A) \[{{\sin }^{-1}}\,\left( \frac{2\sqrt{2}}{3} \right)\]
B) \[{{\cos }^{-1}}\,\left( \frac{2\sqrt{2}}{3} \right)\]
C) \[{{\tan }^{-1}}\,\left( \frac{2\sqrt{2}}{3} \right)\]
D) \[{{\cot }^{-1}}\,\left( \frac{2\sqrt{2}}{3} \right)\]
Correct Answer: A
Solution :
The plane is \[2x-y+z=4\] and the line is \[\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z+1}{1}\] \[\therefore \,\,\,\sin \theta =\frac{2+1+1}{\sqrt{6}\sqrt{3}}=\frac{4}{\sqrt{18}}=\frac{2\sqrt{2}}{3}\Rightarrow \theta ={{\sin }^{-1}}\left( \frac{2\sqrt{2}}{3} \right)\].
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