A) \[2\]
B) \[4\]
C) \[6\]
D) \[8\]
Correct Answer: B
Solution :
Since,\[\alpha =\beta =\gamma \Rightarrow {{\cos }^{2}}\alpha +{{\cos }^{2}}\alpha +{{\cos }^{2}}\alpha =1\] \[\Rightarrow \] \[\Rightarrow \alpha +{{\cos }^{-}}\left( \pm \frac{1}{\sqrt{3}} \right)\] So, there are four line whose \[DC's\] are \[\left( \frac{1}{\sqrt{3}},\,\,\frac{1}{\sqrt{3}},\,\,\frac{1}{\sqrt{3}} \right),\,\,\left( \frac{-1}{\sqrt{3}},\,\,\frac{1}{\sqrt{3}},\,\,\frac{1}{\sqrt{3}} \right),\,\,\left( \frac{1}{\sqrt{3}},\,\,\frac{-1}{\sqrt{3}},\,\,\frac{1}{\sqrt{3}} \right)\]\[\left( \frac{1}{\sqrt{3}},\,\,\frac{1}{\sqrt{3}},\,\,\frac{-1}{\sqrt{3}} \right)\]
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