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I am writing a 3D solar panel positioning programme and have a section of code where I use the Gram-Schmidt Orthogonalization process to go from 3D to 2D for easier calculations.

(For reference, here is the process in more detail: 3D to 2D rotation matrix)

My simple question is this.... after applying the orthogonalization processs, is one unit in 2D still the same length as a unit in 3D?

Am I right to assume that there is no distortion as the plane is simply rotated by the Gram-Schmidt process?

Many thanks for your time, Kelvin.

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up vote 1 down vote accepted

Based on the answers to your previous question, I suppose you implemented the mapping from 3D to 2D. Whether or not this mapping preserves distances depends on how you did the implementation.

If your 3D-to-2D mapping is just a rotation, then as you suspected, it will preserve distances (and angles).

If you really want to be sure, experiment by mapping some points. Take two points that are some known distance $d$ apart, apply your mapping to each of them to get two new points, and check that these new points are again a distance $d$ apart.

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Thanks bubba, your advice of experimenting and testing my implementation was a great idea. Using the Gram-Schmidt from… results in only a rotation where z-values become 0 and therefore all distances are preserved. Thank you! – Kel196 Aug 19 '12 at 14:08

I don't see any reason why this should be true. For example: $$v_1 = \{1,1\}, v_2 = \{1,0\}$$ Following the Gram-Schmidt process, you get: $$u_1 = v_1, u_2 = v_2 - \frac{v_2\cdot u_1}{u_1\cdot u_1}u_1 = \{1,0\} - \{1/2,1/2\} = \{1/2,-1/2\} $$

Which do not have the same length as the original vectors.

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