A) \[{{\cos }^{-1}}\frac{1}{\sqrt{14}}\]
B) \[{{\sin }^{-1}}\frac{2}{\sqrt{14}}\]
C) \[{{\cos }^{-1}}\frac{2}{\sqrt{14}}\]
D) \[{{\sin }^{-1}}\frac{3}{\sqrt{14}}\]
Correct Answer: C
Solution :
Here direction of light is given by normal vector\[\vec{n}=\hat{i}+2\hat{j}+3\hat{k}\] \[\therefore \] angle made by the \[\vec{n}\] with y-axis is given by \[\cos \beta =\frac{2}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}}=\frac{2}{\sqrt{14}}\]
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