Three Dimensional Geometry Question 226
Question: The straight lines whose direction cosines are given by $ al+bm+cn=0,fmn+gnl+hlm=0 $ are perpendicular, if
Options:
A) $ \frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0 $
B) $ \sqrt{\frac{a}{f}}+\sqrt{\frac{b}{g}}+\sqrt{\frac{c}{h}}=0 $
C) $ \sqrt{af}=\sqrt{bg}=\sqrt{ch} $
D) $ \sqrt{\frac{a}{f}}=\sqrt{\frac{b}{g}}=\sqrt{\frac{c}{h}} $
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Answer:
Correct Answer: A
Solution:
From the first relation, $ n=-( \frac{al+bm}{c} ) $ Put the value of n in second relation,
$ fm,( -\frac{(al+bm)}{c} )+gl,( -\frac{(al+bm)}{c} )+hlm=0 $
or $ afml+bfm^{2}+agl^{2}+bglm-chlm=0 $
$ ag\frac{l^{2}}{m^{2}}+\frac{l}{m}(af+bg-ch)+bf=0 $ …..(i)
Now if $ l_1,m_1,n_1 $ and $ l_2,m_2,n_2 $ be direction cosines of two lines, then from (i) $ \frac{l_1l_2}{m_1m_2}=\frac{bf}{ag} $ , $ [ \text{Since roots of (i) are }\frac{l_1}{m_1},\frac{l_2}{m_2} ] $ or $ \frac{l_1l_2}{f/a}=\frac{m_1m_2}{g/b} $
Similarly, elimination of l will yield $ \frac{m_1m_2}{g/b}=\frac{n_1n_2}{h/c} $
$ \therefore ,\frac{l_1l_2}{f/a}=\frac{m_1m_2}{g/b}=\frac{n_1n_2}{h/c}=q $
(Say) We know that the lines are perpendicular, if $ l_1l_2+m_1m_2+n_1n_2=0 $
i.e., $ ( \frac{f}{a} ),q+( \frac{g}{b} ),q+( \frac{h}{c} ),q=0 $ or $ \frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0 $ .
Note: Student should remember this question as a fact.
BETA
