Find the shortest distance between the lines whose vector equatio… - Vidyadip

Question

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

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Answer

We have,
$\vec r=\hat i - 2\hat j + 3\hat k + t( - \hat i + \hat j - 2\hat k)$
$\vec r = \hat i - \hat j - \hat k + s(\hat i + 2\hat j - 2\hat k)$
${\vec a_1} = \hat i - 2\hat j + 3\hat k$
${\vec b_1} = - \hat i + \hat j - 2\hat k$
${\vec a_2} = \hat i - \hat j - \hat k$
${\vec b_2} = \hat i + 2\hat j - 2\hat k$
${\vec a_2} - {\vec a_1} = \hat j - 4\hat k$
${\vec b_1} \times {\hat b_2} = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ { - 1}&1&{ - 2} \\ 1&2&{ - 2} \end{array}} \right|$
$ = 2\hat i - 4\hat j - 3\hat k$
$\left| {{{\vec b}_1} \times {{\vec b}_2}} \right| = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( { - 3} \right)}^2}} $
$ = \sqrt {29} $
$d = \left| {\frac{{\left( {{{\vec a}_2} - {{\vec a}_1}} \right)\left( {{b_1} \times {b_2}} \right)}}{{\left| {{{\vec b}_1} \times {{\vec b}_2}} \right|}}} \right|$
$ = \frac{8}{{\sqrt {29} }}$

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