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PO
 

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    <p>We have to solve this exact same problem. Here's how I do it:</p> <p>Call the points P and W, so we have P1..P3 and W1..W3</p> <p>Construct three vectors in each space like so</p> <pre><code>A1 = P2-P1 A2 = P3-P1 A3 = A1 x A2 </code></pre> <p>and</p> <pre><code>B1 = W2-W1 B2 = W3-W1 B3 = B1 x B2 </code></pre> <p>These two pairs of three vectors each constitute a non-orthogonal basis, and you want to find how to represent your cartesian axes (x y and z) in one space so that you can find them in another. To do this construct a matrix so that its columns are the three vectors found above. Then invert this matrix, if this inversion fails, then the non-orthogonal basis does not span the space and the problem cannot be solved.</p> <p>Then pull the three columns out of the inverted matrix. These columns are the cartesian axes in terms of your non-orthogonal basis (V1, V2 and V3). From this we can reconstruct an orthogonal basis that will serve as a transformation matrix from the first space to the second. </p> <p>If we call this matrix R, and denote R[row, column] as our notation, then the rows (or columns, depending on how you use the matrix) of the final transformation matrix will be:</p> <pre><code>B1 * R[0,0] + B2* R[1,0] + B3 * R[2,0] B1 * R[0,1] + B2* R[1,1] + B3 * R[2,1] B1 * R[0,2] + B2* R[1,2] + B3 * R[2,2] </code></pre> <p>Now, because one of the columns of the original matrix before the inversion was the cross of the other two columns, it's probably possible to optimise the inversion of the matrix. I haven't bothered to do this - especially because in our case the three points P1..P3 don't change, and so the inverted matrix can be cached.</p> <p>This method has the advantage that if you have a half-decent matrix/vector library it's very simple to implement. And it doesn't use angles, which is always a good thing.</p>
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    1. PODeriving axis/angle rotation from two pairs of three points (or, two pairs of two vectors)
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    1. USDave Branton
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    1. COI must admit that I still can't 100% follow, even after reading it 3 times, giving it a month break, and reading it again (the last code section is just one column of the matrix?). What you describe seems to be similar to the "triad method" as outlined in the question, except you do not orthonormalize the base (which means you can represent shear and scale, but the inverse is not as simple as transposing).
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      1. USDamon
    2. COIt's possible that my explanation isn't as clear as it might be. What I'm effectively saying is that the answer is in the question. That is, you want to transform from one basis to another, and you have both bases. The only complication is that your two bases are not orthogonal, and orthonormalising them is one approach that will work. I personally think it's conceptually simpler to calculate the representation of the cartesian axes in your first basis so that you can transform them into your other basis, but they are probably equivalent methods.
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      1. USDave Branton
 

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