question_answer

If \[{{l}_{1}},{{m}_{1}},{{n}_{1}}\] and \[{{l}_{2}},{{m}_{2}},{{n}_{2}}\] are direction consines of the two lines inclined to each other at an angle \[\theta \], the direction cosines of the bisector of the angle between these lines are

A) \[\frac{{{l}_{1}}-{{l}_{2}}}{2\sin \frac{\theta }{2}},\frac{{{m}_{1}}-{{m}_{2}}}{2\sin \frac{\theta }{2}},\frac{{{n}_{1}}-{{n}_{2}}}{2\sin \frac{\theta }{2}}\]

B) \[\frac{{{l}_{1}}-{{l}_{2}}}{2\cos \frac{\theta }{2}},\frac{{{m}_{1}}-{{m}_{2}}}{2\cos \frac{\theta }{2}},\frac{{{n}_{1}}-{{n}_{2}}}{2\cos \frac{\theta }{2}}\]

C) \[\frac{{{l}_{1}}-{{l}_{2}}}{2\sin \frac{\theta }{2}},\frac{{{m}_{1}}-{{m}_{2}}}{2\sin \frac{\theta }{2}},\frac{{{n}_{1}}-{{n}_{2}}}{2\sin \frac{\theta }{2}}\]

D) \[\frac{{{l}_{1}}-{{l}_{2}}}{2\cos \frac{\theta }{2}},\frac{{{m}_{1}}-{{m}_{2}}}{2\cos \frac{\theta }{2}},\frac{{{n}_{1}}-{{n}_{2}}}{2\cos \frac{\theta }{2}}\]

Correct Answer: C

Solution :

[c]

Let the lines \[{{L}_{1}}\]and \[{{L}_{2}}\] intersect at 0 say origin, Consider points P and Q on these lines such that OP=OQ=1. Then coordinates of P and Q are \[({{l}_{1}},{{m}_{1}},{{n}_{1}})\] and \[({{l}_{2}},{{m}_{2}},{{n}_{2}})\]. Their mid-point\[\left( \frac{{{l}_{1}}+{{l}_{2}}}{2},\frac{{{m}_{1}}+{{m}_{2}}}{2},\frac{{{n}_{1}}+{{n}_{2}}}{2} \right)\] lies on the bisector L ...

So, direction ratios of L are

\[\frac{{{l}_{1}}+{{l}_{2}}}{2},\frac{{{m}_{1}}+{{m}_{2}}}{2},\frac{{{n}_{1}}+{{n}_{2}}}{2}\]

Also, \[OR=OP\cos \frac{\theta }{2}=\cos \frac{\theta }{2}\]

\[\therefore \] Direction cosines of L are

\[\frac{{{l}_{1}}+{{l}_{2}}}{2\cos \frac{\theta }{2}},\frac{{{m}_{1}}+{{m}_{2}}}{2\cos \frac{\theta }{2}},\frac{{{n}_{1}}+{{n}_{2}}}{2\cos \frac{\theta }{2}}\]

Similarly for other bisector we can replace

\[{{l}_{2}},{{m}_{2}},{{n}_{2}},\] by \[-{{l}_{2}},-{{m}_{2}},-{{n}_{2}}\] and \[\theta \] by \[\pi -\theta \]