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Help in equation of a plane?

  1. May 7, 2012 #1
    An equation for the plane that contains the three points $${(0, 0, 0)}$$, $${(1, 1, 1)}$$ and $${(2, 3, 4)}$$.

    A$${x}$$ - B$${y}$$ +C $${z}$$ =0

    Now I evaluated it, C comes out to be eual to A and B=2A. But that just means that A=B=C=0 which is just wrong.

    Then I used the http://www.ehow.com/how_8072475_equations-planes.html
    and evaluated A,B,C to be 1, 2, 1 respectively. But where was I making a mistake initially?
     
  2. jcsd
  3. May 7, 2012 #2
    plane equation is :ax+by+cz=d
    we substitute 3 points (0,0,0), (1,1,1) and (2,3,4) in the plane equation, therefore:
    (0,0,0) -> 0=d (1)
    (1,1,1) -> a+b+c=d (2)
    (2,3,4) -> 2a+3b+4c=d (3)

    by solving equations (1), (2) and (3) relative to a we have:
    d=0
    b=-2a
    c=a

    therefore the plane eq. is : ax-2ay+az=0 ->

    plane equation is: " x-2y+z=0"
     
  4. May 7, 2012 #3
    I don't think you got what I was asking. I arrived at the same equation you got
    ax-2ay+az=0 but how do we find a=1?
     
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