# Finding a plane

1. Nov 11, 2009

### doublemint

1. The problem statement, all variables and given/known data

Find the equation of the plane which contains the line:

(x,y,z) = (3,3,-3) + t(0,-3,-3)

and is parallel to the line:

(x,y,z) = (4,2,0) + t (3,-2,1)

Write your answer in the form 2 x + B y + C z = D, and give the values of B, C and D as your answer

2. Relevant equations
Cross Product
(x-x0, y-y0, z-z0) . (a, b, c) = 0

3. The attempt at a solution

So I need a point and a normal to determine the equation of the plane. Point: (3,3,-3). The normal I found was the cross product of (0,-3,-3) and (3,-2,1) = (-9,9,-9).
I then used the equation (x-x0, y-y0, z-z0) . (a, b, c) = 0 and got -9x+9y-9z=27.
Then i multiplied it by -2/9 since the answer needed to start with 2x, I get 2x-2y+2z=-6
but its wrong..

So any help would be appreciated!
Thank You!

2. Nov 11, 2009

### sutupidmath

i don't konw whether you did a mistake or me, when finding the normal vector on your plane because i got (-9,-9,9)

and for the final answer i'm getting x+y-z=9. what's the answer on yor book/notes?
look you might be getting a parallel plane, depending on what point you chose.

Last edited: Nov 11, 2009
3. Nov 11, 2009

### doublemint

4. Nov 11, 2009

### sutupidmath

yes, i got x+y-z=9, which is a parallel plane with yours.

5. Nov 11, 2009

### doublemint

Nice! Thanks sutupidmath!