Calculate Rotation matrix from 3 points

I have 3 base coordinate points

p0 = [533,-422 -1];
px = [219 -57 -993];
py = [5 -70 -23];

I would like to calculate the transformation matrix above these points. How would I go about doing so? because I am quite lost...

Answer 1

[Matt J]

You can use this File Exchange submission,

https://www.mathworks.com/matlabcentral/fileexchange/26186-absolute-orientation-horn-s-method

but it is not enough to have 3 points. You need 3 points both before and after the transformation (actually only 2 if you are only estimating rotation and not translation).

Answer 2

[William Rose]

@Jimmy Neutron,

A rotation in three dimensions has three degrees of freedom, so you need to know three "before" points and three "after" points.

I asssume you have given us the "after" points:

p0a = [533;-422;-1]; pxa = [219;-57;-993]; pya = [5;-70;-23];

I assume from the names of the ponts that the "before" points are

p0b=[0;0;0]; pxb=[1;0;0]; pyb=[0;1;0];

But those cannot be the real before points, because the "after" points are separated by more than 1 unit. Also, as @Matt J pointed out, the origin has been translated. You should check the "after" points to see that they form two equal-length legs of a right triangle. You may do this as follows:

fprintf('Length(pxa-p0a)=%.2f\n',norm(pxa-p0a));

Length(pxa-p0a)=1102.67

fprintf('Length(pya-p0a)=%.2f\n',norm(pya-p0a));

Length(pya-p0a)=634.96

fprintf('angle(pxa-p0a-pya)=%.2f radians\n',...
    acos((pya-p0a)'*(pxa-p0a)/(norm(pya-p0a)*norm(pxa-p0a))));

angle(pxa-p0a-pya)=1.10 radians

We see that the sides are of unequal length and are not at right angles. Therefore you must reevaluate the original problem statement.