A) \[\frac{2\pi }{3}\]
B) \[\frac{\pi }{6}\]
C) \[\frac{5\pi }{6}\]
D) \[\frac{\pi }{3}\]
Correct Answer: D
Solution :
Given, \[l+m+n=0\] ...(i) and \[{{l}^{2}}+{{m}^{2}}-{{n}^{2}}=0\] ...(ii) We know that, \[{{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1\] ...(iii) From Eqs. (ii) and (iii), \[2{{n}^{2}}=1\] \[\Rightarrow \] \[n=\pm \frac{1}{\sqrt{2}}\] From Eq. (i) \[l+m=-n\] \[\Rightarrow \] \[{{(l+m)}^{2}}={{(-n)}^{2}}\] \[\Rightarrow \] \[{{l}^{2}}+{{m}^{2}}+2lm={{n}^{2}}\] \[\Rightarrow \] \[{{l}^{2}}+{{m}^{2}}+2lm={{l}^{2}}+{{m}^{2}}\][from Eq. (ii)] \[\Rightarrow \] \[2lm=0\] \[\Rightarrow \] \[l=0\]or \[m+n=0\] \[\Rightarrow \] \[m-n=\mp \frac{1}{\sqrt{2}}\] \[\therefore \]Direction cosines of first line are \[0,\mp \frac{1}{\sqrt{2}},\pm \frac{1}{\sqrt{2}}\] If\[m=0\] \[\Rightarrow \] \[l+n=0\] \[\Rightarrow \] \[l=-n=\mp \frac{1}{\sqrt{2}}\] \[\therefore \]Direction cosines of second line are \[\mp \frac{1}{\sqrt{2}},0,\pm \frac{1}{\sqrt{2}}\] \[\therefore \] \[\cos \theta ={{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}\] \[\Rightarrow \] \[\theta ={{\cos }^{-1}}\left[ 0+0+\left( \pm \frac{1}{\sqrt{2}} \right)\left( \pm \frac{1}{\sqrt{2}} \right) \right]\] \[\Rightarrow \] \[\theta ={{\cos }^{-1}}\left( \frac{1}{2} \right)\] \[\Rightarrow \] \[\theta =\frac{\pi }{3}\]
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