Creating a 3D Plane using the normal and point vector I'm not understanding the relationship of a normal vector and a position vector that makes it into a 3D plane, and how I can visualize what that 3D plane is going to look like in 3D space.
Say I wanted a 3D plane that extends along the X and Z axis from the origin out to some points like 30x and 30z, this leaves the plane sitting entirely flat on the y axis such as the image below:

Would the point vector just be (0,0,0) and the normal vector be (0,30,0)? 
I don't want to necessarily use the 3 point equation to obtain the plane's normal. Imagine I didn't know the 3 points other than the origin and I just wanted a REALLY big plane in the same x-z axis flat against y? What is the perpendicular vector actually saying about the plane's size?
 A: It's not entirely clear what you're asking, but the normal vector is the vector perpendicular to the plane. So for a plane that spans the $x$ and $y$ directions, the normal vector is in the $z$ direction, $(0,0,1)$ (note that by convention the normal vector is normalized, e.g. has unit norm). A plane by definition is infinite (though in CAD programs we usually just show a piece of it) and so the normal vector has nothing to do with its "size."
The way to move between spanning vectors and normal vectors is with the cross product. If $v_1$ and $v_2$ are two vectors tangent to the plane, the normal vector is given by
$$\frac{v_1 \times v_2}{\|v_1 \times v_2\|}$$
so for your example, using $v_1 = (30,0,0)$ and $v_2=(0,30,0)$ would give you a normal vector for your $xy$ plane.
You can also go from a normal vector $n$ to a spanning vector. Pick (almost) any vector $v$; then $v\times n$ is tangent to the plane, since the cross product of two vectors is always perpendicular to both vectors; similarly $(v\times n)\times n$ is also tangent, and these two tangent vectors span the plane.
