JEE Main 12 Jan 2019 Evening Question 25
Question: 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}}( \frac{2\sqrt{2}}{3} ), $ then a value of k is
[JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) $ \sqrt{\frac{5}{3}} $
B) $ -\frac{3}{5} $
C) $ \sqrt{\frac{3}{5}} $
D) $ -\frac{5}{3} $
Show Answer
Answer:
Correct Answer: A
Solution:
- D.R’s of line are $ 2,1,-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{4k^{2}}{9(k^{2}+5)}+\frac{8}{9}=1 $
[Using (i) & (ii)]
$ \Rightarrow $ $ \frac{4k^{2}}{k^{2}+5}+8=9\Rightarrow 4k^{2}=k^{2}+5 $
$ \Rightarrow $ $ k^{2}=\frac{5}{3}\Rightarrow k=\sqrt{\frac{5}{3}} $
