Three Dimensional Geometry Question 409
Question: If $ l_1,m_1,n_1 $ and $ l_2,m_2,n_2 $ are direction consines of the two lines inclined to each other at an angle $ \theta $ , the direction cosines of the bisector of the angle between these lines are
Options:
A) $ \frac{l_1-l_2}{2\sin \frac{\theta }{2}},\frac{m_1-m_2}{2\sin \frac{\theta }{2}},\frac{n_1-n_2}{2\sin \frac{\theta }{2}} $
B) $ \frac{l_1-l_2}{2\cos \frac{\theta }{2}},\frac{m_1-m_2}{2\cos \frac{\theta }{2}},\frac{n_1-n_2}{2\cos \frac{\theta }{2}} $
C) $ \frac{l_1-l_2}{2\sin \frac{\theta }{2}},\frac{m_1-m_2}{2\sin \frac{\theta }{2}},\frac{n_1-n_2}{2\sin \frac{\theta }{2}} $
D) $ \frac{l_1-l_2}{2\cos \frac{\theta }{2}},\frac{m_1-m_2}{2\cos \frac{\theta }{2}},\frac{n_1-n_2}{2\cos \frac{\theta }{2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Let the lines $ L_1 $ and $ L_2 $ intersect at 0 say origin, Consider points P and Q on these lines such that OP=OQ=1. Then coordinates of P and Q are $ (l_1,m_1,n_1) $ and $ (l_2,m_2,n_2) $ . Their mid-point $ ( \frac{l_1+l_2}{2},\frac{m_1+m_2}{2},\frac{n_1+n_2}{2} ) $ lies on the bisector L … So, direction ratios of L are $ \frac{l_1+l_2}{2},\frac{m_1+m_2}{2},\frac{n_1+n_2}{2} $ Also, $ OR=OP\cos \frac{\theta }{2}=\cos \frac{\theta }{2} $
$ \therefore $ Direction cosines of L are $ \frac{l_1+l_2}{2\cos \frac{\theta }{2}},\frac{m_1+m_2}{2\cos \frac{\theta }{2}},\frac{n_1+n_2}{2\cos \frac{\theta }{2}} $ Similarly for other bisector we can replace $ l_2,m_2,n_2, $ by $ -l_2,-m_2,-n_2 $ and $ \theta $ by $ \pi -\theta $
BETA
