Question

The square of the distance of the point (-2, -8, 6) from the line \(\frac{x-1}{1} = \frac{y-1}{2} = \frac{z}{1}\) along the line \(\frac{x+5}{1} = \frac{y+5}{1} = \frac{z}{2}\) is equal to:

• A. 3
• B. 6
• C. 8
• D. 12

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to find the distance between a given point \(P\) and a given line \(L_1\), but this distance must be measured parallel to another given line \(L_2\).
Step 2: Key Formula or Approach:
Given point \(P(-2, -8, 6)\).
Line \(L_1\): \(\frac{x-1}{1} = \frac{y-1}{2} = \frac{z}{-1}\).
Line \(L_2\): \(\frac{x+5}{1} = \frac{y+5}{-1} = \frac{z}{2}\), which has direction ratios \(\vec{v} = (1, -1, 2)\).
We write the equation of a line \(L_3\) passing through \(P\) and parallel to \(L_2\):
\[ \frac{x+2}{1} = \frac{y+8}{-1} = \frac{z-6}{2} = t \] Any point \(Q\) on \(L_3\) is \(Q(t-2, -t-8, 2t+6)\).
Step 3: Detailed Explanation:
The point \(Q\) must lie on \(L_1\). Substitute the coordinates of \(Q\) into the equation of \(L_1\):
\[ \frac{(t-2)-1}{1} = \frac{(-t-8)-1}{2} = \frac{(2t+6)}{-1} \] \[ \frac{t-3}{1} = \frac{-t-9}{2} = \frac{2t+6}{-1} \] Take the first two parts:
\[ 2(t - 3) = -t - 9 \implies 2t - 6 = -t - 9 \implies 3t = -3 \implies t = -1 \] Check with the third part for consistency:
\[ \frac{t-3}{1} = -4 \] \[ \frac{2(-1)+6}{-1} = \frac{4}{-1} = -4 \] Since they match, the intersection point \(Q\) is found using \(t = -1\):
\[ Q = (-1-2, -(-1)-8, 2(-1)+6) = (-3, -7, 4) \] We need to calculate the square of the distance \(PQ^2\).
\[ PQ^2 = (-3 - (-2))^2 + (-7 - (-8))^2 + (4 - 6)^2 \] \[ PQ^2 = (-1)^2 + (1)^2 + (-2)^2 = 1 + 1 + 4 = 6 \] Step 4: Final Answer:
The square of the required distance is 6.