A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{2}\]
Correct Answer: C
Solution :
Given,\[l+m+n=0\],\[\Rightarrow \] \[l=-m-n\] and \[{{l}^{2}}+{{m}^{2}}-{{n}^{2}}=0\] \[\therefore \] \[{{(-m-n)}^{2}}+{{m}^{2}}-{{n}^{2}}=0\] \[\Rightarrow \] \[2{{m}^{2}}+2mn=0\] \[\Rightarrow \] \[2m(m+n)=0\] \[\Rightarrow \] \[m=0\]or\[m+n=0\] If\[m=0\], then\[l=-n\] \[\therefore \] \[\frac{{{l}_{1}}}{-1}=\frac{{{m}_{1}}}{0}=\frac{n}{1}\] and if\[m+n=0\Rightarrow m=-n\], then\[l=0\] \[\therefore \] \[\frac{{{l}_{2}}}{0}=\frac{{{m}_{2}}}{-1}=\frac{{{n}_{2}}}{1}\] \[ie\], \[({{l}_{1}},\,\,{{m}_{1}},\,\,{{n}_{1}})=(-1,\,\,0,\,\,1)\] and \[({{l}_{2}},\,\,{{m}_{2}},\,\,{{n}_{2}})=(0,\,\,-1,\,\,1)\] \[\therefore \]\[\cos \theta =\frac{0+0+1}{\sqrt{1+0+1}\sqrt{0+1+1}}=\frac{1}{2}\] \[\Rightarrow \] \[\theta =\frac{\pi }{3}\]
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