Orthogonal projection of point onto line/plane












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Upon looking through the definitions of different types of projections in $mathbb{R}^3$, I stumbled upon the definition of orthogonal projection of a point onto a line/plane. The textbook gives the following statement but does not provide a proof of it: given a point $P$ and a line or plane $ltext{/}pi$ there exists exactly one point $P^{prime}in ltext{/}pi$ such that $vec{PP^{prime}}$ is orthogonal to $ltext{/}pi$.



If anyone could provide a proof of the statement or an outline of one, it'd be much appreciated.










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  • 1




    $begingroup$
    If you could provide full context, or enough context, it'd be much appreciated. Please read How to ask a good question, and then edit your post to improve it.
    $endgroup$
    – amWhy
    Dec 25 '18 at 18:14






  • 1




    $begingroup$
    Using translation, WLOG we can assume the line or the plane provided is a closed subspace. Then we can apply the Hilbert projection theorem (see the proof in Wikipedia) to obtain the desired result.
    $endgroup$
    – Alex Vong
    Dec 25 '18 at 20:02










  • $begingroup$
    @AlexVong It seems like the proof you provided lies beyond the scope of the course I'm taking. It's an introductory course to linear algebra. Do you know of some proof that falls within those boundaries?
    $endgroup$
    – S. Doe
    Dec 27 '18 at 16:25










  • $begingroup$
    @S.Doe Which vector space are you working in? Is it $mathbb{R}^2, mathbb{R}^3, mathbb{R}^n$ or $mathbb{C}^n$?
    $endgroup$
    – Alex Vong
    Dec 27 '18 at 20:35












  • $begingroup$
    @AlexVong $mathbb{R}^3$. I'll add that to my question.
    $endgroup$
    – S. Doe
    Dec 27 '18 at 21:09


















0












$begingroup$


Upon looking through the definitions of different types of projections in $mathbb{R}^3$, I stumbled upon the definition of orthogonal projection of a point onto a line/plane. The textbook gives the following statement but does not provide a proof of it: given a point $P$ and a line or plane $ltext{/}pi$ there exists exactly one point $P^{prime}in ltext{/}pi$ such that $vec{PP^{prime}}$ is orthogonal to $ltext{/}pi$.



If anyone could provide a proof of the statement or an outline of one, it'd be much appreciated.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If you could provide full context, or enough context, it'd be much appreciated. Please read How to ask a good question, and then edit your post to improve it.
    $endgroup$
    – amWhy
    Dec 25 '18 at 18:14






  • 1




    $begingroup$
    Using translation, WLOG we can assume the line or the plane provided is a closed subspace. Then we can apply the Hilbert projection theorem (see the proof in Wikipedia) to obtain the desired result.
    $endgroup$
    – Alex Vong
    Dec 25 '18 at 20:02










  • $begingroup$
    @AlexVong It seems like the proof you provided lies beyond the scope of the course I'm taking. It's an introductory course to linear algebra. Do you know of some proof that falls within those boundaries?
    $endgroup$
    – S. Doe
    Dec 27 '18 at 16:25










  • $begingroup$
    @S.Doe Which vector space are you working in? Is it $mathbb{R}^2, mathbb{R}^3, mathbb{R}^n$ or $mathbb{C}^n$?
    $endgroup$
    – Alex Vong
    Dec 27 '18 at 20:35












  • $begingroup$
    @AlexVong $mathbb{R}^3$. I'll add that to my question.
    $endgroup$
    – S. Doe
    Dec 27 '18 at 21:09
















0












0








0





$begingroup$


Upon looking through the definitions of different types of projections in $mathbb{R}^3$, I stumbled upon the definition of orthogonal projection of a point onto a line/plane. The textbook gives the following statement but does not provide a proof of it: given a point $P$ and a line or plane $ltext{/}pi$ there exists exactly one point $P^{prime}in ltext{/}pi$ such that $vec{PP^{prime}}$ is orthogonal to $ltext{/}pi$.



If anyone could provide a proof of the statement or an outline of one, it'd be much appreciated.










share|cite|improve this question











$endgroup$




Upon looking through the definitions of different types of projections in $mathbb{R}^3$, I stumbled upon the definition of orthogonal projection of a point onto a line/plane. The textbook gives the following statement but does not provide a proof of it: given a point $P$ and a line or plane $ltext{/}pi$ there exists exactly one point $P^{prime}in ltext{/}pi$ such that $vec{PP^{prime}}$ is orthogonal to $ltext{/}pi$.



