A) \[\left| \,\begin{matrix} {{l}_{1}} & {{n}_{1}} & {{m}_{1}} \\ {{l}_{2}} & {{n}_{2}} & {{m}_{2}} \\ {{l}_{3}} & {{n}_{3}} & {{m}_{3}} \\ \end{matrix}\, \right|=0\]
B) \[\left| \,\begin{matrix} {{l}_{1}} & {{m}_{2}} & {{n}_{3}} \\ {{l}_{2}} & {{m}_{3}} & {{n}_{1}} \\ {{l}_{3}} & {{m}_{1}} & {{n}_{2}} \\ \end{matrix}\, \right|=0\]
C) \[{{l}_{1}}{{l}_{2}}{{l}_{3}}+{{m}_{1}}{{m}_{2}}{{m}_{3}}+{{n}_{1}}{{n}_{2}}{{n}_{3}}=0\]
D) None of these
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 \[l{{l}_{1}}+m{{m}_{1}}+n{{n}_{1}}=0\] ?..(i) \[l{{l}_{2}}+m{{m}_{2}}+n{{n}_{2}}=0\] ?..(ii) and \[l{{l}_{3}}+m{{m}_{3}}+n{{n}_{3}}=0\] ?..(iii) Solving (ii) and (iii), we get \[\frac{l}{{{m}_{2}}{{n}_{3}}-{{n}_{2}}{{m}_{3}}}=\frac{m}{{{n}_{2}}{{l}_{3}}-{{n}_{3}}{{l}_{2}}}=\frac{n}{{{l}_{2}}{{m}_{3}}-{{l}_{3}}{{m}_{2}}}=k\] Þ \[l=k({{m}_{2}}{{n}_{3}}-{{n}_{2}}{{m}_{3}}),\,m=k({{n}_{2}}{{l}_{3}}-{{n}_{3}}{{l}_{2}}),\,n=k({{l}_{2}}{{m}_{3}}-{{l}_{3}}{{m}_{2}})\] Substituting in (i), we get \[{{l}_{1}}({{m}_{2}}{{n}_{3}}-{{n}_{2}}{{m}_{3}})+{{m}_{1}}({{n}_{2}}{{l}_{3}}-{{n}_{3}}{{l}_{2}})+\,{{n}_{1}}({{l}_{2}}{{m}_{3}}-{{l}_{3}}{{m}_{2}})=0\] \[\Rightarrow \]\[\left| \,\begin{matrix} {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\ {{l}_{3}} & {{m}_{3}} & {{n}_{3}} \\ \end{matrix} \right|=0\] \[\Rightarrow \] ?\[\left| \,\begin{matrix} {{l}_{1}} & {{n}_{1}} & {{m}_{1}} \\ {{l}_{2}} & {{n}_{2}} & {{m}_{2}} \\ {{l}_{3}} & {{n}_{3}} & {{m}_{3}} \\ \end{matrix} \right|=0\].
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