Question

The sine of the angle between the straight line $\frac{x-2}{3}=\frac{3-y}{-4}=\frac{z-4}{5}$ and the plane 2x - 2y + z = 5 is

• A. $\frac{3}{\sqrt{50}}$
• B. $\frac{3}{50}$
• C. $\frac{4}{5\sqrt2}$
• D. $\frac{\sqrt2}{10}$

Answer 1

Answer: (D)

$\frac{\sqrt2}{10}$

Answer 2

The correct answer is (D) : $\frac{\sqrt2}{10}$.