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:

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

Answer 1

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

DR's of line are 2, 1, -2
normal vector of plane is $\hat{i}-2\hat{j}-k\hat{k}$

$sin\left(\alpha \right)=\frac{(2\hat{i}+\hat{j}-2\hat{k}).(\hat{i}-2\hat{j}-k\hat{k})}{3\sqrt{1+4+k^2}}$

$sin\left(\alpha \right)=\frac{2k}{3\sqrt{k^2+5}}$ ..............(1)

$cos\alpha =\frac{2\sqrt{2}}{3}$ .............. (2)
$(1{)}^2+(2{)}^2=1willgive{k}^2=\frac{5}{3}$

$k=\sqrt{\frac{5}{3}}$