A) parallel
B) perpendicular
C) inclined at \[{{\cot }^{-1}}\left( \frac{1}{6} \right)\]
D) none of these
Correct Answer: C
Solution :
[c] \[3l+m+5n=0\] ...(i) \[6mn-2nl+5ml=0\] ...(ii) Substituting the value of n from (i) in (ii), we get \[6{{l}^{2}}+9lm-6{{m}^{2}}=0\] Or \[6{{\left( \frac{l}{m} \right)}^{2}}+9\left( \frac{l}{m} \right)-6=0\] \[\therefore \frac{{{l}_{1}}}{{{m}_{1}}}=\frac{1}{2}\] and \[\frac{{{l}_{2}}}{{{m}_{2}}}=-2\] From Eq. (i), we get \[\frac{{{l}_{1}}}{{{n}_{1}}}=-1\] and \[\frac{{{l}_{2}}}{{{n}_{2}}}=-2\] \[\therefore \frac{{{l}_{1}}}{1}=\frac{{{m}_{1}}}{2}=\frac{{{n}_{1}}}{-1}=\sqrt{\frac{l_{1}^{2}+m_{1}^{2}+n_{1}^{2}}{1+4+1}}=\frac{1}{\sqrt{6}}\] and \[\frac{{{l}_{2}}}{2}=\frac{{{m}_{2}}}{-1}=\frac{{{n}_{2}}}{-1}=\frac{\sqrt{l_{2}^{2}+m_{2}^{2}+n_{2}^{2}}}{\sqrt{4+1+1}}=\frac{1}{\sqrt{6}}\] If \[\theta \] be the angle between the lines, then \[\cos \theta =\left( \frac{1}{\sqrt{6}} \right)\left( \frac{2}{\sqrt{6}} \right)+\left( \frac{2}{\sqrt{6}} \right)\left( -\frac{1}{\sqrt{6}} \right)+\left( -\frac{1}{\sqrt{6}} \right)\left( -\frac{1}{\sqrt{6}} \right)\]\[=\frac{1}{6}\] or \[\theta ={{\cos }^{-1}}\left( \frac{1}{6} \right)\]
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