Find the shortest distance between the lines whose vector equations are r = i + 2 j + 3 k + ( i - 3 j + 2 k ) and r = 4 i + 5 j + 6 k + ( 2 i + 3 j + k )

On comparing the given equations with: r= a _ 1 + b _ 1 , and r= a _ 2 + b _ 2 , we get: a_1 = i + 2 j + 3 k, b_1 = i - 3 j + 2 k, and a_2 = 4 i + 5 j + 6 k, b_2 = 2 i + 3 j + k S.D = | ( b_1 b_2 ).( a_2 - a_2 ) | b_1 b_2 | | = | ( - 9 i + 3 j + 9 k).(3 i + 3 j + 3 k) 171 | = | - 27 + 9 + 27 319 | = | 9 319 | = 3 19

Find the shortest distance between the lines whose vector equatio…

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Question

Find the shortest distance between the lines whose vector equations are $\vec r = \hat i + 2\hat j + 3\hat k$ + $\lambda \left( {\hat i - 3\hat j + 2\hat k} \right)$ and $\vec r = 4\hat i + 5\hat j + 6\hat k$ + $\mu \left( {2\hat i + 3\hat j + \hat k} \right)$

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Answer

On comparing the given equations with: $\overrightarrow{r}=\overrightarrow{{{a}_{1}}}+\lambda \overrightarrow{{{b}_{1}}},$ and $\overrightarrow{r}=\overrightarrow{{{a}_{2}}}+\mu \overrightarrow{{{b}_{2}}}$
, we get:
$\overrightarrow {{a_1}} = \hat i + 2\hat j + 3\hat k,\overrightarrow {{b_1}} = \hat i - 3\hat j + 2\hat k,$
and $\overrightarrow {{a_2}} = 4\hat i + 5\hat j + 6\hat k,\overrightarrow {{b_2}} = 2\hat i + 3\hat j + \hat k$
$\therefore S.D = \left| {\frac{{(\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} ).(\overrightarrow {{a_2}} - \overrightarrow {{a_2}} )}}{{\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right|}}} \right|$
$= \left| {\frac{{( - 9\hat i + 3\hat j + 9\hat k).(3\hat i + 3\hat j + 3\hat k)}}{{\sqrt {171} }}} \right|$
$= \left| {\frac{{ - 27 + 9 + 27}}{{3\sqrt {19} }}} \right| = \left| {\frac{9}{{3\sqrt {19} }}} \right| = \frac{3}{{\sqrt {19} }}$

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