Question

Let the line passing through the points \( (-1, 2, 1) \) and parallel to the line \[ \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4} \] intersect the line \[ \frac{x + 2}{3} = \frac{y - 3}{2} = \frac{z - 4}{1} \] at the point P. Then the distance of P from the point Q(4, -5, 1) is:

• A. 5
• B. 10
• C. \( 5\sqrt{6} \)
• D. \( 5\sqrt{5} \)

Hint:

To find the distance between two points in 3D, use the distance formula: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \). Always check your solution by verifying the parametric equations of the lines.

Answer 1

The Correct Option is D

To address this problem, we must determine the intersection point of two lines and subsequently compute the distance from this intersection to a specified point \( Q(4, -5, 1) \).

1. Identify the equations for both lines.
   • The first line is defined by the point \( (-1, 2, 1) \) and is parallel to the line with direction ratios \(\langle 2, 3, 4 \rangle\). Its parametric equation is: \(x = -1 + 2t, \, y = 2 + 3t, \, z = 1 + 4t\)
   • The second line is defined by the ratios: \(\frac{x + 2}{3} = \frac{y - 3}{2} = \frac{z - 4}{1} = \lambda\)
   • Consequently, the parametric equation for the second line is: \(x = 3\lambda - 2, \, y = 2\lambda + 3, \, z = \lambda + 4\)
2. The next critical step is to locate the intersection by solving the following system of equations:
   • \(-1 + 2t = 3\lambda - 2\)
   • \(2 + 3t = 2\lambda + 3\)
   • \(1 + 4t = \lambda + 4\)
3. Solve these equations simultaneously:
   • From the first equation:
     \(2t + 1 = 3\lambda\)
     Thus, \(\lambda = \frac{2t + 1}{3}\)
   • From the second equation:
     \(3t - 1 = 2\lambda\)
     Thus, \(\lambda = \frac{3t - 1}{2}\)
   • Equating the expressions for \(\lambda\):
     \(\frac{2t + 1}{3} = \frac{3t - 1}{2}\)
     Cross-multiplication yields \(4t + 2 = 9t - 3\)
     Solving for \(t\) gives \(5t = 5\), so \(t = 1\).
   • Substitute \(t = 1\) into either \(\lambda = \frac{2t + 1}{3}\) or \(\lambda = \frac{3t - 1}{2}\) to determine \(\lambda\).
     Using the first expression: \(\lambda = \frac{2(1) + 1}{3} = 1\) (verification with the second expression confirms consistency).
4. Determine the coordinates of the intersection point \( P \) by substituting \( t = 1 \) or \( \lambda = 1 \) into the respective parametric equations.
   • Using the parametric equation for the first line: \(x = 1, y = 5, z = 5\)
5. Now, calculate the distance between points \( P(1, 5, 5) \) and \( Q(4, -5, 1) \) utilizing the distance formula:
   \(d = \sqrt{(4 - 1)^2 + (-5 - 5)^2 + (1 - 5)^2}\)
   = \(\sqrt{3^2 + (-10)^2 + (-4)^2} = \sqrt{9 + 100 + 16} = \sqrt{125} = 5\sqrt{5}\).

The distance from point \( P \) to point \( Q \) is therefore \(5\sqrt{5}\).