Find shortest distance between lines in 3D Find shortest distance between lines given by
$$\frac{x-2}{3}=\frac{y-6}{4}=\frac{z+9}{-4}$$
and
$$\frac{x+1}{2}=\frac{y+2}{-6}=\frac{z-3}{1}$$
Is there any shortcut method for this problems?

Hint:
write the equations of the two lines in the form $\vec x=\vec p+t\vec q$:
$$
r_1) \qquad \begin{pmatrix}
x\\y\\z
\end{pmatrix}=\begin{pmatrix}
2\\6\\-9
\end{pmatrix}+t\begin{pmatrix}
3\\4\\-4
\end{pmatrix}
$$
$$
r_2) \qquad \begin{pmatrix}
x\\y\\z
\end{pmatrix}=\begin{pmatrix}
-1\\-2\\3
\end{pmatrix}+t\begin{pmatrix}
2\\-6\\1
\end{pmatrix}
$$
than, noted the the two lines are not parallel nor intersecting,  use the formula from here.
