Question-31 If the angle \theta between the line \frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2} and the plane 2x-y+\sqrt{\lambda}z+4=0 is such that \sin\theta=\frac{1}{3}, then \lambda+1=
(A) \frac{5}{3}
(B) -\frac{5}{3}
(C) \frac{8}{3}
(D) -\frac{8}{3}
Answer: (C)
Explanation:
The d.r.s of line and plane are 1, 2, 2 and 2, −1, \sqrt{\lambda}
∴ \sin\theta=\frac{2(1)+(-1)(2)+(\sqrt{\lambda})(2)}{\sqrt{1^2+2^2+2^2}\ \sqrt{2^2+(-1)^2+(\sqrt{\lambda})^2}}
=\frac{1}{3}=\frac{2\sqrt{\lambda}}{3\sqrt{5+\lambda}}⇒ \sqrt{5+\lambda}=2\sqrt{\lambda}
⇒ \lambda=\frac{5}{3}
⇒ \lambda+1=\frac{8}{3}