Find the shortest distance between the lines whose vector equations are $\vec r = (1 - t)\hat i + (t - 2)\hat j + (3 - 2t)\hat k$ and $\vec r = (s + 1)\hat i + (2s - 1)\hat j - (2s + 1)\hat k$

.: From the given equations,

we get
$\vec r = \hat i - 2\hat j + 3\hat k + t( - \hat i + \hat j - 2\hat k)$

...(1)
and $\vec r = \hat i - \hat j - \hat k + s(\hat i + 2\hat j - 2\hat k)$

….(2)
Equations (1) and (2) are of the form
$\vec r = {\vec a_1} + t{\vec b_1}$ and $\vec r = {\vec a_2} + s{\vec b_2}$

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

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

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

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

$= ( - 2 + 4)\hat i - (2 + 2)\hat j + ( - 2 - 1)\hat k = 2\hat i - 4\hat j - 3\hat k$.

$|{\vec b_1} \times {\vec b_2}| = \sqrt {{2^2} + {{( - 4)}^2} + {{( - 3)}^2}} = \sqrt {4 + 16 + 9} = \sqrt {29}$

Required shortest distance $= \left| {\cfrac{{({{\vec b}_1} \times {{\vec b}_2}) \cdot ({{\vec a}_2} - {{\vec a}_1})}}{{|{{\vec b}_2} \times {{\vec b}_1}|}}} \right|$

$= \left| {\cfrac{{(2\hat i - 4\hat j - 3\hat k) \cdot (\hat j - 4\hat k)}}{{\sqrt {29} }}} \right| = \left| {\cfrac{{ - 4 + 12}}{{\sqrt {29} }}} \right|$ $= \cfrac{8}{{\sqrt {29} }}$ units

Exercise - 11.3