Pi Day Madness

You have a right-handed Cartesian coordinate basis of a three-dimensional Euclidean space, with axes x, y, and z. You’re given some spatial-directional vectors of the form (x,y,z,\rho,\eta), where \rho and \eta are, respectively, the inclination angle (from the positive z-axis zenith) and the azimuth angle (measured right-handed from the positive x-axis) of an associated direction, which is unrelated to the direction of vector (x,y,z). How do you tell if the direction (\rho,\eta) of such a vector falls inside the exterior half-space defined by the plane normal to (x,y,z)?

If it helps, visualize it this way. Travel to a point in space (x,y,z). Then, find a point on the unit sphere surrounding (x,y,z) defined by (\rho,\eta) (like latitude and longitude). Now, cut all of space in half with a plane passing through (x,y,z) and perpendicular to the line between the origin and (x,y,z). Is the point on the unit sphere in the same half of space as the origin?

If x = y = 0, it’s dead simple: \rho \geq \pi/2. Doesn’t even matter what \eta is.

If y = 0 and x > 0, there’s no effect on the azimuth, so it’s still pretty straightforward: \rho \geq \pi/2 + \sin\eta\arctan(x/z). The \arctan(x/z) accounts for the tilt of the plane, and the \sin\eta accounts for the azimuth.

In the general case where x, y \not= 0, 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 (x,y) for the \arctan bit:

r = \sqrt{x^2 + y^2}

Now, for z > 0, what we want is \rho \geq \pi/2 + \cos(\eta - \theta)\arctan(r/z), where \theta is the right-handed angle of (x,y) from the positive x-axis. Conceptually, subtracting \theta is like transforming the azimuth angle into a new coordinate system where the plane tilt is back in the xz plane (as if y = 0 and x = r). We could leave it like this, and have the reader calculate \theta as:

A Giant Ugly Mess

But, with a well-known trigonometric identity we can make it a bit prettier:

\rho \geq \frac{\pi}{2} + \left(\frac{y}{r}\sin\eta + \frac{x}{r}\cos\eta\right) \arctan\left(\frac{r}{z}\right)

More than a few scrap half-pages were harmed during the making of this inequality.

Mar 14th, 2010
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