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 $x$ $z$ plane (as if $y = 0$ and $x = r$). We could leave it like this, and have the reader calculate $\theta$ as: 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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