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Math Help - Find point on the plane ...

  1. #1
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    Find point on the plane ...

    Find the point on the plane x-y+z=4 that is closest to the point (1,2,3).



    .... Not sure how to go about this. Thanks!
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  2. #2
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    Hello, jlt1209!

    Find the point on the plane x-y+z\:=\:4 that is closest to the point (1,2,3).

    The normal direction of the plane is: . \vec{n} \:=\:\langle 1,-1,1\rangle

    The line through (1,2,3) with \vec{n} = \langle1,-1,1\rangle has equations: . \begin{array}{ccc}x &=& 1 + t \\ y &=& 2 - t \\ z &=&3 + t \end{array}

    Substitute into the equation of the plane:

    . . (1+t) - (2-t) + (3+t) \:=\:4 \quad\Rightarrow\quad t \:=\:\tfrac{2}{3}

    Then: . \begin{Bmatrix}x &=& 1 + \frac{2}{3} &=&\frac{5}{3} \\ \\[-4mm] y &=& 2 - \frac{2}{3} &=& \frac{4}{3} \\ \\[-4mm] z &=& 3 + \frac{2}{3} &=& \frac{11}{3} \end{Bmatrix} \quad\Rightarrow\quad \left(\frac{5}{3},\:\frac{4}{3},\:\frac{11}{3}\rig  ht) <br />

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