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Finding cartesian equation of plane from 3 points

  1. Jun 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Find a Cartesian equation of the plane P containing A (2, 0, −3) , B(1, −1, 6) and C(5, 5, 0) , and determine if point D(3, 2, 3) lies on P.

    2. Relevant equations

    vector cross product
    ax + by + cz = 0

    3. The attempt at a solution

    Take the cross product of AB and AC to get normal vector.

    AB = -i -j + 9k
    AC = 31 + 5j + 3k

    I used the determinant method at got:
    AB X AC = -48i + 30j -2k

    Now as A, B and C lie on P, take a point say A(2, 0, -3)

    -48(x - 2) +30(y) -2(z + 3) = 0

    rearranging that gives:

    -48x + 30y -2z = -90

    Then putting in the x, y and z values for D the equation holds.

    The question I have is that the answer for the plane given is:

    24x − 15y + z = 45

    Is there a more common method to follow to get this equation rather than the one I got?
  2. jcsd
  3. Jun 12, 2009 #2


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    The answer is the same one you got. It's just been divided by (-2). That doesn't change the plane. E.g. the equation 9x=3 has exactly the same solutions as 3x=1, doesn't it?
    Last edited: Jun 12, 2009
  4. Jun 13, 2009 #3
    Yes, but why divide by -2? My question was more, how would they get that answer instead? If I am calculating a different (but equivalent) answer then how?
  5. Jun 13, 2009 #4


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    Hi username12345! :smile:
    Same reason as why we write 3/2 and not 6/4 or 21/14. :wink:

    (and the minus because it's customary to start with x, so it makes sense for the x coefficient to be positive)
  6. Jun 13, 2009 #5


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    in other words, you do it in the simplest form, about the minus sign,don’t you think [guys] it’s because the many parts in the equation being negative, so we multiply it by the sign [-]?!
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