If anyone could provide a proof of the statement or an outline of one, it'd be much appreciated.







linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 8:06







S. Doe

















asked Dec 25 '18 at 18:06









S. DoeS. Doe

62




62








  • 1




    $begingroup$
    If you could provide full context, or enough context, it'd be much appreciated. Please read How to ask a good question, and then edit your post to improve it.
    $endgroup$
    – amWhy
    Dec 25 '18 at 18:14






  • 1




    $begingroup$
    Using translation, WLOG we can assume the line or the plane provided is a closed subspace. Then we can apply the Hilbert projection theorem (see the proof in Wikipedia) to obtain the desired result.
    $endgroup$
    – Alex Vong
    Dec 25 '18 at 20:02










  • $begingroup$
    @AlexVong It seems like the proof you provided lies beyond the scope of the course I'm taking. It's an introductory course to linear algebra. Do you know of some proof that falls within those boundaries?
    $endgroup$
    – S. Doe
    Dec 27 '18 at 16:25










  • $begingroup$
    @S.Doe Which vector space are you working in? Is it $mathbb{R}^2, mathbb{R}^3, mathbb{R}^n$ or $mathbb{C}^n$?
    $endgroup$
    – Alex Vong
    Dec 27 '18 at 20:35












  • $begingroup$
    @AlexVong $mathbb{R}^3$. I'll add that to my question.
    $endgroup$
    – S. Doe
    Dec 27 '18 at 21:09
















  • 1




    $begingroup$
    If you could provide full context, or enough context, it'd be much appreciated. Please read How to ask a good question, and then edit your post to improve it.
    $endgroup$
    – amWhy
    Dec 25 '18 at 18:14






  • 1




    $begingroup$
    Using translation, WLOG we can assume the line or the plane provided is a closed subspace. Then we can apply the Hilbert projection theorem (see the proof in Wikipedia) to obtain the desired result.
    $endgroup$
    – Alex Vong
    Dec 25 '18 at 20:02










  • $begingroup$
    @AlexVong It seems like the proof you provided lies beyond the scope of the course I'm taking. It's an introductory course to linear algebra. Do you know of some proof that falls within those boundaries?
    $endgroup$
    – S. Doe
    Dec 27 '18 at 16:25










  • $begingroup$
    @S.Doe Which vector space are you working in? Is it $mathbb{R}^2, mathbb{R}^3, mathbb{R}^n$ or $mathbb{C}^n$?
    $endgroup$
    – Alex Vong
    Dec 27 '18 at 20:35












  • $begingroup$
    @AlexVong $mathbb{R}^3$. I'll add that to my question.
    $endgroup$
    – S. Doe
    Dec 27 '18 at 21:09










1




1




$begingroup$
If you could provide full context, or enough context, it'd be much appreciated. Please read How to ask a good question, and then edit your post to improve it.
$endgroup$
– amWhy
Dec 25 '18 at 18:14




$begingroup$
If you could provide full context, or enough context, it'd be much appreciated. Please read How to ask a good question, and then edit your post to improve it.
$endgroup$
– amWhy
Dec 25 '18 at 18:14




1




1




$begingroup$
Using translation, WLOG we can assume the line or the plane provided is a closed subspace. Then we can apply the Hilbert projection theorem (see the proof in Wikipedia) to obtain the desired result.
$endgroup$
– Alex Vong
Dec 25 '18 at 20:02




$begingroup$
Using translation, WLOG we can assume the line or the plane provided is a closed subspace. Then we can apply the Hilbert projection theorem (see the proof in Wikipedia) to obtain the desired result.
$endgroup$
– Alex Vong
Dec 25 '18 at 20:02












$begingroup$
@AlexVong It seems like the proof you provided lies beyond the scope of the course I'm taking. It's an introductory course to linear algebra. Do you know of some proof that falls within those boundaries?
$endgroup$
– S. Doe
Dec 27 '18 at 16:25




$begingroup$
@AlexVong It seems like the proof you provided lies beyond the scope of the course I'm taking. It's an introductory course to linear algebra. Do you know of some proof that falls within those boundaries?
$endgroup$
– S. Doe
Dec 27 '18 at 16:25












$begingroup$
@S.Doe Which vector space are you working in? Is it $mathbb{R}^2, mathbb{R}^3, mathbb{R}^n$ or $mathbb{C}^n$?
$endgroup$
– Alex Vong
Dec 27 '18 at 20:35






$begingroup$
@S.Doe Which vector space are you working in? Is it $mathbb{R}^2, mathbb{R}^3, mathbb{R}^n$ or $mathbb{C}^n$?
$endgroup$
– Alex Vong
Dec 27 '18 at 20:35














$begingroup$
@AlexVong $mathbb{R}^3$. I'll add that to my question.
$endgroup$
– S. Doe
Dec 27 '18 at 21:09






$begingroup$
@AlexVong $mathbb{R}^3$. I'll add that to my question.
$endgroup$
– S. Doe
Dec 27 '18 at 21:09












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