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Math Help - Points in the plane.

  1. #1
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    Points in the plane.

    I want to find some arbitrary point in the plane

    2x+4y-2z + 10 =0

    Thank you.
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  2. #2
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    Quote Originally Posted by MatteNoob View Post
    I want to find some arbitrary point in the plane

    2x+4y-2z + 10 =0

    Thank you.
    Since 3 non-colinear points determine a plane, then try two coordinates and solve or the 3rd coordinate.

    2x +4y -2z +10 = 0

    Example, try x=1 and y=1,
    2(1) +4(1) -2z +10 = 0
    -2z = -16
    z = 8
    Hence, (1,1,8) is a point on the given plane.
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  3. #3
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    Thank you very much. :]
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  4. #4
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    Hello, MatteNoob!

    I want to find some arbitrary point in the plane: . 2x+4y-2z + 10 \:=\:0
    The easiest points are the intercepts. .You can "eyeball" the answers!

    . . \begin{array}{cccccccc} \text{Let }y = 0,\:z = 0\!: & 2x + 10 &=& 0 & \Rightarrow &  x \:=\: \text{-}5 & \Rightarrow & (\text{-}5,0,0) \\ \\[-4mm] \text{Let }x = 0,\:z = 0\!: & 4y + 10&=& 0 & \Rightarrow & y \:=\: \text{-}\frac{5}{2} & \Rightarrow& \left(0,\:\text{-}\frac{5}{2},\:0\right) \\ \\[-4mm] \text{Let }x = 0,\:y=0\!: & \text{-}2z+10 &=& 0 & \Rightarrow &  z \:= \:5 & \Rightarrow & (0,\:0,\:5) \end{array}

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  5. #5
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    or...

    \frac{x}{-5}\;+\;\frac{y}{-5/2}\;+\;\frac{z}{5}\;=\;1

    They're just kind of staring at you in this form.
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  6. #6
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    Thanks to all of you. I now understand how finding (or checking if) points (are in) the plane. :]
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