A) \[\sqrt{\frac{5}{3}}\]
B) \[-\frac{3}{5}\]
C) \[\sqrt{\frac{3}{5}}\]
D) \[-\frac{5}{3}\]
Correct Answer: A
Solution :
D.R?s of line are \[2,\text{ }1,\text{ }-2\]and normal vector to the plane is \[\hat{i}-2\hat{j}-k\hat{k}.\] Let \[\alpha \] be the angle between the line and the plane. So, \[\sin \alpha =\frac{(2\hat{i}+\hat{j}-2\hat{k}).(\hat{i}-2\hat{j}-k\hat{k})}{3\sqrt{1+4+{{k}^{2}}}}\] \[\Rightarrow \]\[\sin \alpha =\frac{2k}{3\sqrt{{{k}^{2}}+5}}\] ?(i) \[\cos \alpha =\frac{2\sqrt{2}}{3}\] [Given] ?(ii) \[\because \]\[{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1\] \[\Rightarrow \]\[\frac{4{{k}^{2}}}{9({{k}^{2}}+5)}+\frac{8}{9}=1\] [Using (i) & (ii)] \[\Rightarrow \]\[\frac{4{{k}^{2}}}{{{k}^{2}}+5}+8=9\Rightarrow 4{{k}^{2}}={{k}^{2}}+5\] \[\Rightarrow \]\[{{k}^{2}}=\frac{5}{3}\Rightarrow k=\sqrt{\frac{5}{3}}\]
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