A) \[\sqrt{3}\]
B) \[\sqrt{6}\]
C) 3
D) None of these
Correct Answer: B
Solution :
[b] If \[({{l}_{1}},{{m}_{1}},{{n}_{1}})\] and \[({{l}_{2}},{{m}_{2}},{{n}_{2}})\] are the direction ratios then angle between the lines is \[\cos q=\frac{{{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}}{\sqrt{l_{1}^{2}+m_{1}^{2}+n_{1}^{2}}\sqrt{l_{2}^{2}+m_{2}^{2}+n_{2}^{2}}}\] Here \[{{l}_{1}}=1,{{m}_{1}}=1,{{n}_{1}}=1\] and \[{{l}_{2}}=1,{{m}_{2}}=-1,{{n}_{2}}=n\] and \[q=60{}^\circ \] \[\therefore \cos 60{}^\circ =\frac{1\times 1+1\times (-1)+1\times n}{\sqrt{{{1}^{2}}+{{1}^{2}}+{{1}^{2}}}\times \sqrt{{{1}^{2}}+{{1}^{2}}+{{n}^{2}}}}\] \[\Rightarrow \frac{1}{2}=\frac{n}{\sqrt{3}\sqrt{2+{{n}^{2}}}}\Rightarrow {{n}^{2}}=6\Rightarrow n=\pm \sqrt{6}\]
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