Cartesian Equation of Perpendicular Plane to 3x-3z+3=0 at P(2,9,-3)

[Brendanphys]

Homework Statement

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)

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

 

[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 anyone you like. If you mean (tan(2x))^3, I would use the chain rule to differentiate it.