Equation of a plane perpendicular to 2 other planes

[XtremeThunder]

Greetings, I am have a bit of a problem getting started on the following problem. The only thing that is confusing me is I am use to seeing a problem like this be perpendicular to 1 plane and not 2.

The problem I am working with is as follows:

Find the equation of a plane through the origin and perpendicular to the planes given by x-y+z=5 and 2x+y-2z=7.

I just need some direction on where to start.

Edit:
I am not quite sure what to do with both planes given.

The final equation should look similar to the following, correct?:
?(x-0)+?(y-0)+?(z-0)=0. Again, I am not sure how to come up with the coefficients using two given planes.

I spent about 4 hours on this problem and feel really lost and stupid!

-Joe

 

[Quinzio]

Are you familiar with the concept of normal vector ?

 

[XtremeThunder]

Yes, I am familiar with the concept of a normal vector.

Would I be correct in saying each plane's normal vector is i-j+k and 2i+j-2k, or is that totally off base?

Or would I use the cross product between the two?

 

[HallsofIvy]

Yes,exactly. If the new plane is perpendicular to both the given planes, its normal vector must be perpendicular to the normal vectors of both given planes- and thus parallel to the cross product of the two normal vectors

 

[XtremeThunder]

The coefficients of the equation of my first post (?(x-0)+?(y-0)+?(z-0)=0) would be the coefficients of the result of the cross product between the two normal vectors i-j+k and 2i+j-2k, correct, which would be the final answer?