# Find parametric question for the plane

1. May 17, 2007

### bonbon

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?

2. May 17, 2007

### Dick

The plane has two linearly independent 'direction vectors'. Find any two independent vectors perpendicular to your normal.

3. May 18, 2007

### HallsofIvy

Staff Emeritus
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.

4. May 18, 2007

### Dick

That's a much more direct way. I was fixated on the 'direction vector' (basis) picture.