A) \[\frac{\pi }{3}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{2}\]
D) 0
Correct Answer: A
Solution :
Given that, \[l+m+n=0\] ...(i) and \[lm=0\] ...(ii) \[\therefore \]From Eq. (i), \[l=-(m+n)\] and from Eq. (ii), \[-(m+n)m=0\] \[\Rightarrow \] \[m,\text{ }m+n=0\] if\[m=0,l+m+n=0,\]then \[\frac{{{l}_{1}}}{-1}=\frac{{{m}_{1}}}{0}=\frac{{{n}_{1}}}{1}\] and if \[l+m+n=0\] then, \[\frac{{{l}_{2}}}{0}=\frac{{{m}_{2}}}{-1}=\frac{{{n}_{2}}}{1}\] \[({{l}_{1}},{{m}_{1}},{{n}_{1}})=(-1,0,1)\] and \[({{l}_{2}},{{m}_{2}},{{n}_{2}})=(0,-1,1)\] \[\therefore \]Angle between them \[\cos \theta =\frac{0+0+1}{\sqrt{1+0+1}\sqrt{0+1+1}}=\frac{1}{2}\] \[\Rightarrow \] \[\theta =60{}^\circ =\frac{\pi }{3}\]
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