Question

Find the angle \( \theta \) between the line \( \frac{x-1}{1} = \frac{y+2}{-2} = \frac{z-3}{2} \) and the plane \( 2x - y + 2z = 7 \).

• A. \( \sin^{-1}\left(\frac{8}{9}\right) \)
• B. \( \cos^{-1}\left(\frac{8}{9}\right) \)
• C. \( \frac{\pi}{2} \)
• D. \( \sin^{-1}\left(\frac{2}{3}\right) \)

Hint:

Remember the unique rule for mixed line-and-plane problems: finding the angle between a line and a plane uses \( \sin\theta \), whereas finding the angle between two lines or two planes uses \( \cos\theta \).

Answer 1

The Correct Option is A

Concept: The angle \( \theta \) between a straight line with a directional heading vector \( \vec{b} \) and a flat plane with a surface normal vector \( \vec{n} \) is calculated using the sine projection product formula: \[ \sin\theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}||\vec{n}|} \]

Step 1: Extract the directional components from the line and plane equations.

• Line direction vector coefficients (from denominators): \( \vec{b} = 1\hat{i} - 2\hat{j} + 2\hat{k} \)
• Plane normal vector coefficients (from variable modifiers): \( \vec{n} = 2\hat{i} - 1\hat{j} + 2\hat{k} \)

Step 2: Calculate the scalar dot product and vector magnitudes.
1. Compute the dot product: \[ \vec{b} \cdot \vec{n} = (1)(2) + (-2)(-1) + (2)(2) = 2 + 2 + 4 = 8 \] 2. Compute the magnitude of the line's direction vector: \[ |\vec{b}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \] 3. Compute the magnitude of the plane's normal vector: \[ |\vec{n}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

Step 3: Substitute these components into the sine formula to solve for \( \theta \).
\[ \sin\theta = \frac{8}{3 \times 3} = \frac{8}{9} \quad \Rightarrow \quad \theta = \sin^{-1}\left(\frac{8}{9}\right) \]