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 ${\cos }^{-1}\left(\frac{2\sqrt{2}}{3}\right)$, then a value of $k$ is :

• A. $\sqrt{\frac{5}{3}}$
• B. $\sqrt{\frac{3}{5}}$
• C. $-\frac{3}{5}$
• D. $-\frac{5}{3}$

Answer 1

The correct option is A $\sqrt{\frac{5}{3}}$

Let $\alpha$ be the angle between the line and the plane.
Then $\sin \alpha =\frac{2\left(1\right)+1(-2)-2(-k)}{\sqrt{2^2+1^2+(-2{)}^2}\sqrt{1^2+(-2{)}^2+(-k{)}^2}}$
$\Rightarrow \sqrt{1-{\cos }^2\alpha }=\frac{2k}{3\sqrt{5+k^2}}$
$\Rightarrow 1-\frac{8}{9}=\frac{4k^2}{9(5+k^2)}$
​​​​​​​$\Rightarrow 5+k^2=4k^2$
​​​​​​​$\Rightarrow k=\pm \sqrt{\frac{5}{3}}$