/wsr/ - Worksafe Requests » Thread #1549739

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Anonymous Mon 24 Nov 2025 22:22:16 No.1549739 View View Reply Original Report

Is there a trivial way to prove this condition is the minimum for the the distance between u and cv? If I've boxed myself into an algebraic mess, I'd like to know why.

Anonymous

Anonymous Sat 29 Nov 2025 04:36:04 No.1550208 Report

Quoted By:

>>1549739
I don't know, but it looks nice. Have a bump.

Anonymous

Anonymous Sun 30 Nov 2025 08:15:23 No.1550329 Report

Quoted By: >>1550372

>>1549739
You want to minimize ||cv-u||, or equivalently, <cv-u,cv-u>.
Expand using linearity: c^2<v,v> - 2c<u,v> + <u,u>. This is a quadratic equation in c.
Recall that the extremum of a parabola px^2 + qx + r is found at x = -q / 2p.
So in this case, the minimum is found at c = <u,v> / <v,v>.

Anonymous

Anonymous Sun 30 Nov 2025 08:16:23 No.1550330 Report

Quoted By:

We can check that (<u,v>/<v,v>)v is indeed the orthogonal projection:
<u - (<u,v>/<v,v>)v , v>
=
<u,v> - (<u,v>/<v/v>)<v,v> = <u,v> - <u,v> = 0.

Anonymous

Anonymous Sun 30 Nov 2025 08:17:24 No.1550331 Report

Quoted By: >>1550372

We can check that (<u,v>/<v,v>)v is indeed the orthogonal projection:
<u - (<u,v>/<v,v>)v , v>
=
<u,v> - (<u,v>/<v,v>)<v,v> = <u,v> - <u,v> = 0.

Anonymous

Anonymous Sun 30 Nov 2025 17:11:53 No.1550372 Report

Quoted By:

>>1550329
>>1550331
Or, shorter: if p is the projection of u, so that <p,u-p>=0 hence <p,u>=<p,p>.
So minimizing <cp-u,cp-u> = c^2<p,p> - 2c<p,u> + <u,u>
comes down to minimizing (c^2 - 2c)<p,p> which happens at c=1.

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