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May 30, 2020 at 21:32 comment added amd You could’ve saved yourself a lot of work by projecting orthogonally onto $A^\perp$ instead, which is only one-dimensional, and then subtracting that from $x$. You can read a vector that spans $A^\perp$ directly from the equation that defines $A$.
May 30, 2020 at 21:18 history edited José Carlos Santos
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May 30, 2020 at 21:06 comment added lambdaepsilon working the real space would the minimization occur when the interior portion is equal to zero. Also its squared because you are projecting the distance between x and an arbitrary point on itself right?
May 30, 2020 at 20:48 answer added José Carlos Santos timeline score: 2
May 30, 2020 at 20:48 comment added Alexey Burdin Once you have an orthonormal basis of $A$, say $e_1,e_2,e_3$ you can express an arbitrary point of $A$ as $te_1+ue_2+ve_3$ and compute $(x-(te_1+ue_2+ve_3))^2$, then take it to the minimum.
May 30, 2020 at 20:42 history asked lambdaepsilon CC BY-SA 4.0