Question

Let the line \(\frac{x}{1}=\frac{6−y}{2}=\frac{z+8}{5}\)  intersect the lines  \(\frac{x-5}{1}=\frac{y-7}{3}=\frac{z+2}{1}\) and \(\frac{x+3}{6}=\frac{3−y}{3}=\frac{z-6}{1}\) at the points A and B respectively. Then the distance of the mid-point of the line segment AB from the plane 2x – 2y + z = 14 is 

• A. \(\frac{10}{3}\)
• B. \(\frac{11}{3}\)
• C. 4
• D. 3

Answer 1

The Correct Option is C

Solution: The given equations are: \[ \frac{x}{1} = \frac{y - 6}{-2} = \frac{z + 8}{5} = \lambda \quad \text{(1)} \] \[ \frac{x - 5}{4} = \frac{y - 7}{3} = \frac{z + 2}{1} = \mu \quad \text{(2)} \] \[ \frac{x + 3}{6} = \frac{3 - y}{3} = \frac{z - 6}{1} = \gamma \quad \text{(3)} \] For intersection of (1) and (2), solving the system gives: \[ \lambda = -1, \mu = -1, \quad A(1, 4, -3). \] For intersection of (1) and (3), solving the system gives: \[ \lambda = 3, \gamma = 1, \quad B(0, 7, 7). \] The midpoint of \( A(1, 4, -3) \) and \( B(0, 7, 7) \) is: \[ \left( \frac{1+0}{2}, \frac{4+7}{2}, \frac{-3+7}{2} \right) = (0.5, 5.5, 2) \] Now, calculate the perpendicular distance from the plane \( 2x - 2y + z = 14 \): \[ \frac{|2(0.5) - 2(5.5) + 2 - 14|}{\sqrt{2^2 + (-2)^2 + 1^2}} = \frac{|1 - 11 + 2 - 14|}{\sqrt{4 + 4 + 1}} = \frac{|-22|}{3} = 4. \] Thus, the distance is \( 4 \).