nz) you need to define an origin on the plane, as well as two orthogonal directions for x and y. For example if your origin is at r_O = (ox, oy, oz) and your two coordinate axis in the plane are defined by e_1 = (ex_1,ey_1,ez_1), e_2 = (ex_2,ey_2,ez_2) then orthogonality has that Dot(n,e_1)=0, Dot(n,e_2)=0, Dot(e_1,e_2)=0 (vector dot product). Note that all the direction vectors should be normalized (magnitude should be one).

Your target point P must obey the equation:

r_P = r_O + t_1*e_1 + t_2*e_2 + s*n 

where t_1 and t_2 are your 2D coordinates along e_1 and e_2 and s the normal separation (distance) between the plane and the point.

There scalars are found by projections:

s = Dot(n, r_P-r_O) t_1 = Dot(e_1, r_P-r_O)     t_2 = Dot(e_2, r_P-r_O) 

Example with a plane origin r_O = (-1,3,1) and normal:

n = r_O/|r_O| = (-1/√11, 3/√11, 1/√11) 

You have to pick orthogonal directions for the 2D coordinates, for example:

e_1 = (1/√2, 0 ,1/√2) e_2 = (-3/√22, -2/√22, 3/√22) 

such that Dot(n,e_1) = 0 and Dot(n,e_2) = 0 and Dot(e_1, e_2) = 0.

The 2D coordinates of a point P r_P=(1,7,-3) are:

t_1 = Dot(e_1, r_P-r_O) = ( 1/√2,0,1/√2)·( (1,7,-3)-(-1,3,1) ) =  -√2 t_2 = Dot(e_2, r_P-r_O) = (-3/√22, -2/√22, 3/√22)·( (1,7,-3)-(-1,3,1) ) = -26/√22 

and the out of plane separation:

s = Dot(n, r_P-r_O) = 6/√11