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: 
Think of it this way: the plane described by the equation $ax+by+cz = 0$ is the set of all zeros of the function $f(x,y,z) = ax+by+cz$, i.e. the plane is an equipotential of $f$ with the potential $0$. The gradient $\nabla f$ of $f$ at some point is orthogonal to the equipotential at that point, and $\nabla f(x,y,z) = (a,b,c)$. Therefore, $(a,b,c)$ is orthogonal to the plane.

You can, of course, also employ some linear algebra: Let $u=(x\_1,y\_1,z\_1)$ and $v=(x\_2,y\_2,z\_2)$ be two points on the plane. Then we know that $ax\_1+by\_1+cz\_1=0$ and $ax\_2+by\_2+cz\_2=0$. This can be seen as a dot product with the vector $n=(a,b,c)$, i.e. $n\cdot u = 0$ and $n\cdot v = 0$. This then implies $n\cdot (u-v)=0$, i.e. the connecting vector of any two points on the plane is orthogonal to $n$, and therefore $n$ is orthogonal to the entire plane.

Note that this is only true in 3-dimensional space. In 2-dimensional space, it is lines that can be described by a single normal vector; in an $n$-dimensional space, it is $(n-1)$-dimensional hyperplanes.
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.