Why can a plane be defined with its normal line? 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!
 A: 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

A: 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. 
A: 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.
A: 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. 
