Question

The angle between the pair of lines
$ \frac{x-2}{2} = \frac{y-1}{5} = \frac{z+3}{-3}$ and $\frac{x+2}{-1} = \frac{y-4}{8}= \frac{z-5}{4} $ is

• A. $ \cos^{-1} \left(\frac{21}{9\sqrt{38}}\right) $
• B. $ \cos^{-1} \left(\frac{23}{9\sqrt{38}}\right) $
• C. $ \cos^{-1} \left(\frac{24}{9\sqrt{38}}\right) $
• D. $ \cos^{-1} \left(\frac{25}{9\sqrt{38}}\right) $
• E. $ \cos^{-1} \left(\frac{26}{9\sqrt{38}}\right) $

Answer 1

Answer: (E)

Given lines are
$\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}$
and $\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$
dr' of above lines are $<2,5,-3>$ and $<-1,8,4>$ respectively.
$ \therefore \cos \,\theta =\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
$=\frac{(2)(-1)+(5)(8)+(-3)(4)}{\sqrt{(2)^{2}+(5)^{2}+(-3)^{2}} \sqrt{(-1)^{2}+(8)^{2}+(4)^{2}}} $
$=\frac{-2+40-12}{\sqrt{4+25+9} \sqrt{1+64+16}} $
$=\frac{26}{\sqrt{38} \sqrt{81}}$
$=\frac{26}{9 \sqrt{38}}$
$\therefore \theta=\cos ^{-1}\left(\frac{26}{9 \sqrt{38}}\right)$