Intersection between rotating 3D line and 3D line I am trying to find the intersection between the surface of revolution described by rotating a 3D line $\vec{r}_1(s)$ an angle $\theta$ around the $z$ axis and another 3D line $\vec{r}_2$. In parametric form, I know that the solution to this problem is described by the solutions to
\begin{equation}
\begin{pmatrix} x_1(s)\cos{\theta} + y_1(s)\sin{\theta} \\ -x_1(s)\sin{\theta} + y_1(s)\cos{\theta}\\ z_1(s) \end{pmatrix} = \begin{pmatrix} x_2(t) \\ y_2(t) \\ z_2(t) \end{pmatrix}
\end{equation}
I don't know how to manipulate the equations to find the values possible values of $s,t,\theta$.
I am sure that if $\vec{r}_1$ pases through the z-axis, rotating it describes a cone with the z-axis as axis and finding the intersection is straight forward. However, in the case where $\vec{r}_1(s)$ doesn't intersect the $z$ axis I am no entirely sure of the surface that it describes. It is not necesarrily a cone for example, if $\vec{r}_2$ is parallel to the z-axis, the surface is a circular cylinder.
Using cylindrical coordinates, I have found that I have found the following equation for the surface: let $\vec{q}_1 = \vec{r}_1(0)= (x_1,y_1,z_1)$ and $\vec{vq}_1 =\vec{r}_1(1) - \vec{r}_1(0) = (vqx_1,vqy_1,vqz_1)$:
\begin{equation}
\rho^2 = (x_{1}^2 + y_{1}^2) + f(z)^2 (vqx_1^2 + vqy_1^2)
\end{equation}
With $f(z) = (z - z_1)/(vqz_1)$
I am not sure how to use that equation. I guess I could substitute $\rho^2 = x_2^2(t) + y_2^2(t)$ and $z = z_2(t)$ and try to solve for $t$ in order to find the intersection but I am not sure this would work.
 A: This is a comment, just too long to fit in the comment box.
Rotating a line segment $(x_0, y_0, z_0) - (x_1, y_1, z_1)$ (where $z_0 \ne z_1$) around the $z$ axis yields a surface of revolution
$$r(z) = \sqrt{ Z_2 z^2 + 2 Z_1 z + Z_0 } \tag{1a}\label{G1a}$$
where
$$\begin{aligned}
Z_2 &= \displaystyle \frac{(x_1 - x_0)^2 + (y_1 - y_0)^2}{(z_1 - z_0)^2} \\
Z_1 &= \displaystyle \frac{(x_1 - x_0)(x_0 z_1 - x_1 z_0) + (y_1 - y_0)(y_0 z_1 - y_1 z_0)}{(z_1 - z_0)^2} \\
Z_0 &= \displaystyle \frac{(x_0 z_1 - x_1 z_0)^2 + (y_0 z_1 - y_1 z_0)^2}{(z_1 - z_0)^2}
\end{aligned} \tag{1b}\label{G1b}$$
This is because each point on the line defines the radius for that $z$. Parametrising the line using $0 \le t \le 1$,
$$\begin{aligned}
x(t) &= (1-t) x_0 +  t x_1 \\
y(t) &= (1-t) y_0 +  t y_1 \\
z(t) &= (1-t) z_0 +  t z_1 \\
r(t) &= \sqrt{x(t)^2 + y(t)^2} \\
\end{aligned}$$
and solving $z(t) = z$ for $t$ yields
$$t = \frac{z - z_0}{z_1 - z_0}$$
Substituting this $t$ into $r(t)$ yields $\eqref{G1a}$ and $\eqref{G1b}$.
Note that you will be evaluating $r^2(z) = Z_2 z^2 + 2 Z_1 + Z_0$, really.

To calculate the intersection between a surface of revolution described by $r(z)$ and an (unrelated to above) line or line segment between points $(x_A, y_A, z_A)$ and $(x_B, y_B, z_B)$, we parametrise the line (segment) the same way as above, say using $\lambda$,
$$\begin{aligned}
x(\lambda) = (1 - \lambda) x_A + \lambda x_B \\
y(\lambda) = (1 - \lambda) y_A + \lambda y_B \\
z(\lambda) = (1 - \lambda) z_A + \lambda z_B \\
\end{aligned}$$
and the intersection(s) occur when $z(\lambda)$ is within the $z_0$ and $z_1$, and
$$x(\lambda)^2 + y(\lambda)^2 = r^2\bigl(z(\lambda)\bigr)$$
It turns out this has zero, one, or two real algebraic solutions.
First, calculate
$$\begin{aligned}
D &= Z_2 (z_B - z_A)^2 - (x_B - x_A)^2 - (y_B - y_A)^2 \\
R &= Z_2 ( (x_A z_B - x_B z_A)^2 + (y_A z_B - y_B z_A)^2 - Z_0 (z_B - z_A)^2 ) \\
~ &+ Z_1^2 (z_B - z_A)^2 - (x_A y_B - x_B y_A)^2 \\
~ &+ 2 Z_1 ( (x_B z_A - z_B x_A) (x_B - x_A) + (y_B z_A - y_A z_B)(y_B - y_A) ) \\
~ &+ Z_0 ( (x_B - x_A)^2 + (y_B - y_A)^2 ) \\
\end{aligned}$$
If $D = 0$ or $R \lt 0$, there is no intersection.
Otherwise, there is one or two real solutions
$$\lambda = \frac{S \pm \sqrt{R}}{D}$$
where
$$S = x_A (x_B - x_A) + y_A (y_B - y_A) - (Z_1 + Z_2 z_A) (z_B - z_A)$$
Note that $0 \le \lambda \le 1$ is on the line segment between the points, $\lambda \lt 0$ is before point $(x_A, y_A, z_A)$, and $\lambda \gt 1$ is after point $(x_B, y_B, z_B)$.
A: Hint.
Solving
$$
\cases{
\cos\theta x_1+\sin\theta y_1 = x_2\\
-\sin\theta x_1 + \cos\theta y_1 = y_2
}
$$
for $\{\sin\theta,\cos\theta\}$ we have
$$
\cases{
\sin\theta = \frac{x_2y_1-x_1 y_2}{x_1^2+y_1^2}\\
\cos\theta = \frac{x_1x_2+y_1y_2}{x_1^2+y_1^2}
}
$$
so
$$
\left(\frac{x_2y_1-x_1 y_2}{x_1^2+y_1^2}\right)^2+\left(\frac{x_1x_2+y_1y_2}{x_1^2+y_1^2}\right)^2 = 1
$$
but
$$
\cases{
(x_1,y_1,z_1) = p_{01}+\lambda_1\vec v_1\\
(x_2,y_2,z_2) = p_{02}+\lambda_2\vec v_2
}
$$
so making the pertinent substitutions we get two equations and two unknowns
$$
\cases{
\left(\frac{x_2y_1-x_1 y_2}{x_1^2+y_1^2}\right)_{\lambda_1,\lambda_2}^2+\left(\frac{x_1x_2+y_1y_2}{x_1^2+y_1^2}\right)_{\lambda_1,\lambda_2}^2 = 1\\
z_1(\lambda_1) =z_2(\lambda_2)
}
$$
now depending on $p_{01},p_{02},\vec v_1,\vec v_2$ the ruled surface and the line can or cannot intersect. If them do not intersect the system of a quadratic an a linear equation will have not real solutions for $\lambda_1,\lambda_2$
