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 \[\pi -\theta \] |
You need to login to perform this action.
You will be redirected in
3 sec