Parametrization of skew lines for saddle like surface

Question

Given 2 skew lines $m_1$ and $m_2$ which do not lie in $\pi$. The surface is the set of all lines that intersect both $m_1$ and $m_2$ that are parallel to $\pi$.

To make things easier we can consider $\pi$ as the xz-axis. Since the combination of all lines can generate lines that are not parallel to $\pi$, we can restrict to only have the defining points at the same height.

Answer 1

Each line must join points on $m_1$ and $m_2$ having the same $y$ co-ordinate.

Let the equation of $m_k$ be $l_k = \vec r_k + \vec v_k t$

with $\vec r_k = (r_{kx}, r_{ky}, r_{kz} )$

and $v_k = (v_{kx}, v_{ky}, v_{kz} )$

Let $\vec p_1(a)$ be the point on $m_1$ with $\vec p_1.(0,1,0) = a$

$$\vec p_1(a) = \vec r_1 + \vec v_1 \bigg ( \frac{ a-r_{1y}}{ v_{1y}} \bigg ) $$

Let $\vec p_2(a)$ be the point on $m_2$ with $\vec p_2.(0,1,0) = a$

$$\vec p_2(a) = \vec r_2 + \vec v_2 \bigg ( \frac{ a-r_{2y}}{ v_{2y}} \bigg ) $$

So the surface $\pi$ can be parameterized as... $$ \pi(a,t) = \vec p_1(a) + t \bigg(\vec p_2(a) - \vec p_1(a) \bigg) $$

Answer 2

The two skew lines $\,m_i\,$ are parametrized by

$$ p_i + t\,\vec v_i \quad \text{ for } \quad i=1,2. $$

The planes parallel to the plane $\,\pi\,$ are parametrized by

$$ p + t\,\vec v \quad \text{ where } \quad \vec v \perp \vec v_\pi. $$

Suppose that the $\,v_i\,$ are such that $\,\vec v_i\cdot \vec v_\pi = 1 \,$ (which can be ensured by scaling the $\,\vec v_i\,$). Also suppose that the $\,p_i\,$ are in a plane parallel to $\,\pi\,$ (which can be ensured by using the two points of intersection of the skew lines with such a plane). Define the lines $\,\pi_t\,$ by

$$ \pi_t(\lambda) := \lambda(p_1 +t\,\vec v_1) + (1-\lambda)(p_2 + t\,\vec v_2). $$

For each $\,t,\,$ the line $\,\pi_t\,$ intersects $\,m_1\,$ and $\,m_2.\,$ Each line $\,\pi_t\,$ is in a plane parallel to $\,\pi.$ Thus, the $\,\pi_t\,$ form a ruled surface.