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Thread: Vectors Help

  1. #1
    Jan 2009

    Vectors Help

    am looking over a solution to an answer and i dont get some things

    Question is

    Find the distance from the point A(1,-2,3) to the plane 2x-3x+6z-8=0

    Then they picked a random point
    and AP=(3,2,-3)

    However they also got a vector n=(2,-3,6)

    That vector is included in
    |AH|=proj n(PA)

    =PA*n/n*n |n|

    That turns out to be
    (3,2,-3)*(2,-3,6) ALL OVER (2,-3,6)*(2,-3,6)
    all times sqrt 2^2 +(-3)^2+6^2

    My question is how did they get that point n?
    Where could that have come from?
    Thank you for your help
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  2. #2
    MHF Contributor
    Nov 2008

    Here is the situation

    n is a normal vector to the plane
    For a plane ax+by+cz+d=0, we know that a normal vector has coordinates (a,b,c)

    P is a point picked at random on the plane
    H is the projection of A on the plane
    The distance from A to the plane is AH

    The dot product of AP and n is
    AP.n = (AH+HP).n = AH.n + HP.n = AH.n since HP and n are perpendicular
    |AP.n| = |AH.n| = AH ||n|| since AH and n are parallel

    AH = |AP.n| / ||n||
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