Three Dimensional Geometry Question 323
Question: The direction cosines l, m, n of two lines are connected by the relations l + m + n = 0, lm = 0, then the angle between them is:
Options:
A) $ \pi /3 $
B) $ \pi /4 $
C) $ \pi /2 $
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Answer:
Correct Answer: A
Solution:
[a] Given d?c?s of two lines are $ \ell ,m,n $ connected by the relation $ \ell +m+n=0 $ and $ \ell m=0 $ Now, $ \ell +m+n=0\Rightarrow \ell =-m-n $
$ \Rightarrow \ell =-(m+n) $ And $ \ell m=0\Rightarrow -(m+n)m=0\Rightarrow -m^2 -mn=0 $ $ mn=-m^2; $ Therefore m and $ m+n=0 $ Then $ \frac{{\ell_1}}{-1}=\frac{m_1}{0}=\frac{n_1}{1} $ and if $ \ell +m+n=0 $ then $ \frac{{\ell_2}}{0}=\frac{m_2}{-1}=\frac{n_2}{1} $ $ ({\ell_1},m_1,n_1)=(-1,0,1) $ and $ ({\ell_2},m_2,n_2)=(0,-1,1) $ We know that angle between them $ \cos \theta =\frac{0+0+1}{\sqrt{1+0+1}\sqrt{0+1+1}}=\frac{1}{\sqrt{2}\sqrt{2}}=\frac{1}{2} $ $ \cos \theta =\frac{1}{2}=\cos 60{}^\circ \Rightarrow \theta =60{}^\circ \Rightarrow \theta =\frac{\pi }{3} $
BETA
