A) \[{{15}^{0}}\]
B) \[{{30}^{0}}\]
C) \[{{60}^{0}}\]
D) \[{{45}^{0}}\]
Correct Answer: C
Solution :
Let \[{{l}_{1}},{{m}_{1}},{{n}_{1}}\]and \[{{l}_{2}},{{m}_{2}},{{n}_{2}}\]be the d. c of line 1 and 2 respectively, then as given \[{{l}_{1}}+{{m}_{1}}+{{n}_{1}}=0\] and \[{{l}_{2}}+{{m}_{2}}+{{n}_{2}}=0\] and \[l_{1}^{2}+m_{1}^{2}-n_{1}^{2}=0\]and \[l_{2}^{2}+m_{2}^{2}-n_{2}^{2}=0\] (\[\because \]\[l+m+n=0\]and \[{{l}^{2}}+{{m}^{2}}-{{n}^{2}}=0\]) Angle between lines, \[\theta \]is \[\cos \theta ={{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}\] ?(1) As given \[{{l}^{2}}+{{m}^{2}}={{n}^{2}}\]and \[l+m=-n\] \[\Rightarrow \]\[{{(-n)}^{2}}-2lm={{n}^{2}}\Rightarrow 2lm=0\]or \[lm=0\] So \[{{l}_{1}}{{m}_{1}}=0,{{l}_{2}}{{m}_{2}}=0\] If \[{{l}_{1}}=0,{{m}_{1}}\ne 0\]the \[{{l}_{1}}{{m}_{2}}=0\] If \[{{m}_{1}}=0,{{l}_{1}}\ne 0\]then \[{{l}_{2}}{{m}_{1}}=0\] If \[{{l}_{2}}=0,{{m}_{2}}\ne 0\]then \[{{l}_{2}}{{m}_{1}}=0\] If \[{{m}_{2}}=0,{{l}_{2}}\ne 0\]then \[{{l}_{1}}{{m}_{2}}=0\] Also \[{{l}_{1}}{{l}_{2}}=0\]and \[{{m}_{1}}{{m}_{2}}=0\] \[{{l}^{2}}+{{m}^{2}}-{{n}^{2}}={{l}^{2}}+{{m}^{2}}+{{n}^{2}}-2{{n}^{2}}=0\] \[\Rightarrow \]\[1-2{{n}^{2}}=0\Rightarrow n=\pm \frac{1}{\sqrt{2}}\] \[\therefore \] \[{{n}_{1}}=\pm \frac{1}{\sqrt{2}},{{n}_{2}}=\pm \frac{1}{\sqrt{2}}\] \[\therefore \]\[\cos \theta =\frac{1}{2}\theta ={{60}^{o}}\](acute angle)
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