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

  1. Dec 1, 2014 #1
    1. The problem statement, all variables and given/known data
    Let [itex] \Delta [/itex] be a line and [itex]A[/itex] a point in space.

    delta_point_A.jpg
    [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].

    2. Relevant equations
    [itex] \| \vec{AH} \| \leq \| \vec{AB} \| [/itex]

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

    3. 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?
     
  2. jcsd
  3. Dec 1, 2014 #2
    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...
     
  4. Dec 1, 2014 #3
    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!
     
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