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Math Help - Distance:Line and Point

  1. #1
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    Distance:Line and Point

    Given are the line l:m*(6/2/3) and the point P(2/1/2). Sought is the point Q on l with the shortest distance to P.
    I know the solution of inventing a plane E with normal vector N which equals (6/2/3), but I'd liek to know why the following approach doesn't work as well:

    The vector QP must be normal to the direction vector of the line, so (6/2/3)*QP = 0. Point Q is given with (m6/m2/m3), thus the vector QP is (2-6m)/(1-2m)/(2-3m).
    Idea
    (2-6m)/(1-2m)/(2-3m)*(6/2/3)=0
    m=no m valid for all three

    Why doesn't that lead to a correct solution?

    Can anyone help?
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Schdero View Post
    Given are the line l:m*(6/2/3) and the point P(2/1/2). Sought is the point Q on l with the shortest distance to P.
    I know the solution of inventing a plane E with normal vector N which equals (6/2/3), but I'd liek to know why the following approach doesn't work as well:

    The vector QP must be normal to the direction vector of the line, so (6/2/3)*QP = 0. Point Q is given with (m6/m2/m3), thus the vector QP is (2-6m)/(1-2m)/(2-3m).
    Idea
    (2-6m)/(1-2m)/(2-3m)*(6/2/3)=0
    m=no m valid for all three
    I cannot quite parse this: what value of m do you get? In an earlier version of this question you wrote m=1/3, which is wrong. You should get m=20/49.

    Why doesn't that lead to a correct solution?
    The basic idea is correct, but the solution m=1/3 of your equation was wrong.
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  3. #3
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    Yes I see, I also get your result when I do it, I think i have to have made some quite stupid error in the earlier version; again: thanks failure
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