Find the distance between the lines l1 and l2 given by r = i + 2 … - Vidyadip

Question

Find the distance between the lines l₁and l₂ given by
$\vec r = \hat i + 2\hat j - 4\hat k + \lambda (2\hat i + 3\hat j + 6\hat k)$
and $\vec r = 3\hat i + 3\hat j - 5\hat k + \mu (2\hat i + 3\hat j + 6\hat k)$

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Answer

${\vec a_1} = \hat i + 2\hat j - 4\hat k$
${\vec a_2} = 3\hat i + 3\hat j - 5\hat k$
${\vec b_1} = {\vec b_2}=\vec b$$=2\hat i+3\hat j+6\hat k$
$|\vec b|=\sqrt{4+9+36}=\sqrt {49}$
Hence lines are parallel.
${\vec a_2} - {\vec a_1} = 2\hat i + \hat i - \hat k$
$\vec b \times \left( {{{\vec a}_2} - {{\vec a}_1}} \right) = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 2&3&6 \\ 2&1&{ - 1} \end{array}} \right|$
$=-9\vec i+14\vec j-4\vec k$
$d = \left| {\frac{{\vec b \times \left( {{{\vec a}_2} - {{\vec a}_1}} \right)}}{{\left| {\vec b} \right|}}} \right|$
$ = \left| {\frac{{ - 9\hat i + 14\hat j - 4\hat k}}{{\sqrt {49} }}} \right|$
$=\frac{{\sqrt {81+196+16} }}{{\sqrt {49} }} = \frac{{\sqrt {293} }}{7}$

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