Finding a point on a line closest to a point in space

[richghost]

Homework Statement

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 $A$.

Homework Equations

$\| \vec{AH} \| \leq \| \vec{AB} \|$

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

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?

 

[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...

 

[richghost]

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...

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!