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Math Help - vectors - find equidistant point

  1. #1
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    vectors - find equidistant point

    Using vectors:

    Find the point on the y axis that is equidistant from the points (2,-1,1) and (0,1,3)
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi

    The point P we're looking for is on the y axis hence P(0,\,y,\,0)

    The distance between P(0,\,y,\,0) and A(2,\,-1,\,1) is \|\vec{AP}\|=\sqrt{(2-0)^2+(-1-y)^2+(1-0)^2}=\sqrt{5+(1+y)^2}

    The distance between P(0,\,y,\,0) and B(0,\,1,\,3) is \|\vec{BP}\|=\sqrt{(0-0)^2+(1-y)^2+(3-0)^2}=\sqrt{9+(1-y)^2}

    We want these two distances to be equal thus \|\vec{AP}\|=\|\vec{BP}\| \Leftrightarrow \sqrt{9+(1-y)^2}=\sqrt{5+(1+y)^2}

    Squaring both sides gives 9+(1-y)^2=5+(1+y)^2 which you can solve for y.

    Does it help ?
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