# Homework Help: Finding a point on a line closest to a point in space

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1. Dec 1, 2014

### richghost

1. The problem statement, all variables and given/known data
Let $\Delta$ be a line and $A$ a point in space.

[itex ] B [/itex] is a point on $\Delta$ ans H is a point on $\Delta$ such that $\newcommand{\vect}[1]{\vec{#1}} \vec{AH} \perp \Delta$.
a) Show that $\| \vec{AH} \| \leq \| \vec{AB} \|$ (a geometric reasoning should be used)
b) Using the inequality proven in a), explain how you can find a point on $\Delta$ closest to a given point $A$.

2. Relevant equations
$\| \vec{AH} \| \leq \| \vec{AB} \|$

Pythagoras's theorem (I would think...)

3. The attempt at a solution
a) $\| \vec{AB} \|^2 = \| \vec{AH} \|^2 + \| \vec{HB} \|^2$, according to Pythagoras's theorem
$\| \vec{AH} \|^2 \leq \| \vec{AH} \|^2 + \| \vec{HB} \|^2$ , the hypotenuse is the longest side and \| \vec{HB} \|geq 0
$\| \vec{AH} \|^2 \leq \| \vec{AB} \|^2$, Pythagoras's theorem again
$\| \vec{AH} \| \leq \| \vec{AB} \|$
So... I think I've successfully showed that the inequality is true. I find it sort of questionable however.

b) I've found many ways to find the closest point on a line, all of which use some sort of an orthogonal projection like $\vec{OR} = \vec{OP} - proj_\vec{v}\vec{AP}$ , where $\vec{v}$ is the direction vector of the line or even $proj_\vec{v}\vec{AH} = 0$ . None of which, I think you count as using the inequality given in a). So I thought maybe it has something to do with making $\| \vec{AH} \| = \| \vec{AB} \|$ . That's all I really may have figured out... Could you guys help me out on this one?

2. Dec 1, 2014

### Joffan

I suppose you have to actually make the small step to infer that H is indeed the closest point on the line to A.

Then, you can express any point on the line as a function of some reference point on the line and a multiple $k$ of an along-line vector (i.e. $\vec v$)
Then you can express the vector from that point to A (also as a function of $k$)

Now you have to find some distinctive property of vectors at right angles that you can solve for...

3. Dec 1, 2014

### richghost

Thank you so much! Yeah, makes so much sense now! Simply by remembering to infer that H is the closest point on the line to A got me going. Thanks again!