A) \[{{0}^{o}}\]
B) \[{{90}^{o}}\]
C) \[{{45}^{o}}\]
D) None of these
Correct Answer: B
Solution :
Given lines can be rewritten as \[\frac{x-2}{-1}=\frac{y}{2}=\frac{z+3}{1}\] and \[\frac{x-4}{4}=\frac{y-1}{1}=\frac{z-5}{2}\] Here, \[({{a}_{1}},{{b}_{1}},{{c}_{1}})=(1,2,1)\] and \[({{a}_{2}},{{b}_{2}},{{c}_{2}})=(4,1,2)\] \[\therefore \] \[\cos \theta =\frac{{{a}_{1}}{{a}_{2}}+{{b}_{2}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\] \[=\frac{-1\times 4+2\times 1+1\times 2}{\sqrt{{{(-1)}^{2}}+{{2}^{2}}+{{1}^{2}}}\sqrt{{{4}^{2}}+{{1}^{2}}+{{2}^{2}}}}\] \[=\frac{-4+2+2}{\sqrt{6}\,\sqrt{21}}=0\] \[\Rightarrow \] \[\theta ={{90}^{o}}\]
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