A) \[{{l}_{1}}+{{l}_{2}}+{{l}_{3}},{{m}_{1}}+{{m}_{2}}+{{m}_{3}},{{n}_{1}}+{{n}_{2}}+{{n}_{3}}\]
B) \[\frac{{{l}_{1}}+{{l}_{2}}+{{l}_{3}}}{\sqrt{3}},\frac{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}{\sqrt{3}},\frac{{{n}_{1}}+{{n}_{2}}+{{n}_{3}}}{\sqrt{3}}\]
C) \[\frac{{{l}_{1}}+{{l}_{2}}+{{l}_{3}}}{3},\frac{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}{3},\frac{{{n}_{1}}+{{n}_{2}}+{{n}_{3}}}{3}\]
D) None of these
Correct Answer: B
Solution :
Since the three lines are mutually perpendicular, \[\therefore \] \[{{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}=0\] \[{{l}_{2}}{{l}_{3}}+{{m}_{2}}{{m}_{3}}+{{n}_{2}}{{n}_{3}}=0\] \[{{l}_{3}}{{l}_{1}}+{{m}_{3}}{{m}_{1}}+{{n}_{3}}{{n}_{1}}=0\] Also,\[l_{1}^{2}+m_{1}^{2}+n_{1}^{2}=1,\,l_{2}^{2}+m_{2}^{2}+n_{2}^{2}=1,l_{3}^{2}+m_{3}^{2}+n_{3}^{2}=1\] Now, \[{{({{l}_{1}}+{{l}_{2}}+{{l}_{3}})}^{2}}+{{({{m}_{1}}+{{m}_{2}}+{{m}_{3}})}^{2}}+{{({{n}_{1}}+{{n}_{2}}+{{n}_{3}})}^{2}}\] = \[(l_{1}^{2}+m_{1}^{2}+n_{1}^{2})+(l_{2}^{2}+m_{2}^{2}+n_{2}^{2})+(l_{3}^{2}+m_{3}^{2}+n_{3}^{2})\] +\[2({{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}})+2({{l}_{2}}{{l}_{3}}+{{m}_{2}}{{m}_{3}}+{{n}_{2}}{{n}_{3}})\] \[+2({{l}_{3}}{{l}_{1}}+{{m}_{3}}{{m}_{1}}+{{n}_{3}}{{n}_{1}})\] = 3 Þ \[{{({{l}_{1}}+{{l}_{2}}+{{l}_{3}})}^{2}}+{{({{m}_{1}}+{{m}_{2}}+{{m}_{3}})}^{2}}+{{({{n}_{1}}+{{n}_{2}}+{{n}_{3}})}^{2}}=3\] Hence, direction cosines of required line are : \[\left( \frac{{{l}_{1}}+{{l}_{2}}+{{l}_{3}}}{\sqrt{3}},\,\frac{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}{\sqrt{3}},\frac{{{n}_{1}}+{{n}_{2}}+{{n}_{3}}}{\sqrt{3}} \right)\] Note: Students should remember it as a fact.
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