In $\mathbb{R}^n$ two lines pass through $A$ and $B$, respectively, and their directions $r_1$ and $r_2$ (which are non-zero and non-parallel)
I know for $n=3$ the minimal euclidean distance between points on each line can be written
$$\frac{|(r_1\times r_2)\cdot AB|}{||r_1\times r_2||_2}$$
If $n$ is larger, is it possible to write something with determinants that works for all larger $n$?
Indicating by $\mathbf a$, $\mathbf b$, $\mathbf {p_1}$ and $\mathbf {p_2}$ the position vectors of the point $A$, $B$, generic point on line 1 and on line 2, then the parametric equations of the two lines will be $$ {\bf p}_{\,1} = {\bf a} + \lambda \,{\bf r}_{\,1} \quad {\bf p}_{\,2} = {\bf b} + \mu \,{\bf r}_{\,2} $$
Now impose that the difference vector between the two generic point be normal to each line $$ \eqalign{ & \left\{ \matrix{ \left( {{\bf p}_{\,1} - {\bf p}_{\,2} } \right) \cdot {\bf r}_{\,1} = 0 \hfill \cr \left( {{\bf p}_{\,1} - {\bf p}_{\,2} } \right) \cdot {\bf r}_{\,2} = 0 \hfill \cr} \right.\quad \Rightarrow \quad \left\{ \matrix{ \left( {{\bf a} + \lambda {\bf r}_{\,1} - {\bf b} - \mu {\bf r}_{\,2} } \right) \cdot {\bf r}_{\,1} = 0 \hfill \cr \left( {{\bf a} + \lambda {\bf r}_{\,1} - {\bf b} - \mu {\bf r}_{\,2} } \right) \cdot {\bf r}_{\,2} = 0 \hfill \cr} \right.\quad \Rightarrow \cr & \Rightarrow \quad \left\{ \matrix{ {\bf r}_{\,1} \cdot {\bf r}_{\,1} \,\lambda - {\bf r}_{\,2} \cdot {\bf r}_{\,1} \,\mu = \left( {{\bf b} - {\bf a}} \right) \cdot {\bf r}_{\,1} \hfill \cr {\bf r}_{\,2} \cdot {\bf r}_{\,1} \,\lambda - {\bf r}_{\,2} \cdot {\bf r}_{\,2} \,\mu = \left( {{\bf b} - {\bf a}} \right) \cdot {\bf r}_{\,2} \hfill \cr} \right. \cr} $$ you get a simple linear system in the unknowns $\lambda$ and $\mu$. After solving that, it is easy to find the modulus of ${\bf p}_{\,1} - {\bf p}_{\,2} $.
---- Addendum - Matrix notation -----
Indicating $$ \eqalign{ & {\bf P} = \left( {{\bf p}_{\,1} |{\bf p}_{\,2} } \right)\quad {\bf R} = \left( {{\bf r}_{\,1} |{\bf r}_{\,2} } \right)\quad {\bf C} = \left( {{\bf a}|{\bf b}} \right) \cr & {\bf \Lambda } = \left( {\matrix{ \lambda & 0 \cr 0 & \mu \cr } } \right)\quad {\bf u} = \left( {\matrix{ 1 \cr { - 1} \cr } } \right) \cr} $$ then
the parametric equations become $$ {\bf P = C + R}\;{\bf \Lambda } $$
the vector connecting generic points on the two lines is $$ {\bf d} = {\bf P}\,{\bf u} $$
the orthogonality condition becomes $$ \eqalign{ & {\bf 0} = {\bf R}^{\bf T} {\bf P}\,{\bf u} = {\bf R}^{\bf T} {\bf C}\,{\bf u + R}^{\bf T} {\bf R}\;{\bf \Lambda }\,{\bf u}\quad \Rightarrow \quad \cr & \Rightarrow \quad \;{\bf \Lambda }\,{\bf u} = - \;\left( {{\bf R}^{\bf T} {\bf R}} \right)^{\, - \,{\bf 1}} {\bf R}^{\bf T} {\bf C}\,{\bf u} \cr} $$
which gives as the ${\bf d}$ normal to both lines $$ \eqalign{ & {\bf d}_ \bot = {\bf C}\,{\bf u + R}\;{\bf \Lambda }\,{\bf u} = {\bf C}\,{\bf u} - {\bf R}\;\left( {{\bf R}^{\bf T} {\bf R}} \right)^{\, - \,{\bf 1}} {\bf R}^{\bf T} {\bf C}\,{\bf u} = \cr & = \left( {{\bf I} - {\bf R}\;\left( {{\bf R}^{\bf T} {\bf R}} \right)^{\, - \,{\bf 1}} {\bf R}^{\bf T} } \right){\bf C}\,{\bf u} \cr} $$
whose squared modulus is (the core matrix in ${\bf R}$ is idempotent)
It is interesting to note that for the said matrix we have $$ \left( {{\bf I} - {\bf R}\;\left( {{\bf R}^{\bf T} {\bf R}} \right)^{\, - \,{\bf 1}} {\bf R}^{\bf T} } \right){\bf R} = {\bf R}^{\bf T} \left( {{\bf I} - {\bf R}\;\left( {{\bf R}^{\bf T} {\bf R}} \right)^{\, - \,{\bf 1}} {\bf R}^{\bf T} } \right) = {\bf 0} $$ i.e. that its columns form a basis for the space normal to $ {\bf r}_{\,1} $ and $ {\bf r}_{\,2} $.