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Cartesian Equation of a plane containing 3 points

  1. Aug 27, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the Cartesian Eqationn of the plane containing the points (0,-2,6), (1,8,-2) and (2,0,5).



    2. Relevant equations



    3. The attempt at a solution

    OA = 0,-2,6


    AB = 1,10,8


    AC = 2,2,-1

    Cross product of AB AC = 6,-15,-18

    X - 0 * 6
    Y - (-2) * -15
    Z - 6 * -18

    Which becomes,

    6x - 15y + 30 - 18z + 108 = 0

    Clean it up

    6x - 15y - 18z = -138

    Proper answer is

    2x - 5y - 6z = -26

    Now, I know how you get the LHS to = the proper answer (divide by 3), but how do I go about getting the RHS to equate properly?

    Or have I screwed up some of my arithmetic?
     
  2. jcsd
  3. Aug 27, 2012 #2

    SammyS

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    Hello Blackh4wk. Welcome to PF !

    No, your arithmetic looks OK, but you found the constant incorrectly.

    Once you find [itex]\displaystyle \stackrel{\rightarrow}{\text{AB}}\times\stackrel{ \rightarrow}{\text{AC}}=\left\langle 6,\,-15,\,-18 \right\rangle\,,[/itex] and notice that each of the components is a multiple of 3, you should know that the equation of the plane can be written as:
    [itex]\displaystyle 2x-5y-6z=d\,,[/itex]​
    where d is a constant. Plugging in the coordinates of any point in the plane should give you d.
     
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