Question

Let the foot of perpendicular from the point \((\lambda, 2, 3)\) on the line \(\frac{x-4}{1} = \frac{y-9}{2} = \frac{z-5}{1}\) be the point \((1, \mu, 2)\). Then the distance between the lines \(\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}\) and \(\frac{x-\lambda}{2} = \frac{y-\mu}{3} = \frac{z+5}{6}\) is equal to:

• A. \(\frac{12}{7} \)
• B. \(\frac{\sqrt{145}}{7} \)
• C. \(\frac{\sqrt{146}}{7} \)
• D. \(\frac{\sqrt{143}}{7} \)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
First, we find \(\lambda\) and \(\mu\) using the property that the line joining a point to its foot of perpendicular is perpendicular to the given line. Then, we calculate the distance between the two resulting parallel lines.

Step 2: Key Formula or Approach:
1. Perpendicular condition: \[ a_1a_2 + b_1b_2 + c_1c_2 = 0 \] 2. Distance between parallel lines: \[ d = \frac{|\vec{BA} \times \vec{b}|}{|\vec{b}|} \] where \(A, B\) are points on the lines and \(\vec{b}\) is the direction vector.

Step 3: Detailed Explanation:
1. Find \(\mu\):
The point \((1, \mu, 2)\) lies on: \[ \frac{x-4}{1} = \frac{y-9}{2} = \frac{z-5}{1} \] \[ \frac{1-4}{1} = \frac{\mu-9}{2} \Rightarrow -3 = \frac{\mu-9}{2} \Rightarrow \mu - 9 = -6 \Rightarrow \mu = 3 \] 2. Find \(\lambda\):
Vector from \((\lambda, 2, 3)\) to \((1, 3, 2)\): \[ (1-\lambda, 1, -1) \] This is perpendicular to direction \((1,2,1)\): \[ 1(1-\lambda) + 2(1) + 1(-1) = 0 \] \[ 1 - \lambda + 2 - 1 = 0 \Rightarrow \lambda = 2 \] 3. Distance between parallel lines:
Points: \[ A(1,2,-4), \quad B(2,3,-5) \] Direction vector: \[ \vec{b} = (2,3,6) \] \[ \vec{AB} = (1,1,-1) \] Cross product: \[ \vec{AB} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & -1 \\ 2 & 3 & 6 \end{vmatrix} \] \[ = \mathbf{i}(9) - \mathbf{j}(8) + \mathbf{k}(1) = (9, -8, 1) \] Magnitude: \[ |\vec{AB} \times \vec{b}| = \sqrt{9^2 + (-8)^2 + 1^2} = \sqrt{146} \] \[ |\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{49} = 7 \] Distance: \[ d = \frac{\sqrt{146}}{7} \] 
Step 4: Final Answer:
\[ \boxed{\frac{\sqrt{146}}{7}} \]