Answer:
Given, \[3l+m+5n=0\] ?(i) and \[6mn-2nl+5lm=0\] ?(ii) From Eq. (i), we have \[m=-\,3l-5n\] On putting the value of m in Eq. (ii), we get \[6(-\,5n\,-3l)n\,-2nl\,+5l(-\,3l-5n)=0\] \[\Rightarrow \] \[-\,30{{n}^{2}}-18\ln \,-2nl-25\ln \,-15{{l}^{2}}=0\] \[\Rightarrow \] \[-\,30{{n}^{2}}-45nl\,-15{{l}^{2}}=0\] \[\Rightarrow \] \[2{{n}^{2}}+3nl+{{l}^{2}}=0\] [dividing both sides by \[(-\,15)\]] \[\Rightarrow \] \[(n+l)(2n+l)=0\] Either \[n=-\,l\,\,\,\text{or}\,\,l=-\,2n\] Now, if \[l=-\,n,\] then \[m=-\,2n\] and if \[l=-\,2n,\] then m = n Thus, the direction ratios of two lines are proportional to \[-\,n,\] \[-\,2n,\] \[n\] and \[-\,2n,\] \[n\], \[n\] i.e. \[-\,1,\,\,-\,2,\,\,1\] and \[-\,2,\,\,1,\,\,1.\] Let the required angle be \[\theta .\] then, \[\cos \theta =\frac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\] \[\cos \theta =\frac{-\,1(-\,2)-\,2(1)+1(1)}{\sqrt{{{(-\,1)}^{2}}+{{(-\,2)}^{2}}+{{(1)}^{2}}}\sqrt{{{(-\,2)}^{2}}+{{1}^{2}}+{{1}^{2}}}}\] \[=\frac{1}{\sqrt{6}\cdot \sqrt{6}}=\frac{1}{6}\] \[\therefore \] \[\theta ={{\cos }^{-1}}\left( \frac{1}{6} \right)\]
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