Orientation of a normal vector of a plane I found this question:
vector normal to a plane
It seems to be related to a problem I'm struggling with, but I need to know what is the rule for the normal vector's orientation. Assuming we want a normal unit vector to a plane, it could still go two ways. How is this solution oriented and why?
 A: A plane divides 3D space into two halfspaces (halves of the full space; one half is on one side of the plane, the other is on the other).
Because the surface normal is perpendicular to the plane, it always points to one halfspace.  Essentially, the normal distinguishes the two halfspaces, by always pointing to exactly one of them.

Let's say we have a plane described by its unit normal vector $\hat{n}$, and the signed distance $d$ from origin to the plane.  Point $\vec{p}$ is on the plane if and only if
$$\bbox{ \vec{p} \cdot \hat{n} - d = 0 }$$
This is just the special case for the signed distance between any point $\vec{p}$ and the plane described by $\hat{n}$ and $d$:
$$\bbox{ L = \vec{p} \cdot \hat{n} - d }$$
or, if you prefer the traditional Cartesian component form, $L = x n_x + y n_y + z n_z - d$, where $\hat{n} = ( n_x , n_y , n_z )$ and $\vec{p} = ( x , y , z )$.
Compare those signed distances $L$ to the real line. Effectively, the direction of the plane normal determines where the positive axis of the signed distances to that plane is.
(And $d$ is just the offset from coordinate system origin to the zero point on that line, measured along that line. In fact, if $\hat{n}$ is not an unit vector, $d$ measures the distance in units of $\left\lVert\hat{n}\right\rVert_2$, i.e. in units of the Euclidean length of the normal vector.  Which is why I think the analog I used above is apt.)
If we only knew the locus of points on the plane (i.e., where the plane is), without the normal, we could only have positive distances to the plane, since we'd have no way of telling one side from another.
