Question

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

• A. \(\frac{\sqrt{2}}{10}\)
• B. $\frac{2}{5\sqrt{2}}$
• C. $\frac{3}{50}$
• D. $\frac{3}{\sqrt{50}}$

Answer 1

The Correct Option is A

1. Identify the direction vector of the line and normal vector of the plane:

Given the line equation \(\frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5}\), its direction vector is:

\[ \vec{d} = (3, 4, 5) \]

Given the plane equation \(2x - 2y + z = 5\), its normal vector is:

\[ \vec{n} = (2, -2, 1) \]

2. Compute the angle between the line and the plane:

The angle \(\theta\) between the line and the plane satisfies:

\[ \sin \theta = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}| \cdot |\vec{n}|} \]

where \(\vec{d} \cdot \vec{n}\) is the dot product, and \(|\vec{d}|\), \(|\vec{n}|\) are the magnitudes of the vectors.

3. Calculate the dot product and magnitudes:

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

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

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

4. Compute \(\sin \theta\):

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

Correct Answer:  \(\frac{\sqrt{2}}{10}\)

Answer 2

The direction vector of the line $ \frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5} $ is $ \mathbf{v} = \langle 3, 4, 5 \rangle $. 

The normal vector of the plane $ 2x - 2y + z = 5 $ is $ \mathbf{n} = \langle 2, -2, 1 \rangle $.

The sine of the angle $ \theta $ between the line and the plane is given by:

$$ \sin \theta = \frac{|\mathbf{v} \cdot \mathbf{n}|}{\|\mathbf{v}\| \|\mathbf{n}\|} $$

where $ \cdot $ denotes the dot product and $ \|\cdot\| $ denotes the magnitude.

Calculating the dot product and magnitudes:

• $ \mathbf{v} \cdot \mathbf{n} = (3)(2) + (4)(-2) + (5)(1) = 3 $
• $ \|\mathbf{v}\| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{50} = 5\sqrt{2} $
• $ \|\mathbf{n}\| = \sqrt{2^2 + (-2)^2 + 1^2} = 3 $

Therefore:

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

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