Question
Find the angle between the pair of lines given by $\vec r = 3\hat i + 2\hat j - 4\hat k + \lambda (\hat i + 2\hat j + 2\hat k)$ and $\vec r = 5\hat i - 2\hat j + \mu (3\hat i + 2\hat j + 6\hat k)$
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Answer
Given that, ${\vec b_1} = \hat i + 2\hat j + 2\hat k$
${\vec b_2} = 3\hat i + 2\hat j + 6\hat k$
$\cos \theta = \left| {\frac{{{{\vec b}_1}.{{\vec b}_2}}}{{\left| {{{\vec b}_1}} \right|\left| {{{\vec b}_2}} \right|}}} \right| = \frac{{19}}{{21}}$
$=\left| \frac{(\hat i +2\hat j+2\hat k).(3\hat i+2\hat j+6\hat k)}{\sqrt{1+4+4}\sqrt{9+4+36}}\right|$
$ =\frac{{3+4+12}}{3×7}$
$=\frac{19}{21}$
$\therefore \theta=cos^{-1}\frac{19}{21}$
${\vec b_2} = 3\hat i + 2\hat j + 6\hat k$
$\cos \theta = \left| {\frac{{{{\vec b}_1}.{{\vec b}_2}}}{{\left| {{{\vec b}_1}} \right|\left| {{{\vec b}_2}} \right|}}} \right| = \frac{{19}}{{21}}$
$=\left| \frac{(\hat i +2\hat j+2\hat k).(3\hat i+2\hat j+6\hat k)}{\sqrt{1+4+4}\sqrt{9+4+36}}\right|$
$ =\frac{{3+4+12}}{3×7}$
$=\frac{19}{21}$
$\therefore \theta=cos^{-1}\frac{19}{21}$
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