# Cartesian Equation of a plane containing 3 points

1. Aug 27, 2012

### Blackh4wk

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

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. Aug 27, 2012

### SammyS

Staff Emeritus
Hello Blackh4wk. Welcome to PF !

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

Once you find $\displaystyle \stackrel{\rightarrow}{\text{AB}}\times\stackrel{ \rightarrow}{\text{AC}}=\left\langle 6,\,-15,\,-18 \right\rangle\,,$ 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:
$\displaystyle 2x-5y-6z=d\,,$​
where d is a constant. Plugging in the coordinates of any point in the plane should give you d.