Show that through each point $(\cos\theta, \sin\theta, 0)$ on the hyperboloid of one sheet $$x^2+y^2=z^2+1$$
there are two lines which lie entirely within the hyperboloid. Two families of lines on the hyperboloid can be generated by varying $\theta$. Show that lines of the same family do not intersect, but that each line of the first family meets each line of the second.
Think I managed to get the two lines as
$$\begin{align}
r &= ( \cos\theta, \sin\theta, 0) + t\,(\phantom{-}\sin\theta,-\cos\theta,1) \\
r &= ( \cos\theta, \sin\theta, 0) + t\,(-\sin\theta,\phantom{-}\cos\theta, 1 )
\end{align}$$
but couldn't show the second part of the question. Any tips would be great.
Best Answer
You could show that given two different angles $\theta$ and $\phi$, that the lines from the same family generated by $\theta$ and $\phi$ are skew, i.e. they don't intersect. You could do this by setting the lines equal to each other (with different parameters, say $s$ and $t$) and attempting to solve the equations for the parameters of lines. There should be a contradiction or no solution somewhere.
Alternatively note that for each $\theta$ the two lines on the hyperboloid define a plane. Given two different angles $\theta$ and $\phi$, if those lines given by $\theta$ and $\phi$ are to intersect, they must intersect in the intersection of the two planes given by $\theta$ and $\phi$. If you can find these planes, then find their intersection, you've got a good start. Now note that the lines given by $\theta$ and $\phi$ must also intersect on the hyperboloid. Thus, the any intersection point happens on the intersection of the two planes and the hyperboloid. This should give only two possible points, which you could then exclude by plugging them into the equations for the lines.