Three Dimensional Geometry Question 155
Question: The pair of lines whose direction cosines are given by the equations $ 3l+m+5n=0 $ and $ 6mn-2nl+5lm=0 $ are
Options:
A) parallel
B) perpendicular
C) inclined at $ {{\cot }^{-1}}( \frac{1}{6} ) $
D) none of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ 3l+m+5n=0 $ …(i) $ 6mn-2nl+5ml=0 $ …(ii) Substituting the value of n from (i) in (ii), we get $ 6l^{2}+9lm-6m^{2}=0 $ Or $ 6{{( \frac{l}{m} )}^{2}}+9( \frac{l}{m} )-6=0 $
$ \therefore \frac{l_1}{m_1}=\frac{1}{2} $ and $ \frac{l_2}{m_2}=-2 $ From Eq. (i), we get $ \frac{l_1}{n_1}=-1 $ and $ \frac{l_2}{n_2}=-2 $
$ \therefore \frac{l_1}{1}=\frac{m_1}{2}=\frac{n_1}{-1}=\sqrt{\frac{l_1^{2}+m_1^{2}+n_1^{2}}{1+4+1}}=\frac{1}{\sqrt{6}} $ and $ \frac{l_2}{2}=\frac{m_2}{-1}=\frac{n_2}{-1}=\frac{\sqrt{l_2^{2}+m_2^{2}+n_2^{2}}}{\sqrt{4+1+1}}=\frac{1}{\sqrt{6}} $ If $ \theta $ be the angle between the lines, then $ \cos \theta =( \frac{1}{\sqrt{6}} )( \frac{2}{\sqrt{6}} )+( \frac{2}{\sqrt{6}} )( -\frac{1}{\sqrt{6}} )+( -\frac{1}{\sqrt{6}} )( -\frac{1}{\sqrt{6}} ) $ $ =\frac{1}{6} $ or $ \theta ={{\cos }^{-1}}( \frac{1}{6} ) $
BETA
