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
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.