# Show that the minimized distance from a point to a line is projection onto line

• Aug 31st 2012, 01:48 PM
x3bnm
Show that the minimized distance from a point to a line is projection onto line
I've a math problem that I like to solve.

Let \$\displaystyle L\$ be a line and let \$\displaystyle p\$ be a point. Show that the orthogonal projection
of \$\displaystyle p\$ onto \$\displaystyle L\$ is the unique point \$\displaystyle w\$ on \$\displaystyle L\$ for which the distance from \$\displaystyle p\$
to \$\displaystyle w\$ is minimized.

Is it possible to tell me how I can prove this using differential geometry?
• Sep 1st 2012, 01:04 AM
Vlasev
Re: Show that the minimized distance from a point to a line is projection onto line
I'm not sure how the differential geometry proof will work since I have not studied it yet but what will happen if you take a regular geometric proof and try to translate it into differential geometry lingo?
• Sep 1st 2012, 09:46 AM
x3bnm
Re: Show that the minimized distance from a point to a line is projection onto line
Quote:

Originally Posted by Vlasev
I'm not sure how the differential geometry proof will work since I have not studied it yet but what will happen if you take a regular geometric proof and try to translate it into differential geometry lingo?

Yes regular geometric proof is okay. I'll try to translate it into differential geometry.
• Sep 1st 2012, 10:55 AM
HallsofIvy
Re: Show that the minimized distance from a point to a line is projection onto line
Why would one even consider "differential geometry" when the Pythagorean theorem gives it easily? In any right triangle, \$\displaystyle c^2= a^2+ b^2\$ so the hypotenuse is longer than either of the two legs.
• Sep 1st 2012, 12:09 PM
x3bnm
Re: Show that the minimized distance from a point to a line is projection onto line
Quote:

Originally Posted by HallsofIvy
Why would one even consider "differential geometry" when the Pythagorean theorem gives it easily? In any right triangle, \$\displaystyle c^2= a^2+ b^2\$ so the hypotenuse is longer than either of the two legs.

HallofIvy, sorry I didn't understand from your last post. Is it possible to kindly explain how "the hypotenuse is longer than either of the two legs" is related to prove my above stated problem?
• Sep 1st 2012, 04:15 PM
x3bnm
Re: Show that the minimized distance from a point to a line is projection onto line
Quote:

Originally Posted by HallsofIvy
Why would one even consider "differential geometry" when the Pythagorean theorem gives it easily? In any right triangle, \$\displaystyle c^2= a^2+ b^2\$ so the hypotenuse is longer than either of the two legs.

Yes, I got it. Thanks for the answer.