= Solution
Assume the direction vectors are nonzero. The <cross product> equation is equivalent to
$$
(\mathbf x-\mathbf a)\times\mathbf u=0,
$$
so $\mathbf x-\mathbf a$ is parallel to $\mathbf u$ and the locus is the line $\mathbf x=\mathbf a+\lambda\mathbf u$.
For nonparallel directions, $\mathbf u_1\times\mathbf u_2$ is normal to both lines. Projecting their displacement onto the corresponding <unit vector> gives the <distance between skew lines>
$$
\boxed{d=
\frac{|(\mathbf a_2-\mathbf a_1)\mathbin\cdot
(\mathbf u_1\times\mathbf u_2)|}
{|\mathbf u_1\times\mathbf u_2|}}.
$$
If the directions are parallel, the corresponding formula is
$$
\boxed{d=\frac{|(\mathbf a_2-\mathbf a_1)\times\mathbf u_1|}{|\mathbf u_1|}}.
$$
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