Coordinate system transformation, 3d projection to 2d plane

Question

I have a global coordinates system (X, Y, Z), and a triangle with points (A, B, C and Center). I know all coordinates of those points.

1. I need to move global coordinates system from (0; 0; 0) to the triangle center, so that all the points: A, B, C and Center would have new coordinates where Z = 0. After that i need to know new coordinates of those points in the relation with new coordinates system. The orientation of new coordinate system is not important.
2. Also, if there is a possibility to convert a 3D points (triangle points) to 2D plane without loosing its geometry (size). It shouldn’t be a projection to 2D plane.

>> A=[10.63307; -7.72528; 21.26636];
B=[4.06139; -12.49988; 21.26636];
C=[-6.57172; -20.22529; 13.14344];
Centr=[-4.38113; -13.48349; 18.55872];

>> V1=(B-A)/(norm(B-A))

V1 =

   -0.8090
   -0.5878
         0

>> V2=((C-A)-(dot((C-A),V1)*V1))/(norm((C-A)-(dot((C-A),V1)*V1)))

V2 =

    0.0000
   -0.0000
   -1.0000

>> V3=cross(V1,V2)

V3 =

    0.5878
   -0.8090
    0.0000

>> M=[V1,V2,V3]

M =

   -0.8090    0.0000    0.5878
   -0.5878   -0.0000   -0.8090
         0   -1.0000    0.0000

>> Anew=inv(M)*(A-Centr)

Anew =

  -15.5313
   -2.7076
    4.1666

>> Bnew=inv(M)*(B-Centr)

Bnew =

   -7.4083
   -2.7076
    4.1666

>> Cnew=inv(M)*(C-Centr)

Cnew =

    5.7350
    5.4153
    4.1666

This is what i got: From this

To this

Answer 1

The problem can be expressed as finding the 3-by-3 matrix M such that the coordinates of a point P can be converted between the old coordinate system (P_old, 3 rows) and the new coordinate system (P_new, 3 rows). This is an affine transformation:

P_old = Center + M * P_new     (1)

The (matrix-vector) multiplication with M orientates it back to the old system, and adding Center's coordinates translates it back to the old origin.

The equation (1) can then be turned into:

P_new = M^{-1} * (P_old - Center)     (2)

where M^{-1} is the inverse of M, to compute the new coordinates from the old ones (the third row will have a 0 if the point belongs to the plane of the triangle).

The matrix M is made of the coordinates of the new basis in the old system, one basis vector in each column. One must now find such a basis.

This basis can be taken from (this is all pseudo-code):

1. Renormalizing AB

          AB
   V1 = ______
       || AB ||
   

   • AB here is meant as the vector AB (with an arrow on top):

     |b_x - a_x|
     |b_y - a_y|
     |b_z - a_z|
     

   • || . || is the Euclidian norm (^2 means the square, not xor):

     || V || = sqrt(V_x^2 + V_y^2 + V_z^2)
     

2. AC (also a vector, defined like AB), but minus its projection on V1 to make it orthogonal to V1, and renormalized (this will fail with a division by zero if the triangle is not really a triangle):

           AC - (AC.V_1) V1
   V2 = _______________________
        || AC - (AC.V_1) V1 ||
   

   • M.N is the scalar product:

     M.N = M_x * N_x + M_y * N_y + M_z * N_z
     

   • (AC.V_1) V1 is simply the product of a scalar, (AC.V_1), with a vector, V1

3. A third vector that can be taken as the cross product to get a Cartesian coordinate system:

   V3 = V1 x V2
   

   • The cross product is defined as:

               |V1_y*V2_z - V1_z*V2_y|
     V1 x V2 = |V1_z*V2_x - V1_x*V2_z|
               |V1_x*V2_y - V1_y*V2_x|
     

Then M can be taken as |V1 V2 V3| (each Vx is on 3 rows) and, then its inverse computed to use formula (2).

This transformation (with the inverted M) should both generate new coordinates for the points on the plane of the triangle that have 0 on the third axis (which makes it 2D-coordinates on that plane), and preserve the size in terms of Euclidian norm.