Question

Let the foot of the perpendicular from the point \( (1, 2, 4) \) on the line \( \frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3} \) be \( P \). Then the distance of \( P \) from the plane \( 3x + 4y + 12z + 23 = 0 \) is:

• A. \( 5 \)
• B. \( \frac{50}{13} \)
• C. \( 4 \)
• D. \( \frac{63}{13} \)

Hint:

To calculate the perpendicular distance from a point to a plane, use the formula: \[ \text{Distance} = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}. \]

Answer 1

The Correct Option is A

Let the coordinates of the foot of the perpendicular \( P(x, y, z) \) be represented by the following relation: \[ \frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3} = \lambda. \] From this, the parametric form of the line is: \[ (x, y, z) = (4\lambda - 2, 2\lambda + 1, 3\lambda - 1). \] Next, the vector \( \overrightarrow{AP} \) from the point \( A(1, 2, 4) \) to \( P \) is given by: \[ \overrightarrow{AP} = (4\lambda - 3) \hat{i} + (2\lambda - 1) \hat{j} + (3\lambda - 5) \hat{k}. \] The direction vector of the line is: \[ \overrightarrow{b} = 4\hat{i} + 2\hat{j} + 3\hat{k}. \] Since \( \overrightarrow{AP} \) is perpendicular to \( \overrightarrow{b} \), their dot product must be zero: \[ \overrightarrow{AP} \cdot \overrightarrow{b} = 0. \] Substituting the expressions: \[ 4(4\lambda - 3) + 2(2\lambda - 1) + 3(3\lambda - 5) = 0. \] Simplifying the equation: \[ 16\lambda - 12 + 4\lambda - 2 + 9\lambda - 15 = 0, \] \[ 29\lambda - 29 = 0. \] Solving for \( \lambda \): \[ \lambda = 1. \] Substituting \( \lambda = 1 \) into the parametric equations of the line: \[ P(2, 3, 2). \] Now, we calculate the perpendicular distance from the point \( P(2, 3, 2) \) to the plane \( 3x + 4y + 12z + 23 = 0 \): \[ \text{Distance} = \frac{|3(2) + 4(3) + 12(2) + 23|}{\sqrt{3^2 + 4^2 + 12^2}}. \] Simplifying the numerator: \[ |6 + 12 + 24 + 23| = 65. \] Simplifying the denominator: \[ \sqrt{9 + 16 + 144} = \sqrt{169} = 13. \] Thus, the perpendicular distance is: \[ \text{Distance} = \frac{65}{13} = 5. \] Final Answer: \[ \boxed{5} \]