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Math Help - Question on 3-d vectors

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    Question on 3-d vectors

    A triangle has vertices at the points A(5,0,0), B(0,5,0), and C(0,0,5). Identify the point in the interior of the triangle that is closest to the origin.
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    Quote Originally Posted by dalbir4444 View Post
    A triangle has vertices at the points A(5,0,0), B(0,5,0), and C(0,0,5). Identify the point in the interior of the triangle that is closest to the origin.
    1. The points A, B, C define a plane:

    x,y,z)=(5,0,0)+r(5,-5,0+s(5,0,-5)" alt="px,y,z)=(5,0,0)+r(5,-5,0+s(5,0,-5)" />

    2. p in normal form:

    \vec n = (5,-5,0) \times (5,0,-5) = (5,5,5) Thus

    p: x+y+z=5

    3. The point F lies on the line through the origin perpendicular to p:

    (x,y,z)=r (1,1,1)

    4. Calculate the point of intersection:

    r+r+r = 5~\implies~r=\dfrac53

    Therefore F\left(\dfrac53\ ,\ \dfrac53\ ,\ \dfrac53\right)
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