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Equation of plane given points

  1. Sep 11, 2012 #1
    1. The problem statement, all variables and given/known data

    Find equation of plane containing points: (1,1,5),(3,5,3),(8,8,1),(10,2,2),(18,6,-1),(-1,-3,6)


    2. Relevant equations

    Find 2 vectors given 3 points, using a common point. The cross product of these 2 vectors will be the normal vector of the plane. Use normal vector coords <a,b,c> as coefficients in the ax+by+cz=d formula where x,y,z is any point in the plane and solve for d. This equation better be true for all points.


    3. The attempt at a solution

    P = (1,1,5)
    Q = (3,5,3)
    R = (8,8,1)

    PQ = <2,4,-2>
    PR = <7,7,-4>

    PQ x PR = <-2,-6,14>

    Plug in point P into formula and get:

    -2x-6y+14z = 62

    Test formula by plugging in Q and don't get 62.
     
  2. jcsd
  3. Sep 11, 2012 #2
    Check the z-component of your cross product result.
     
  4. Sep 11, 2012 #3
    Okay I had 14 when should be -14

    PQ x PR = <-2,-6,-14>

    Plug in point P into formula and get:

    -2x-6y+14z = -78

    Now point Q works out to be -78 but the point (10,2,2) turns out to be -60

    Is this a trick question?
     
  5. Sep 11, 2012 #4

    I just used point p and the normal vector we now agree on to form the equation of a plane and got something different than what you have. You might have messed up the algebra.
     
  6. Sep 11, 2012 #5
    I forgot to update the formula with -14:


    -2x-6y-14z = -78

    Still when using (10,2,2) I get -60 and not -78
     
  7. Sep 11, 2012 #6
    Yeah that is weird, are you sure they mean all the points are in the same plane?
     
  8. Sep 11, 2012 #7
    That is how it is worded, reading it word for word. I thought it was weird too given 6 points instead of the common 3
     
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