I came across this while solving some optics related question. Please have a look at the image link.
So, in the diagram, lines AO and BO lie on the XY plane. I know that $\ \angle AOX=\angle BOX=\theta$, $\angle POX=\phi$ and the angle made by the plane containing the lines PO and XO w.r.t. XY plane is $\psi$. I want to find $\angle POA$ and $\angle POB$ as a function of $\theta, \phi\ \text{and}\ \psi$. In the actual problem I am trying to solve, $\theta$ is constant but $\phi$ and $\psi$ are varying. Please help me out here.
Thank you!
It’s easy to see that the the unit vectors along $OA$ and $OB$ are $(\cos\theta, -\sin\theta,0)$ and $(\cos\theta, \sin\theta,0)$ respectively. Let $\hat{OP}=(x,y,z)$. Since it makes an angle $\phi$ with the $x$-axis, $$\cos \phi = (x,y,z)\cdot(1,0,0)=x$$
Now, the normal vector to the plane containing $OX,OP$ is $$\vec{n_1}=\hat{OX}\times \hat{OP} =(0,-z,y)$$ And the normal to the $XY$ plane is $\vec{n_2}=(0,0,1)$. The angle between two planes equals that between their normals, and so $$\cos\psi = \frac{\vec{n_1}\cdot \vec{n_2}}{|\vec{n_1|}}=\frac{y}{\sqrt{y^2+z^2}} =\frac{y}{\sqrt{1-x^2}}\\ \implies y=\sin\phi\cos\psi, z=\sin\phi\sin\psi$$
From here we get $$\cos\angle POA=\hat{OP}\cdot\hat{OA} =\cos\phi\cos\theta-\sin\theta\sin\phi\cos\psi \\\cos\angle POB=\hat{OP}\cdot\hat{OB}=\cos\phi\cos\theta+\sin\theta\sin\phi\cos\psi$$