- #1
richghost
- 4
- 0
Homework Statement
Let [itex] \Delta [/itex] be a line and [itex]A[/itex] a point in space.
[itex ] B [/itex] is a point on [itex] \Delta [/itex] ans H is a point on [itex] \Delta [/itex] such that [itex] \newcommand{\vect}[1]{\vec{#1}} \vec{AH} \perp \Delta [/itex].
a) Show that [itex] \| \vec{AH} \| \leq \| \vec{AB} \| [/itex] (a geometric reasoning should be used)
b) Using the inequality proven in a), explain how you can find a point on [itex] \Delta [/itex] closest to a given point [itex] A [/itex].
Homework Equations
[itex] \| \vec{AH} \| \leq \| \vec{AB} \| [/itex]
Pythagoras's theorem (I would think...)
The Attempt at a Solution
a) [itex]
\| \vec{AB} \|^2 = \| \vec{AH} \|^2 + \| \vec{HB} \|^2 [/itex], according to Pythagoras's theorem
[itex]\| \vec{AH} \|^2 \leq \| \vec{AH} \|^2 + \| \vec{HB} \|^2 [/itex] , the hypotenuse is the longest side and \| \vec{HB} \|geq 0
[itex]\| \vec{AH} \|^2 \leq \| \vec{AB} \|^2 [/itex], Pythagoras's theorem again
[itex]\| \vec{AH} \| \leq \| \vec{AB} \| [/itex]
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 [itex] \vec{OR} = \vec{OP} - proj_\vec{v}\vec{AP} [/itex] , where [itex] \vec{v} [/itex] is the direction vector of the line or even [itex] proj_\vec{v}\vec{AH} = 0 [/itex] . 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 [itex] \| \vec{AH} \| = \| \vec{AB} \| [/itex] . That's all I really may have figured out... Could you guys help me out on this one?