# Thread: Equation of a plane

1. ## Equation of a plane

Determine the equation of the plane that passes through points (1,0,-1), (0,2,3), and (0,0,0).

Letting the equation of the plane be ax+by+cz+d=0

Skipping through the steps...

$\displaystyle cx-\frac {3}{2}cy+cz=0$

$\displaystyle 2x-3y+2z=0$

How do you know c=2?

2. Hi

You could use a normal vector to the plane and ANY point on the plane to get your equation.

Your plane goes through the origin. You have two vectors in the plane:

$\displaystyle (1,0,-1) \mbox{ and } (0,2,3)$

We use the fact that a crossing two vectors (cross-product) produces a vector that is normal to the plane spanned by these two vectors:

$\displaystyle (1,0,-1) \times (0,2,3) = (2,-3,2)$

Now, since these vector is normal to the plane, $\displaystyle \underbrace{(2,-3,2) \cdot (x,y,z)=0}_{\text{Scalarproduct}} \Longleftrightarrow 2x-3y+2z=0$

3. I've never learned cross product, only inner product.

4. Originally Posted by chengbin
Determine the equation of the plane that passes through points (1,0,-1), (0,2,3), and (0,0,0).

Letting the equation of the plane be ax+by+cz+d=0

Skipping through the steps...

$\displaystyle cx-\frac {3}{2}cy+cz=0$
Just divide throughout by c and rearrange to get the equation of the plane ie 2x-3y+2z=0.