Find the normal form of the equation?

[Math9999]

Homework Statement

Find the normal form of the equation of the plane that contains the point P=(0, 1, 0) and has normal vector n=[3, 2, 1].

Homework Equations

None.

The Attempt at a Solution

I know that the general form: ax+by=c since that's written in my textbook. But why is the answer 3x+2y+z=2? I know that a=3, b=2 and c=1 based on the normal vector, but how did they get the 2 on the other side?

 

[fresh_42]

The equation $\vec{n} \cdot \vec{x} = 3x+2y+z$ gives a plane through the origin with the given normal vector. But we are looking for a parallel plane of it, which contains the given point $\vec{p}$. So parallel means $3x+2y+z=c$ for some constant $c$. Now what does it mean for $c$, that $\vec{p}=(0,1,0)$ satisfies this equation?

 

[Math9999]

So you're saying that c=n*p where 3*0+2*1+1*0=2, and c=2?

 

[fresh_42]

Yep. This way you make sure that the point is in the plane and it's orientation in space isn't changed.

 

[Math9999]

Thank you!