KVPY Sample Paper KVPY Stream-SX Model Paper-20

If an angle between the line, \[\frac{x+1}{2}=\frac{y-2}{1}=\frac{z-3}{-2}\]and the plane, \[x-2y-kz=3\]is \[{{\cos }^{-1}}\left( \frac{2\sqrt{2}}{3} \right),\] then a value of k is:

A) \[\sqrt{\frac{5}{3}}\]

B) \[\sqrt{\frac{3}{5}}\]

C) \[-\frac{3}{5}\]

D) \[-\frac{5}{3}\]

Correct Answer: A

Solution :

Direction Ratio of line are\[2,1,-2\]
Normal vector of plane is \[\hat{i}-2\hat{j}-k\hat{k}\]
\[\sin \alpha =\frac{\left( 2\hat{i}+\hat{j}-\hat{k} \right).\left( \hat{i}-2\hat{j}-k\hat{k} \right)}{3\sqrt{1+4+{{k}^{2}}}}\]
\[\sin \alpha =\frac{2k}{3\sqrt{{{k}^{2}}+5}}\]	? (1)
\[\cos \alpha =\frac{2\sqrt{2}}{3}\] (Given)	? (2)
Using, \[{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1\]\[\Rightarrow \] \[{{k}^{2}}=\frac{5}{3}\]\[\Rightarrow \] \[k=\sqrt{\frac{5}{3}}.\]