Finding orthogonal angles / polar coordinate of an n-dimensional vector

Question

Orthogonal angles are the angles used when converting a vector to polar coordinates. So for vector $(1, 1)$, the orthogonal angle is $45$ degrees.

Given a vector $(x_1, x_2, x_3, ..., x_n)$, what is a standard way of finding its orthogonal angles?

Answer 1

Polar coordinates as such are usually on defined the Cartesian plane, that is, strictly speaking your question only makes sense for pairs $(x, y) \in \Bbb R^2$. (One can build analogues of polar coordinates in higher dimensions, but there is more than way to do this.)

Even in this setting, there is an important ambiguity, namely, that two reference angles (what you call "orthogonal angles") can specify the same direction. For example, the angles $\frac{3\pi}{2}$ and $-\frac{\pi}{2}$ both corresponding to the negative $y$-direction. In practice, one usually handles this issue by picking a half-open interval of length $2 \pi$ in which the function sending a nonzero pair $(x, y)$ takes values; both $[0, 2\pi)$ and $(-\pi, \pi]$ are common choices. (The latter is the interval almost universally used in the angle function $\texttt{atan2(x, y)}$ common to many programming languages.) Still, this choice is essentially arbitrary, and in practice one must take care when working with this sort of function.

For points in the open right half-plane $\{x > 0\}$, some simply geometry shows that we can take our angle function to be $$\theta(x, y) = \arctan\left(\frac{y}{x}\right),$$ and the periodicity of tangent means that we can take the angle function to be $$\theta(x, y) = \arctan\left(\frac{y}{x}\right) + \pi$$ on the open left half-plane $\{x < 0\}$. We can extend this function continuously across the positive $y$-axis $\{(0, y) : y > 0\}$ by defining $$\theta(0, y) = \frac{\pi}{2}$$ for $y > 0$ and extend it to a (necessarily noncontinuous!) function $$\theta: \Bbb R^2 - \{(0, 0)\} \to \left(-\frac{\pi}{2}, \frac{3 \pi}{2}\right]$$ by declaring $$\theta(0, y) = \frac{3 \pi}{2}$$ on the negative $y$-axis $\{(0, y) : y < 0\}$.