# Finding cartesian equation of plane from 3 points

1. Jun 12, 2009

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. Jun 12, 2009

### Dick

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
3. Jun 13, 2009

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?

4. Jun 13, 2009