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) \(\frac{\sqrt{2}}{10}\).

Answer 2

$$ \sin \theta = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}| |\vec{n}|}. $$

Compute $ \vec{d} \cdot \vec{n} $:

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

Compute $ |\vec{d}| $:

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

Compute $ |\vec{n}| $:

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

Substitute into the formula for $ \sin \theta $:

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