Prashant Kumar answered . 2024-05-06 13:12:50

planes(:,:,1) = [0 3 3; 0 0 3; 0 3 0; 0 0 0; 0 0 0];
planes(:,:,2) = [0 0 3; 3 0 3; 0 0 0; 3 0 0; 0 0 0];
planes(:,:,3) = [3 0 3; 3 3 3; 3 0 0; 3 3 0; 3 0 0];
planes(:,:,4) = [3 3 3; 0 3 3; 3 3 0; 0 3 0; 0 3 3];
planes(:,:,5) = [0 3 0; 3 3 0; 0 0 0; 3 0 0; 0 0 0];
planes(:,:,6) = [0 3 3; 3 3 3; 0 0 3; 3 0 3; 0 0 3];

6 different planes, represented by points in each plane. First, I'll verify that each set of 5 points you designated do indeed lie in a plane in R^3. (Hey you might have made a mistake.)

for i = 1:6
  rank(planes(:,:,i) - planes(1,:,i))
end

ans = 2
ans = 2
ans = 2
ans = 2
ans = 2
ans = 2

Two planes are parallel if they have the same normal vector (Though you could multiply by -1.) So just compute the normal vectors to each plane.

Nullvecs = zeros(6,3);
for i = 1:6
  nullvecs(:,i) = null(planes(:,:,i) - planes(1,:,i));
end
nullvecs

nullvecs = 3×6
     1     0     1     0     0     0
     0    -1     0     1     0     0
     0     0     0     0     1     1

Now, look at those vectors. See that planes 1 and 3 have the same normal vectors, The signs of the normal vectors are the same. Planes 2 and 4 have a sign flip on the normal vectors. And planes 5 and 6 are also parallel. In the last case again, the normal vectors had the same signs. Could we identify those pairs automatically? Yes, of course. We can compute a corrrelation matrix, then look for elements of the correlation matrix that are exactly either 1 or -1. I'll put a small tolerance on that result.

C = abs(abs(corr(nullvecs) - eye(6)) - 1)<2*eps

C = 6×6 logical array
   0   0   1   0   0   0
   0   0   0   1   0   0
   1   0   0   0   0   0
   0   1   0   0   0   0
   0   0   0   0   0   1
   0   0   0   0   1   0

Note that I subtracted the identity matrix so we don't identify each vector as the same as itself.

[I,J] = find(C);
unique(sort([I,J],2),'rows')

ans = 3×2
     1     3
     2     4
     5     6

So there were 3 pairs of parallel planes identified.

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