Let's imagine in 3D space I have three angle, $\theta_x$,$\theta_y$,$\theta_z$ respectively with X-axis, Y-axis, Z-axis.
Just like $\alpha,\beta,\gamma$ here:
I also have $\theta_{xy},\theta_{yz},\theta_{zx}$ angle respectively with plane-XY, plane-YZ, plane-ZX.
Just like r,$\theta$,$\sigma$ here:
What are relation between them?
Best Answer
Let's name a unit vector $v$ with these angles, so $$ v = \pmatrix{x \\y\\z} $$ with $x^2 + y^2 + z^2 = 1$. The assumption on the axis-angles says that $$ \cos \theta_x = x\\ \cos \theta_y = y\\ \cos \theta_z = z $$ What about $\theta_{xy}$? The angle between $v$ and the $xy$-plane is ... well, that depends. Suppose that $z > 0$. Then the angle is somewhere between $0$ and $\pi/2$, right? What if $z < 0$ --- is the angle between $-\pi/2$ and $0$? Or is it again in the interval $[0, \pi/2]$?
I'm going to assume the first case. Let's start with $z > 0$. Then the angle $\theta_z$ from $v$ to the $z$-axis is $\arccos(z)$, and so the angle from $v$ to the $xy$-plane is $\pi/2$ minus this, i.e., $\theta_{xy} = \pi/2 - \theta_z$.
Now let's look at the case $z < 0$. I'm going to assume that this produces an angle $\theta_{xy}$ that's in the interval $[-\pi/2, 0]$. When $\theta_z = \pi/2$, we want $\theta_{xy} = 0$; when $\theta_z = \pi$, we want $\theta_{xy} = -\pi/2$. Because we know they're linearly related, we see that $\theta_{xy} + \theta_z = \pi/2$, hence $\theta_{xy} = \pi/2 - \theta_z$.
Aha! The same formula applies in both cases, so we have $$ \theta_{xy} =\pi/2 - \theta_z\\ \theta_{xz} =\pi/2 - \theta_y\\ \theta_{yz} =\pi/2 - \theta_x $$