Three Dimensional Geometry — Class 12 Maths Solution

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

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

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

$= \hat i(4 + 4) - \hat j( - 2 - 6) + \hat k( - 2 + 6) = 8\hat i + 8\hat j + 4\hat k$

$\therefore$ $|{\vec b_1} \times {\vec b_2}| = \sqrt {64 + 64 + 16} = \sqrt {144} = 12$

$\therefore$ Shortest distance between the given lines

$= \left| {\cfrac{{({{\vec a}_2} - {{\vec a}_1}) \cdot ({{\vec b}_1} \times {{\vec b}_2})}}{{|{{\vec b}_1} \times {{\vec b}_2}|}}} \right| = \left| {\cfrac{{( - 10\hat i - 2\hat j - 3\hat k) \cdot (8\hat i + 8\hat j + 4\hat k)}}{{12}}} \right|$

$= \left| {\cfrac{{ - 80 - 16 - 12}}{{12}}} \right| = \left| { - \cfrac{{108}}{{12}}} \right| = 9$ units