1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook