# How to prove projection of a point onto a line(perpendicular to the line) is the most minimum distance?

I've a differential geometry math problem that I can't solve.

The problem is:

Prove that the projection of a 2D point on a 2D line which is perpendicular to the line, is the most minimum distance between the point and the line.

Can anyone kindly give me some hints or better if shown the steps necessary to prove this?

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I will prove a related theorem using Cauchy-Schwarz, so that you may attempt to adapt this proof to fit your theorem. If things are still not clear, comment again and I will provide the full details of your proof.

Theorem: Given a point $\vec P$ on a line $L$, every vector $\vec X$ on $L$ has length greater than or equal to the projection of $\vec P$ on a nonzero normal vector to the line.

Note that this shows that the distance between the origin and a point on the line is a minimum when that point is the projection of that point onto the normal vector.

You must set up the problem in a convenient way to apply Cauchy-Schwarz. Start with a point on the line $\vec P$ and a (nonzero) vector normal to the line $\vec N$. Then the line is the set of all points $\vec X=(x,y)$ satisfying $$\vec X\cdot\vec N=\vec P \cdot \vec N$$ (Think about this for a bit, make sure it is consistent with whatever definition of a line you are working with.)

Then by Cauchy-Schwarz we have that $$\lvert \vec P\cdot \vec N\rvert=\lvert \vec X \cdot \vec N\rvert\le\lVert\vec X\rVert\lVert\vec N\rVert$$ And since $\vec N$ is nonzero, we have that $\lVert\vec X\rVert\ge\lvert \vec P \cdot \vec N \rvert/\lVert \vec N \rVert$. Cauchy-Schwarz tells us that equality holds iff $\vec X=t\vec N$ for some scalar $t$, in which case $\vec P \cdot \vec N=\vec X \cdot \vec N = t\vec N \cdot \vec N$ and so $t=\vec P \cdot \vec N / \vec N \cdot \vec N$ and therefore $\vec X$ is the projection of $\vec P$ on $\vec N$.

If you haven't covered Cauchy-Schwarz, and don't want to, you could also treat this using the method of Lagrange's multipliers.

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By Cauchy-Schwarz did you mean Cauchy-Schwarz inequality? –  mvr950 Sep 1 '12 at 2:15
@mvr950 Indeed I did. I updated my answer to contain more details and a proof very similar to the one you want to prove. –  Michael Boratko Sep 1 '12 at 2:19
Thanks a lot for help. I understand now. –  mvr950 Sep 1 '12 at 2:51
@mvr950 No problem! –  Michael Boratko Sep 1 '12 at 3:11
Sorry for my ignorance but I've some questions and confusions. You said $\vec{X} \cdot \vec{N} = \vec{P} \cdot \vec{N}$ why is that? What are these $\vec{X} \text{ and } \vec{P}$ actually? Also you said "Cauchy-Schwarz tells us that equality holds iff $\vec{X} = t \vec{N}$ for some scalar t" how did you find that? –  mvr950 Sep 1 '12 at 18:40