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Equation of plane

  1. Oct 14, 2011 #1
    1. The problem statement, all variables and given/known data

    Show that the points in R3 equidistant from two fixed points, p and q, form a plane and find the equation of the plane.


    2. Relevant equations



    3. The attempt at a solution

    So the equation of the plane takes the form ax + by+ cz = d, where (a,b,c) is a normal vector of the plane.

    assuming p = (x1, y1, z1) and q = (x2, y2, z2), how can i show that the points equidistant from both these points is a plane?
    It doesn't really make sense to me visually I mean. The only way i can think of a point being equidistant is if its on a line in the middle of the two points. Its like an equilateral triangle, where if p and q were bottom left and right corners and the point on the top is equidistant from both, if it wasn't in the middle then the distance to one of the points will be greater than the distance to the other.

    Hopefully someone understands what I mean.
     
  2. jcsd
  3. Oct 14, 2011 #2
    I think the way to go about this is to go from the other direction than what you're trying. Start by defining what it means for a point (x,y,z) to be equidistant from p and q. That means that the distance from p to (x,y,z) is some number, say r, and the distance from q to (x,y,z) is also r. Then set them equal to each other and play with the algebra. The squared terms (i.e. x^2, not x1^2 - x1^2 is a constant) should cancel, leaving you with a plane equation
     
  4. Oct 15, 2011 #3

    HallsofIvy

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    In other words, think geometrically. Geometrically, the line equidistant from two given points in the plane, is the perpendicular bisector of the line segment between the two points. Can you generalize that to three dimensions?
     
    Last edited: Oct 15, 2011
  5. Oct 15, 2011 #4
    Hi thanks for the replies. I'm still a bit lost on this.

    I tried the method suggested by tjackson. Basically if r is the distance from p to x,y,z and from q to x,y,z then they should be equal. Makes sense. However how do i even get this equation without knowing the points p and q? the distance equation gives

    sqrt [(x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2]

    so then in this context i would have the points from p (some x, some y, some z) - (x,y,z) and do the same for q and set them equal.

    how do i derive a plane equation from this though?


    Hmm, visually this doesn't make sense to me. I thought that the only equidistant points from p and q would lie on the line that goes in between them. Why would it be on the perpendicular bisector of that line? If the points lie anywhere other than the middle of p and q, it wouldn't be equidistant anymore?

    Edit: Oh wait, i think i get it. In 3 dimensions there are pretty much an infinite number of lines that lie between p and q which form a plane...am I thinking about this right?
     
    Last edited: Oct 15, 2011
  6. Oct 15, 2011 #5

    HallsofIvy

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    Yes, the set of all points equidistant from two given points in three dimensions is the plane perpendicular to the line between the two points and passing through the midpoint of that line segment.
     
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