1. The problem statement, all variables and given/known data Give parametric questions for the plane : 2x-3y+z-6=0 3. The attempt at a solution i know that the normal is (2,-3,1) how do i find the direction vector of the plane?
The plane has two linearly independent 'direction vectors'. Find any two independent vectors perpendicular to your normal.
Dick is assuming that by 'direction vector', you mean two independent vectors in the plane. I've only seen the term used with lines. I wonder if you are not confusing this with "find the parametric equations for a line". To do that you would find 'direction vector' for the line. Parametric equations for a plane will involve two parameters. Here, since you can write z as a function of x and y, z= 6- 2x+ 3y, you can use x and y themselves as parameters or, if you prefer distinct variables, x= u, y= w, z= 6- 2u+3w.