# Homework Help: Vector Question

1. Jun 21, 2009

### Brendanphys

1. The problem statement, all variables and given/known data
A plane has 3x-3z+3 = 0 as it's cartesian equation. Determine the cartesian equation of a plane that is perpendicular to this plane and contains the point P(2,9,-3)

3. The attempt at a solution
Since (3,0 ,-3) is the normal for the first plane, I figured it to be a dir'n vector for the second.

Now to find the cartesian eq'n I'd need to cross two direction vectors but I only have one. So I figure if I find AP, where A is a point on the plane and P is the point on the new plane, then I'd cross the two to find the new "normal" and thus the cartesian equation.

However, I am unsure of what point to use; should I just plug in random point to plane 1 to find A? How am I certain it is in the same direction as the perpindicular plane... or maybe I'm just approaching the question all wrong.

Any help would be greatly appreciated as I am getting prepared for my final.

p.s. if anyone knows how to find the derivative of (tan^3)2x, it'd be greatly appreciated

2. Jun 21, 2009

### Dick

If all you have is the point P and a direction vector, then you can make any choice of a second direction vector and the resulting plane will still be perpendicular to the original one. I guess you can pick any one you like. If you mean (tan(2x))^3, I would use the chain rule to differentiate it.