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Equation of the plane EQUIDISTANT from two points

  1. Jan 17, 2008 #1
    I got two points that are

    (2, -1, 1) and (3, 1, 5)

    and I need to find the equation of the plane equidistant from those points.

    Anyone?
     
  2. jcsd
  3. Jan 17, 2008 #2

    siddharth

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    Since it's equidistant, finding the midpoint should allow you to calculate the direction ratios of the normal to the plane. Once you find that, you should be able to obtain the equation to the plane. ie, [tex](\vec{r} - \vec{a}) . \hat{n} = 0[/itex]
     
  4. Jan 17, 2008 #3
    Midpoint between what though? Midpoint between the points maybe?

    I'm confused. I was thinking the first thing was to create a vector parallel to the lines or vector equation of the line based on the two points (if I did that part correctly)

    what I got was
    <x,y,z> = <2,-1,1> + t<1,2,4> = <z+t,-1+2t,1+4t>
     
  5. Jan 18, 2008 #4

    siddharth

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    Yes, and that gives you the direction ratios of the normal to the plane.

    Yeah. Since it's equidistant, this midpoint lies on the required plane. Now that you've got a point lying on the plane, and the direction ratios of the normal to the plane, you should be able to find the equation of the plane.
     
  6. Jan 18, 2008 #5

    HallsofIvy

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    The plane equidistant from two points contains the midpoint of the line segment between them, and is perpendicular to that line. That's all you need.
     
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