Three Dimensional Geometry Question 405
Question: What is the value of n so that the angle between the lines having direction ratios (1, 1, 1) and (1, -1, n) is $ 60{}^\circ $ ?
Options:
A) $ \sqrt{3} $
B) $ \sqrt{6} $
C) 3
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] If $ (l_1,m_1,n_1) $ and $ (l_2,m_2,n_2) $ are the direction ratios then angle between the lines is $ \cos q=\frac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{l_1^{2}+m_1^{2}+n_1^{2}}\sqrt{l_2^{2}+m_2^{2}+n_2^{2}}} $ Here $ l_1=1,m_1=1,n_1=1 $ and $ l_2=1,m_2=-1,n_2=n $ and $ q=60{}^\circ $
$ \therefore \cos 60{}^\circ =\frac{1\times 1+1\times (-1)+1\times n}{\sqrt{1^{2}+1^{2}+1^{2}}\times \sqrt{1^{2}+1^{2}+n^{2}}} $
$ \Rightarrow \frac{1}{2}=\frac{n}{\sqrt{3}\sqrt{2+n^{2}}}\Rightarrow n^{2}=6\Rightarrow n=\pm \sqrt{6} $
BETA
