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

Concept:
The given two lines have the same direction ratios, so they are parallel. If these are two opposite sides of a square, then the perpendicular distance between the lines equals the side length of the square. Hence, \[ \text{Perimeter}=4\times \text{side length} \] ip

Step 1: Write points and direction vector for the lines.
For the first line: \[ \frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{4} \] a point on it is: \[ P_1=(1,-2,3) \] For the second line: \[ \frac{x}{2}=\frac{y-1}{3}=\frac{z+1}{4} \] a point on it is: \[ P_2=(0,1,-1) \] Both have direction vector: \[ \vec{d}=(2,3,4) \] ip

Step 2: Use the distance formula for parallel lines.
Distance between two parallel lines is: \[ \text{distance}=\frac{|(\overrightarrow{P_1P_2}\times \vec{d})|}{|\vec{d}|} \] Now, \[ \overrightarrow{P_1P_2}=P_2-P_1=(0-1,\ 1-(-2),\ -1-3)=(-1,3,-4) \] ip

Step 3: Find the cross product.
\[ (-1,3,-4)\times(2,3,4) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 3 & -4 \\ 2 & 3 & 4 \end{vmatrix} \] \[ =24\hat{i}-4\hat{j}-9\hat{k} \] So, \[ |(-1,3,-4)\times(2,3,4)|=\sqrt{24^2+(-4)^2+(-9)^2} \] \[ =\sqrt{576+16+81}=\sqrt{673} \] Also, \[ |(2,3,4)|=\sqrt{2^2+3^2+4^2}=\sqrt{29} \] Therefore, side length: \[ s=\frac{\sqrt{673}}{\sqrt{29}} \] ip

Step 4: Find the perimeter.
\[ \text{Perimeter}=4s=4\cdot \frac{\sqrt{673}}{\sqrt{29}} \] \[ \text{Perimeter}=\frac{4\sqrt{673}}{\sqrt{29}} \] ip Hence, the correct answer is:
\[ \boxed{(B)\ \frac{4\sqrt{673}}{\sqrt{29}} \text{ units}} \]