Given:
- Two lines in $ℝ^3$ described by the vector equations: \begin{array}{lcl} \textbf{r}(t) & = & \textbf{r}_0+t\textbf{v} \\ \textbf{s}(t) & = & \textbf{s}_0+t\textbf{w} \end{array}
- The additional constraint that $\textbf{r}$ and $\textbf{s}$ are skew lines (meaning, neither parallel nor intersecting).
- An arbitrary vector $\textbf{u}$.
How can I construct a third line that...
- has direction $\textbf{u}$, and
- intersects both lines $\textbf{r}$ and $\textbf{s}$?
Can a general vector equation be derived for that third line?
Let the intersection point with the first line be
$\mathbf{P} = \mathbf{r_0} + t \mathbf{v} $
and with the second line be
$\mathbf{Q} = \mathbf{s_0} + s \mathbf{w} $
Then you want
$ \mathbf{Q} - \mathbf{P} = r \mathbf{u} $
The last equation can be written in matrix form as
$ \begin{bmatrix} \mathbf{w} && - \mathbf{v} && - \mathbf{u} \end{bmatrix} \begin{bmatrix} s \\ t \\ r \end{bmatrix} = \mathbf{r_0} - \mathbf{s_0} $
Solve this system for the unknown $s, t, r $, then your line will be
$ \mathbf{R} = \mathbf{P} + t \mathbf{u}, \hspace{25pt} t \in \mathbb{R}$