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Math Help - Geometry 3D Line of Intersection Coordinates

  1. #1
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    Geometry 3D Line of Intersection Coordinates

    Hi attached question and mark scheme answer . I understand everything up to part (d)
    I don't follow how using [ λ ,1/2-1/2 λ , 3-λ] dot product [2,-1,-2]=0 is used . Are these two vectors perpendicular .
    I understand it evaluates to λ=13/9 and gives the coordinates .

    I am also struggling with the alternative answer of differentiation for OP^2.

    Thank you
    Attached Thumbnails Attached Thumbnails Geometry 3D Line of Intersection Coordinates-answer.png   Geometry 3D Line of Intersection Coordinates-question.png  
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  2. #2
    Junior Member
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    Re: Geometry 3D Line of Intersection Coordinates

    Hi attached question and mark scheme answer . I understand everything up to part (d)
    I don't follow how using [ λ ,1/2-1/2 λ , 3-λ] dot product [2,-1,-2]=0 is used . Are these two vectors perpendicular .
    I understand it evaluates to λ=13/9 and gives the coordinates .

    I am also struggling with the alternative answer of differentiation for OP^2.

    Thank you
    Let P be the point on the line, l, closest to the origin, O. Then the vector, OP, is perpendicular to l, which has direction [2,-1,-2]. Assuming P has coordinates (λ , 1/2-1/2 λ, 3-λ), we have OP = [λ , 1/2-1/2 λ, 3-λ]. But OP is perpendicular to [2,-1,-2], or put another way OP.[2,-1,-2] = 0.

    Alternatively, minimizing the magnitude of OP, or the square of it for simplicity, will also yield the same result.

    |OP|2 = λ2 + (1/2 - 1/2 λ)2 + (3 - λ)2
    = λ2 + 1/4 - 1/2 λ + 1/4 λ2 + 9 - 6λ + λ2
    = 9/4 λ2 - 13/2 λ + 37/4
    = 1/4(9λ2 - 26 λ + 37)

    We set the derivative of this to 0 to find the critical value of λ. The derivative is
    1/4(18 λ - 26).

    Setting it to 0 gives the result λ = 13/9.

    Hope this helped.
    Thanks from minicooper58
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