# Homework Help: Equation of a plane

1. Mar 28, 2012

### molly16

1. The problem statement, all variables and given/known data
Determine the equation of the plane that passes through (5,-5,5) and is perpendicular to the lines of intersection of the planes 3x-2z+1=0 and 4x+3y+7=0

2. Relevant equations

Ax+By+Cz+D=0

3. The attempt at a solution

I found the cross product of the normals of the planes given. Then used that as the direction vector of the line of intersection. Then I let z=0 and solved for x and y using the equations of the planes given in order to find a point on the line of intersection. The equation for the line of intersection is r=(-1/3,-17/9,0) + s(6,8,9).

I'm not sure what to do now. Can someone please explain?

2. Mar 28, 2012

### HallsofIvy

Great! Now use the fact that if <A, B, C> is the normal to the plane and the plane contains the point $(x_0, y_0, z_0)$ then the plane is given by $A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0$. Of course, you use [itex](x_0, y_0, z_0)= (5, -5, 5).

The "equation of the line of intersection" is irrelevant. The point you are given is not near that line.

3. Mar 28, 2012

### molly16

Ohhh I see! Thanks for the help!