You have a right-handed Cartesian coordinate basis of a three-dimensional Euclidean space, with axes 
, 
, and 
. You’re given some spatial-directional vectors of the form 
, where 
and 
are, respectively, the inclination angle (from the positive -axis zenith) and the azimuth angle (measured right-handed from the positive -axis) of an associated direction, which is unrelated to the direction of vector 
. How do you tell if the direction 
of such a vector falls inside the exterior half-space defined by the plane normal to ?
If it helps, visualize it this way. Travel to a point in space . Then, find a point on the unit sphere surrounding defined by (like latitude and longitude). Now, cut all of space in half with a plane passing through and perpendicular to the line between the origin and . Is the point on the unit sphere in the same half of space as the origin?
If 
, it’s dead simple: 
. Doesn’t even matter what is.
If 
and 
, there’s no effect on the azimuth, so it’s still pretty straightforward: 
. The 
accounts for the tilt of the plane, and the 
accounts for the azimuth.
In the general case where 
, and thus the azimuth angle relative to the plane’s direction of tilt differs from the absolute azimuth angle, it gets a bit trickier. First, we need the magnitude of 
for the 
bit:
Now, for 
, what we want is 
, where 
is the right-handed angle of from the positive -axis. Conceptually, subtracting is like transforming the azimuth angle into a new coordinate system where the plane tilt is back in the - plane (as if and 
). We could leave it like this, and have the reader calculate as:
But, with a well-known trigonometric identity we can make it a bit prettier:
More than a few scrap half-pages were harmed during the making of this inequality.