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

The Correct Option is D

We need to find the sine of the angle between the straight line $\frac{x-2}{3} = \frac{3-y}{-4} = \frac{z-4}{5}$ and the plane $2x - 2y + z = 5$.

First, rewrite the equation of the line in the standard form: $\frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5}$. So, the direction vector of the line is $\vec{b} = (3, 4, 5)$.

The normal vector to the plane $2x - 2y + z = 5$ is $\vec{n} = (2, -2, 1)$.

Let $\theta$ be the angle between the line and the plane. The angle between the direction vector $\vec{b}$ of the line and the normal vector $\vec{n}$ of the plane is the complement of $\theta$, i.e., $\frac{\pi}{2} - \theta$.

So, $\sin \theta = \cos(\frac{\pi}{2} - \theta) = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$

$\vec{b} \cdot \vec{n} = (3)(2) + (4)(-2) + (5)(1) = 6 - 8 + 5 = 3$

$|\vec{b}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$

$|\vec{n}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$

$\sin \theta = \frac{|3|}{(5\sqrt{2})(3)} = \frac{3}{15\sqrt{2}} = \frac{1}{5\sqrt{2}} = \frac{\sqrt{2}}{10}$

Therefore, the sine of the angle between the line and the plane is $\frac{\sqrt{2}}{10}$.

Thus, the correct option is (D) $\frac{\sqrt{2}}{10}$.