Ruled surface generated by a rod and its shadow The sun is shines down the z-axis direction. A thin rod $OP$ of unit length has one end $O$ hinged at origin. Other end $P$ moves, making  angle $\gamma$ to the z-axis while its  shadow in $xy$ plane makes an angle $\delta$ to the x-axis. 
How to find a ruled surface parameterization described by rod in 3d if $ \mu = \sin \gamma/ \sin \delta $ remains constant?
Started solving with direction cosines.For $P$..
$$ x =\cos \alpha ,\, y= \cos \beta, \, z= \cos \gamma $$
 A: The position of each point in the rod is defined by its distance form the center and by two angles. We can understand one of them, $\gamma$, as the latitude and the other, $\delta$, as the azimuth to relate them to spherical coordinates $(r,\phi,\theta)$
The relation between $\gamma$ and $\delta$ is said to be such that $\mu = \sin \gamma/ \sin \delta$ is constant. Now,
$\gamma=\pi/2-\phi$, with $\phi$ the polar angle and $\delta=\theta$ with $\theta$ the azimuthal angle and the relation is $\mu=\cos\phi/\sin\theta$
So, for all the points in the surface, they are related by $\phi=\arccos(\mu\sin\theta)$ and the rod is along the radial distance r. One parametrization is
$$\begin{cases}
   r=u\\
   \phi=\arccos(\mu\sin v)\\
   \theta=v
\end{cases}$$
I've tried to draw this surface with GeoGebra but this program has not the capability to directly draw in spherical coordinates. The surface in cartesian is:
$$\begin{cases}
   x=u\sqrt{1-\mu^2\sin^2v}\cos v\\
   y=u\sqrt{1-\mu^2\sin^2v}\sin v\\
   z=u\mu\sin v
\end{cases}$$
I've checked it for values $\mu>1$. It has no tangent plane along the $z$ axis.
