Question

The perimeter of a square whose two sides have equations \(\frac{x-1}{2} = \frac{y+2}{3} = \frac{z-3}{4}\) and \(\frac{x}{2} = \frac{y-1}{3} = \frac{z+1}{4}\) is

• A. \(\frac{\sqrt{673}}{\sqrt{29}}\) units
• B. \(\frac{4\sqrt{673}}{\sqrt{29}}\) units
• C. \(\frac{4\sqrt{573}}{\sqrt{29}}\) units
• D. \(\frac{4}{\sqrt{29}}\) units

Hint:

For two parallel lines in 3D: \[ \text{distance}=\frac{|(\vec{P_2}-\vec{P_1})\times \vec{d}|}{|\vec{d}|} \] This formula is very useful in line-distance questions.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The lines are parallel since their DRs are \((2, 3, 4)\). The distance between them is the side length \(s\) of the square.
Step 2: Key Formula or Approach:
Distance from point \(P_1(1, -2, 3)\) on line 1 to line 2 \((P_2(0, 1, -1), \bar{b}(2, 3, 4))\).
\(d = \frac{|(\bar{P_1} - \bar{P_2}) \times \bar{b}|}{|\bar{b}|}\).
Step 3: Detailed Explanation:
Vector \(\bar{a} = \bar{P_1} - \bar{P_2} = (1, -3, 4)\).
\(\bar{a} \times \bar{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & -3 & 4
2 & 3 & 4 \end{vmatrix} = (-24)\hat{i} - (4-8)\hat{j} + (3-(-6))\hat{k} = (-24, 4, 9)\).
Magnitude \(|\bar{a} \times \bar{b}| = \sqrt{576 + 16 + 81} = \sqrt{673}\).
\(|\bar{b}| = \sqrt{4+9+16} = \sqrt{29}\).
Side length \(s = \frac{\sqrt{673}}{\sqrt{29}}\). Perimeter \(= 4s = \frac{4\sqrt{673}}{\sqrt{29}}\).
Step 4: Final Answer:
Perimeter is \(\frac{4\sqrt{673}}{\sqrt{29}}\).