question_answer

The three lines through the origin with direction cosines \[{{l}_{1}},{{m}_{1}},{{n}_{1}};{{l}_{2}},{{m}_{2}},{{n}_{2}};\]and \[{{l}_{3}},{{m}_{3}},{{n}_{3}}\]are coplanar, if

A) \[{{l}_{1}}({{l}_{2}}-{{l}_{3}})+{{m}_{1}}({{m}_{2}}-{{m}_{3}})+{{n}_{1}}({{n}_{2}}-{{n}_{3}})=0\]

B) \[{{l}_{1}}({{m}_{2}}{{n}_{3}}-{{m}_{3}}{{n}_{2}})+{{m}_{1}}({{n}_{2}}{{l}_{3}}-{{n}_{3}}{{l}_{2}})+{{n}_{1}}({{l}_{2}}{{m}_{3}}-{{l}_{3}}{{m}_{2}})=0\]

C)

D) None of these

Correct Answer: C

Solution :

If the given lines are coplanar, then a line which is normal to the plane in which they lie, will be perpendicular to all the three given lines. Let the direction cosines of the normal be \[\lambda ,\mu ,\upsilon \]. \[\therefore \] \[\lambda {{l}_{1}}\mu {{m}_{1}}+\upsilon {{n}_{1}}=0,\lambda {{l}_{2}}+\mu {{m}_{2}}+\upsilon {{n}_{2}}=0,\] \[\lambda {{l}_{3}}+\mu {{m}_{3}}+\upsilon {{n}_{3}}=0.\]Eliminating \[\lambda ,\mu ,\upsilon \]from these equations, we get,