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. \(\frac{63}{13}\)

Answer 1

The Correct Option is A

To solve this problem, we need to find the distance of the foot of the perpendicular from the point \( (1, 2, 4) \) on the given line from the specified plane. We'll start step-by-step by determining the coordinates of the foot of the perpendicular on the line.

1. Identify the equation of the line: The given line is represented as \[ \frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3} = \lambda \] which can be rewritten in parametric form as: \[ x = 4\lambda - 2, \quad y = 2\lambda + 1, \quad z = 3\lambda - 1 \]
2. To find the foot of the perpendicular, we first calculate the direction ratios (DRs) of the line, which are \( \langle 4, 2, 3 \rangle \).
3. The direction ratios of the line connecting the point \( (1, 2, 4) \) to the point on the line \( (4\lambda - 2, 2\lambda + 1, 3\lambda - 1) \) should be perpendicular to the direction ratios of the line. Thus, we have:
   • The direction ratios from \( (1, 2, 4) \) to \( (4\lambda - 2, 2\lambda + 1, 3\lambda - 1) \) are \( \langle 4\lambda - 3, 2\lambda - 1, 3\lambda - 5 \rangle \).
   • Since the line is perpendicular, their dot product with the line's DRs is zero: \[ 4(4\lambda - 3) + 2(2\lambda - 1) + 3(3\lambda - 5) = 0 \]
4. Expanding and simplifying, we get: \[ 16\lambda - 12 + 4\lambda - 2 + 9\lambda - 15 = 0 \implies 29\lambda = 29 \]
5. Solving for \( \lambda \), we find \( \lambda = 1 \).
6. Substitute \( \lambda = 1 \) back into the parametric equations to find \( P \):
   • \( x = 4(1) - 2 = 2 \)
   • \( y = 2(1) + 1 = 3 \)
   • \( z = 3(1) - 1 = 2 \)
7. The coordinates of the foot of the perpendicular \( P \) are \( (2, 3, 2) \).
8. Now, calculate the distance from \( P (2, 3, 2) \) to the plane \( 3x + 4y + 12z + 23 = 0 \) using the formula for the distance of a point from a plane: \[ \text{Distance} = \frac{|3(2) + 4(3) + 12(2) + 23|}{\sqrt{3^2 + 4^2 + 12^2}} \]
9. This evaluates to:
   • Numerator: \(|6 + 12 + 24 + 23| = 65\)
   • Denominator: \(\sqrt{9 + 16 + 144} = 13\)
10. Thus, the distance is \(\frac{65}{13} = 5\).

The distance of \( P \) from the plane is \(5\), verifying that the correct answer is indeed \(5\).