Question

If the angle $\theta$ between the lines $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$ and the plane $2x-y+\sqrt{\lambda z+4}=0$ such that $sin\theta =\frac{1}{3}$, then value of $\lambda$ is

• A. $\frac{3}{4}$
  
• B. $\frac{-4}{3}$
  
• C. $\frac{5}{3}$
  
• D. $\frac{-3}{5}$

Answer 1

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


Since the line makes $\theta$ angle with the plane hence it would make ${90}^{\circ }-\theta$ anglewith the normal to the plane. Hence cos $({90}^{\circ }-\theta )=sin\theta =\frac{1}{3}$ is $\frac{2-2+2\sqrt{\lambda }}{\sqrt{9}\sqrt{5+\lambda }}=\frac{1}{3}\Rightarrow \lambda =\frac{5}{3}$