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$^{-1} \bigg(\frac{2\sqrt{2}}{3}\bigg),$ 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 $\alpha \, =\frac{2k}{3\sqrt{k^2 + 5}} \, \, \, \, \, \, \, \, \, ...(1)$
$cos \, \alpha \, = \, \frac{2\sqrt{2}}{3} \, \, \, \, \, \, \, \, \, ...(2)$
$(1)^2 \, + \, (2)^2 \, = \, 1 \, \Rightarrow \, k^2 \, = \, \frac{5}{3}$