Question: The title is worded a bit confusingly. I apologize, I just couldn't think of how to phrase it.

Either way, say you have the plane $x+2y-4z=8$

The normal line will have direction $x=1, y=2, z=-4$

So they are very closely linked. However, I'm having trouble wrapping my head around *why* that is. How come you can list the normal vector and (along with a scalar) define a plane? I understand that it works, I just wonder how come?

I hope someone can explain this. Thanks!

Answer: 
View it *geometrically*. For example, let the normal be the $z$-axis. Geometrically, this makes no difference, since any line could be picked as the $z$-axis. Now you can *see* what planes have the $z$-axis as a normal. They are just the planes parallel to the $x$-$y$ plane.

Any such plane can be fully identified if you know in addition a single number, namely the height at which the plane meets the $z$-axis.
First, let's work backwards. If we're on the plane, and we head towards the y axis, x and z will head to zero. Then 2y=8, so y=4. So apparently, the point (0,4,0) is on the plane. Notice if we are at any other point in the plane, (x,y,z), then the displacement vector, (x,y,z)-(0,4,0) must be perpendicular to the normal, let's call it n=(a,b,c). Thus the dot product must be zero. 

Then n$\cdot$(x,y-4,z)=0, or ax+b(y-4)+cz=0. Or ax+by+cz=4b. But we know we have x+2y-4z=8, so n=(1,2,-4) works. 

The logic probably isn't exactly what you wanted, but it may have to do. It's better to remember that 

$$n\cdot(r-r\_0)=0,$$

since that is the equation most closely associated with the geometry.