### Pi Day Madness

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.