Question

The distance of the point \( P(3, 8, 2) \) from the line \[ \frac{x}{2} = \frac{y - 3}{4} = \frac{z - 3}{3}, \quad \text{measured parallel to the plane} \quad 3x + 2y - 2z + 15 = 0, \] is

• A. 7 units
• B. 6 units
• C. 8 units
• D. 10 units

Hint:

To find the distance from a point to a line parallel to a plane, use projection methods involving the normal vector of the plane.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find a point \( Q \) on the line such that the line segment \( PQ \) is parallel to the given plane. Then calculate distance \( PQ \).
Step 2: Detailed Explanation:
1. Point \( Q \) on line: \( (2\lambda + 1, 4\lambda + 3, 3\lambda + 2) \).
2. Direction ratios of \( \vec{PQ} = (2\lambda - 2, 4\lambda - 5, 3\lambda) \).
3. Since \( PQ \) is parallel to plane \( 3x + 2y - 2z + 15 = 0 \), its direction is perpendicular to the normal \( (3, 2, -2) \):
\( 3(2\lambda - 2) + 2(4\lambda - 5) - 2(3\lambda) = 0 \)
\( 6\lambda - 6 + 8\lambda - 10 - 6\lambda = 0 \Rightarrow 8\lambda = 16 \Rightarrow \lambda = 2 \).
4. Point \( Q = (5, 11, 8) \).
5. Distance \( PQ = \sqrt{(5-3)^2 + (11-8)^2 + (8-2)^2} = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{49} = 7 \).
Step 3: Final Answer:
The distance is 7 units.