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?

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?

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.

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.

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?

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.