Vector Expression for lines of intersection for degenerate conic I specify a cone with a position vector $\mathbf{Q}_0$ as its vertex, a unit vector $\hat{\mathbf{q}}$ for its axis of rotation, with known angle $\alpha$ as the angle between any generator $\hat{\mathbf{d}}$ on the cone and its axis.  Thus the inner product $\hat{\mathbf{d}} \cdot \hat{\mathbf{q}} = \cos (\alpha)$. A plane with normal $\hat{\mathbf{\eta}}$ contains the vertex of the cone resulting in two lines of intersection with the cone and the plane in the directions of the vectors $\hat{\mathbf{d}}_{1,2}$.  I am looking for expressions for $\hat{\mathbf{d}}_{1,2}$.
Regards,
Cue
 A: I can think of a few ways to approach this. The most straightforward is to solve the system of equations that you have: $\hat{\mathbf d}_k\cdot\hat{\mathbf q}=\cos\alpha$ and $\hat{\mathbf d}_k\cdot\hat{\mathbf\eta}=0$, which are linear equations in the coordinates of $\hat{\mathbf d_k}$, and the quadratic equation $\hat{\mathbf d}_k\cdot\hat{\mathbf d}_k=1$.   
If you don’t want to solve a system that involves a quadratic equation, you can use the Pythagorean theorem instead. Find the intersection $\mathbf p$ of the two planes above and the plane through the origin with normal $\hat{\mathbf q}\times\hat{\mathbf\eta}$. (In homogeneous coordinates, this is the null vector of the matrix with the three planes as rows.) This point is the foot of the perpendicular from $\hat{\mathbf q}$ to the line of intersection of the two planes in the first paragraph. With this point in hand, you can compute the pair of vectors $$\mathbf p \pm \sqrt{\tan^2\alpha-\|\mathbf p-\hat{\mathbf q}\|^2}{\hat{\mathbf q}\times\hat{\mathbf\eta}\over\|\hat{\mathbf q}\times\hat{\mathbf\eta}\|}.$$ These will not be unit vectors, so you will need to normalize them if that’s what you want for $\hat{\mathbf d}_k$. There’s a related method in which you can construct $\mathbf p$ directly by using the angle between $\hat{\mathbf q}$ and the plane given by $\hat{\mathbf\eta}$.
A: Take a coordinate system that has the cone axis along the +x direction. So the axis is $$\hat{\mathbf{k}} = \pmatrix{1\\0\\0} $$
The all points on the cone can be parametrized by a distance $d$ and an angle $t$ as follows:
$$ \mathbf{r} = \pmatrix{x\\y\\z} =  \left| \matrix{1 & & \\ & \cos t & -\sin t \\ & \sin t & \cos t } \right| \pmatrix{d \cos \beta \\ d \sin \beta \\ 0 } $$
You can verify that $$ \mathbf{r} \cdot \hat{\mathbf{k}} = \| \mathbf{r} \| \cos \beta $$
Now consider a plane at an angle $\varphi$ rotated about the z-axis with normal vector $${\boldsymbol \eta} = \pmatrix{-\sin \varphi \\ \cos\varphi \\ 0} $$
The points on the plane are those that ${\boldsymbol \eta} \cdot \mathbf{r} = 0$
The is solved by $$ t= \cos^{-1} \left( \cot \beta \tan \varphi \right) $$ with limits $-\beta \le \varphi \le \beta $. 
On the rotated plane define two local coordinates $(u,v)$ such that
$$\pmatrix{x\\y\\z} = \pmatrix{u \cos\varphi \\ u \sin\varphi \\ v} $$
Here $u$ is along the plane closest to +x axis, and $v$ is along +z axis.
The left hand side of the above, restricted on the cone, is found by substituting $t$ into $\mathbf{r}$.
$$ \pmatrix{d \cos\beta \\ d \cos\beta \tan \varphi \\ \pm d \sqrt{1 - \frac{\cos^2 \beta}{\cos^2 \varphi} } }=\pmatrix{u \cos\varphi \\ u \sin\varphi \\ v}$$
The above is solved with the equation of the line(s)
$$ (\cos \beta) v  \pm (\cos\varphi \sqrt{1-\gamma^2} ) u =0$$ with $\gamma = \frac{\cos\beta}{\cos \varphi}$
You can find the angle $\theta$ between the two lines on the plane by noting that $$\tan \frac{\theta}{2} = \frac{{\rm d}v}{{\rm d}u} = \sqrt{\frac{1}{\gamma^2}-1}$$
