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supppose that we have a vector space $$A=\left(\left[ \begin{array}{} x\\ y\\ w\\ z\\ \end{array} \right] : x-y+w=0 \right)$$ and we wanted to find the closest point to a vector $$x= \left[\begin{array}{} a\\ b\\ c\\ d\\ \end{array}\right] $$ I solved the problem by finding the basis of the vector space A and orthonormalizing it, then taking the projection of x onto the orthonormal basis of A which returned a vector. However I would like a second method to verify that this is in fact the closest point. Is there a way to verify whether or not my solution is correct?

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    $\begingroup$ 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. $\endgroup$ – Alexey Burdin May 30 '20 at 20:48
  • $\begingroup$ 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? $\endgroup$ – lambdaepsilon May 30 '20 at 21:06
  • $\begingroup$ 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 $A$. $\endgroup$ – amd May 30 '20 at 21:32
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Yes: if $p\in A$, then $p$ is the point of $A$ closest to $x$ if and only if $x-p$ is orthogonal to every element of $A$.

For instance, the solution of your problem is$$p=\begin{bmatrix}\frac13(2a+b-c)\\\frac13(a+2b+c)\\\frac13(-a+b+2c)\\d\end{bmatrix}$$and$$x-p=\begin{bmatrix}\frac13(a-b+c)\\\frac13(-a+b-c)\\\frac13(a-b+c)\\0\end{bmatrix}.$$And, indeed, if $x$, $y$, $w$, and $z$ are such that $x-y+w=0$, then$$\begin{bmatrix}x\\y\\w\\z\end{bmatrix}.(x-p)=0.$$

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  • $\begingroup$ interesting I know that the shortest point from a plane to a point is always when there exists an orthogonal relationship, but what is the intuition behind this? $\endgroup$ – lambdaepsilon May 30 '20 at 21:02
  • $\begingroup$ and thanks alot!! $\endgroup$ – lambdaepsilon May 30 '20 at 21:02
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    $\begingroup$ Imagine a point $(x,y,z)\in\Bbb R^3$. It's projection into the $z=0$ plane is $(x,y,0)$. And $(x,y,z)-(x,y,0)=(0,0,z)$, which is orthogonal to every element of the plane $z=0$. $\endgroup$ – José Carlos Santos May 30 '20 at 21:13

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