Find the shortest distance between the lines whose vector equations are $\vec r = (\hat i + 2\hat j + 3\hat k) + \lambda (\hat i - 3\hat j + 2\hat k)$ and $\vec r = 4\hat i + 5\hat j + 6\hat k + \mu (2\hat i + 3\hat j + \hat k)$

. : Equations of the given lines are of the form

$\vec r = {\vec a_1} + \lambda {\vec b_1}$
and $\vec r = {\vec a_2} + \mu {\vec b_2}$

where, ${\vec a_1} = \hat i + 2\hat j + 3\hat k,{\vec a_2} = 4\hat i + 5\hat j + 6\hat k$

and ${\vec b_1} = \hat i - 3\hat j + 2\hat k,{\vec b_2} = 2\hat i + 3\hat j + \hat k$

Here, ${\vec a_2} - {\vec a_1} = (4\hat i + 5\hat j + 6\hat k) - (\hat i + 2\hat j + 3\hat k) = 3\hat i + 3\hat j + 3\hat k$

Also, ${\vec b_1} \times {\vec b_2} = \left| {\begin{array}{cccccccccccccccccccc}{\hat i}&{\hat j}&{\hat k}\\1&{ - 3}&2\\2&3&1\end{array}} \right|$

$= \hat i( - 3 - 6) - \hat j(1 - 4) + \hat k(3 + 6) = - 9\hat i + 3\hat j + 9\hat k$

Required shortest distance
$= \left| {\cfrac{{({{\vec b}_1} - {{\vec b}_2}) \cdot ({{\vec a}_2} - {{\vec a}_1})}}{{|{{\vec b}_1} \times {{\vec b}_2}|}}} \right| = \left| {\cfrac{{ - 9 \times 3 + 3 \times 3 + 9 \times 3}}{{\sqrt {{{( - 9)}^2} + {3^2} + {9^2}} }}} \right|$

$\left| {\cfrac{9}{{\sqrt {{3^2}} \sqrt {{3^2} + 1 + {3^2}} }}} \right| = \left| {\cfrac{9}{{3\sqrt {19} }}} \right| = \cfrac{3}{{\sqrt {19} }}$ units.