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. \(\sqrt{\frac{3}{5}}\)
• B. \(-\frac{5}{2}\)
• C. \(-\frac{3}{2}\)
• D. \(\sqrt{\frac{5}{3}}\)

Hint:

Angle between a line and a plane: \(\sin\theta=\frac{|\vec d\cdot\vec n|}{|\vec d||\vec n|}\). Here \(\cos\theta=\frac{2\sqrt2}{3}\) gives \(\sin\theta=\frac13\).

Answer 1

The Correct Option is D

Direction vector of the line: \(\vec d=(2,1,-2)\), so \(|\vec d|=\sqrt{4+1+4}=3\).
Normal of the plane: \(\vec n=(1,-2,-k)\), so \(|\vec n|=\sqrt{5+k^2}\).
\(\vec d\cdot\vec n=2-2+2k=2k\).
Since the angle between line and plane is \(\theta\) with \(\cos\theta=\frac{2\sqrt2}{3}\), \(\sin\theta=\sqrt{1-\frac{8}{9}}=\frac13\).
\(\sin\theta=\frac{|2k|}{3\sqrt{5+k^2}}=\frac13\Rightarrow 2|k|=\sqrt{5+k^2}\Rightarrow 4k^2=5+k^2\Rightarrow k^2=\frac53\).
So \(k=\pm\sqrt{\frac53}\); the value available in the options is \(\sqrt{\frac{5}{3}}\).
Answer: (4) \(\sqrt{\frac{5}{3}}\)