Projecting 3D points to 2D plane [closed]

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

Answer 2

Find the projection of A onto the normal direction. Then subtract that projection from A. What is left is the projection of A onto the orthogonal plane.

The projection of A onto the unit normal direction n is given by:

(A · n) n

If A = (x, y, z) and the unit normal is given by n = (nx, ny, nz), then the projection of A onto n is

(x*nx + y*ny + z*nz) n

So the projection of A onto the orthogonal plane is

A - (A · n) n
= (x, y, z) - (x*nx + y*ny + z*nz) (nx, ny, nz)

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For example, if A = (1,2,3) and n is the unit normal in direction (4,5,6), then

In [12]: A
Out[12]: array([1, 2, 3])
In [17]: d
Out[17]: array([4, 5, 6])

In [20]: n = d/sqrt(4*4 + 5*5 + 6*6)   # make n a unit vector
In [13]: n
Out[13]: array([ 0.45584231,  0.56980288,  0.68376346])

So the projection of A onto the orthogonal plane is

In [15]: A - np.dot(A,n)*n
Out[15]: array([-0.66233766, -0.07792208,  0.50649351])

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How to find 2D coordinates:

You'll need to define a 2D coordinate system on the orthogonal plane. In other words, you need to define where the x-axis and y-axis are. For example, you could define the x-axis to be the projection of (1,0,0) onto the orthogonal plane (using the computation shown above). This will work except in the degenerate case where (1,0,0) is normal to the plane.

Once you have unit vectors for the x and y axis directions, then you could project A directly onto x and y. The magnitude of those vectors are the 2D coordinates.

For example, this is the projection of (1,0,0) onto the plane. We take this to be the x-axis direction:

In [42]: x = np.array([1,0,0])    
In [45]: x = x - np.dot(x, n) * n
In [52]: x /= sqrt((x**2).sum())   # make x a unit vector    
In [53]: x
Out[53]: array([ 0.89006056, -0.29182313, -0.35018776])

Here we compute the y-axis direction: The y-axis direction must be perpendicular to both the normal direction n and to x. So we could define y to be the cross product of n and x:

In [68]: y = np.cross(n, x)

In [69]: y
Out[69]: array([ -2.77555756e-17,   7.68221280e-01,  -6.40184400e-01])

So here are the coordinates for A in the plane:

In [70]: np.dot(A, x), np.dot(A, y)
Out[70]: (-0.74414898890755965, -0.38411063979868798)