Three Dimensional Geometry Question 324
Question: The direction cosines of three lines passing through the origin are $ l_1,m_1,n_1;,l_2,m_2,n_2 $ and $ l_3,m_3,n_3 $ . The lines will be coplanar, if
Options:
A) $ | ,\begin{matrix} l_1 & n_1 & m_1 \\ l_2 & n_2 & m_2 \\ l_3 & n_3 & m_3 \\ \end{matrix}, |=0 $
B) $ | ,\begin{matrix} l_1 & m_2 & n_3 \\ l_2 & m_3 & n_1 \\ l_3 & m_1 & n_2 \\ \end{matrix}, |=0 $
C) $ l_1l_2l_3+m_1m_2m_3+n_1n_2n_3=0 $
D) None of these
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Answer:
Correct Answer: A
Solution:
Here, three given lines are coplanar if they have common perpendicular Let d.c.’s of common perpendicular be $ l,,m,,n $
Þ $ ll_1+mm_1+nn_1=0 $ ?..(i) $ ll_2+mm_2+nn_2=0 $ ?..(ii) and $ ll_3+mm_3+nn_3=0 $ ?..(iii) Solving (ii) and (iii), we get $ \frac{l}{m_2n_3-n_2m_3}=\frac{m}{n_2l_3-n_3l_2}=\frac{n}{l_2m_3-l_3m_2}=k $
Þ $ l=k(m_2n_3-n_2m_3),,m=k(n_2l_3-n_3l_2),,n=k(l_2m_3-l_3m_2) $ Substituting in (i), we get $ l_1(m_2n_3-n_2m_3)+m_1(n_2l_3-n_3l_2)+,n_1(l_2m_3-l_3m_2)=0 $
$ \Rightarrow $ $ | ,\begin{matrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \\ \end{vmatrix} =0 $
$ \Rightarrow $ ? $ | ,\begin{matrix} l_1 & n_1 & m_1 \\ l_2 & n_2 & m_2 \\ l_3 & n_3 & m_3 \\ \end{vmatrix} =0 $ .
BETA
