# Math Help - Plane, Sphere

1. ## Plane, Sphere

If you know the plane 2x+2y-z-18=0, you know that the normal vector to this plane is 2,2,-1. If you seek know direction vectors of the plane, can you say that (2,2,-1)(x,y,z)=0
-->
2x+2y-z=0 and then invent some points?
Are therefore (1,-1,0),(-1,1,0),(1,1,4) etc. all direction vectors of the plane or am i making some kind of mistake here?

2. Originally Posted by Schdero
If you know the plane 2x+2y-z-18=0, you know that the normal vector to this plane is 2,2,-1. If you seek know direction vectors of the plane, can you say that (2,2,-1)(x,y,z)=0
--> 2x+2y-z=0 and then invent some points?
Are therefore (1,-1,0),(-1,1,0),(1,1,4) etc. all direction vectors of the plane or am i making some kind of mistake here?
The is no such concept as direction vectors of the plane as far as I know.

3. There is. Maybe I give it a wrong name, but if you look at the parametric quation of a plane you of one point + t(directionvector1) + u(directionvector2). These two directionvectors, if that's there name, are normal to the normal vector of the plane, which is in my case 2,2,-1

4. Originally Posted by Schdero
If you know the plane 2x+2y-z-18=0, you know that the normal vector to this plane is 2,2,-1. If you seek know direction vectors of the plane, can you say that (2,2,-1)(x,y,z)=0
-->
2x+2y-z=0 and then invent some points?
Are therefore (1,-1,0),(-1,1,0),(1,1,4) etc. all direction vectors of the plane or am i making some kind of mistake here?
1. In German the vectors you are looking for are called "span vectors" (literally translated).

2. Since P(5,5,2) is a point in the plane the equation of the plane could be:

$\left(\begin{array}{c}x \\y \\ z \end{array}\right) = \left(\begin{array}{c}5 \\5 \\ 2\end{array} \right) + r \cdot \left(\begin{array}{c}1 \\-1 \\ 0\end{array} \right) + s \cdot \left(\begin{array}{c}1 \\1 \\ 4 \end{array} \right)$

which you easily can transform into the parametric equations of the plane.