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: 
For a really down-to-earth explanation, how about this:

Imagine a satellite dish with a central aerial that you aim at a distant point (the aerial is the normal), and imagine that the "dish" is actually rather flat (this is your plane). Any plane passing through the place where you are can be achieved by rotating your dish appropriately, and you can get to any parallel plane just by moving a fixed distance along the direction your aerial points along.

So, you just need 2 pieces of information to specify your plane: 

* the direction of the aerial (= the normal)
* the distance you displace
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.