# Equation of a plane

1. Feb 24, 2010

### nhartung

1. The problem statement, all variables and given/known data

Find the equation of a plane with distance 3 units from the origin and perpendicular to the line through P(1,2,3) and Q(-2,4,1).

2. Relevant equations

$$\vec{n}$$ = $$\frac{\vec{PQ}}\left|{\vec{PQ}}\left|$$

Plane Equation: a(x-x0) + b(y-y0) + c(z-z0)

3. The attempt at a solution

Ok so I think I can solve this all the way up until the end.

We have $$\vec{PQ}$$ = <-3, -2, 2>

so $$\vec{n}$$ = $$\frac{1}{\sqrt{17}}$$<-3,-2,2>

Now if I scale that vector by 3 or -3 I can get a point on the plane that I am looking for, I need to put this into the form of a plane so I use the equation above and end up getting:
$$\frac{-3}{\sqrt{17}}$$x + $$\frac{2}{\sqrt{17}}$$y - $$\frac{2}{\sqrt{}17}$$z = $$\frac{-27}{17}$$ + $$\frac{18}{17}$$ + $$\frac{18}{17}$$

This can be simplified further but this is where mine and my professors work differs. He gets the following on the right side of the equation = $$\frac{27}{17}$$ + $$\frac{12}{17}$$ + $$\frac{12}{17}$$

Any ideas?

2. Feb 24, 2010

### nhartung

nevermind I figured it out.