Three Dimensional Geometry Question 106
Question: The direction cosines of a line equally inclined to three mutually perpendicular lines having direction cosines as $ l_1,m_1,n_1;l_2,m_2,n_2 $ and $ l_3,m_3,n_3 $ are
Options:
A) $ l_1+l_2+l_3,m_1+m_2+m_3,n_1+n_2+n_3 $
B) $ \frac{l_1+l_2+l_3}{\sqrt{3}},\frac{m_1+m_2+m_3}{\sqrt{3}},\frac{n_1+n_2+n_3}{\sqrt{3}} $
C) $ \frac{l_1+l_2+l_3}{3},\frac{m_1+m_2+m_3}{3},\frac{n_1+n_2+n_3}{3} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Since the three lines are mutually perpendicular,
$ \therefore $ $ l_1l_2+m_1m_2+n_1n_2=0 $ $ l_2l_3+m_2m_3+n_2n_3=0 $ $ l_3l_1+m_3m_1+n_3n_1=0 $ Also, $ l_1^{2}+m_1^{2}+n_1^{2}=1,,l_2^{2}+m_2^{2}+n_2^{2}=1,l_3^{2}+m_3^{2}+n_3^{2}=1 $ Now, $ {{(l_1+l_2+l_3)}^{2}}+{{(m_1+m_2+m_3)}^{2}}+{{(n_1+n_2+n_3)}^{2}} $ = $ (l_1^{2}+m_1^{2}+n_1^{2})+(l_2^{2}+m_2^{2}+n_2^{2})+(l_3^{2}+m_3^{2}+n_3^{2}) $ + $ 2(l_1l_2+m_1m_2+n_1n_2)+2(l_2l_3+m_2m_3+n_2n_3) $ $ +2(l_3l_1+m_3m_1+n_3n_1) $ = 3
Þ $ {{(l_1+l_2+l_3)}^{2}}+{{(m_1+m_2+m_3)}^{2}}+{{(n_1+n_2+n_3)}^{2}}=3 $ Hence, direction cosines of required line are : $ ( \frac{l_1+l_2+l_3}{\sqrt{3}},,\frac{m_1+m_2+m_3}{\sqrt{3}},\frac{n_1+n_2+n_3}{\sqrt{3}} ) $ Note: Students should remember it as a fact.
BETA
