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

Answer 1

Answer: (C)

DR's of line are $2,1,-2$
normal vector of plane is $\hat{i}-2\hat{j}-k\hat{k}$
$sin\alpha =\frac{(2\hat{i}+\hat{j}-2\hat{k}).(\hat{i}-2\hat{j}-k\hat{k})}{3\sqrt{1+4+k^2}}$
sin $\alpha =\frac{2k}{3\sqrt{k^2+5}}...(1)$
$cos\alpha =\frac{2\sqrt{2}}{3}...(2)$
$(1{)}^2+(2{)}^2=1\Rightarrow k^2=\frac{5}{3}$