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.
$\therefore \cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
$=\frac{(2)(-1)+(5)(8)+(-3)(4)}{\sqrt{(2{)}^2+(5{)}^2+(-3{)}^2}\sqrt{(-1{)}^2+(8{)}^2+(4{)}^2}}$
$=\frac{-2+40-12}{\sqrt{4+25+9}\sqrt{1+64+16}}$
$=\frac{26}{\sqrt{38}\sqrt{81}}$
$=\frac{26}{9\sqrt{38}}$
$\therefore \theta ={\cos }^{-1}\left(\frac{26}{9\sqrt{38}}\right)$