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Three Dimensional Geometry — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSThree Dimensional Geometry3 Marks
Question
Find the shortest distance between the lines $l_1$ and $l_2$ whose vector equations are
$\vec r = \hat i + \hat j + \lambda (2\hat i - \hat j + \hat k) ...(1)  $
and $\vec r = 2\hat i + \hat j - \hat k + \mu (3\hat i - 5\hat j + 2\hat k) ...(2)$
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Answer

${\vec a_1} = \hat i + \hat j,{\vec b_1} = 2\hat i - \hat j + \hat k$
${\vec a_2} = 2\hat i + \hat j - \hat k,{\vec b_2} = 3\hat i - 5\hat j + 2\hat k$
${\vec a_2} - {\vec a_1} = \hat i - \hat k$
${\vec b_1} \times {\vec b_2} = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 2&{ - 1}&1 \\ 3&{ - 5}&2 \end{array}} \right|$
$=\hat i (-2+5)-\hat j(4-3)+\hat k(-10+3)$
$ = 3\hat i - \hat j - 7\hat k$
$\left| {{{\vec b}_1} \times {{\vec b}_2}} \right|=\sqrt {9+1+49}= \sqrt {59} $
Also, $(\vec b_1×\vec b_2).(\vec a_2-\vec a_1)=(3\hat i-\hat j-7 \hat k)(\hat i-\hat k)=3+7+0=10$
$d=\left|\frac{(\vec b_1×\vec b_2).(\vec a_2-\vec a_1)}{|\vec b_1×\vec b_2|}\right|=\frac{10}{\sqrt{59}}$

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