Question

Find the shortest distance between the two parallel straight lines whose vector position equations are given by: \[ \vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) \] \[ \vec{r} = (3\hat{i} + 3\hat{j} - 5\hat{k}) + \mu(2\hat{i} + 3\hat{j} + 6\hat{k}) \]

• A. \( \sqrt{2} \)
• B. \( \frac{\sqrt{293}}{7} \)
• C. \( 0 \)
• D. \( \frac{5}{7} \)

Hint:

Always double-check the direction vectors before starting. If they match or are direct scalar multiples of each other, the lines are parallel. You must use the parallel cross product formula, not the standard skew-lines formula.

Answer 1

The Correct Option is B

Concept: When two straight lines run completely parallel in a 3D space, they share the exact same directional heading vector \( \vec{b} \). The shortest distance (\( d \)) separating them depends on the cross product of their position difference with this direction vector: \[ d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|} \]

Step 1: Extract the vector parameters and calculate the position difference.
From the provided parallel equations, isolate the coordinates:

• \( \vec{a}_1 = \hat{i} + 2\hat{j} - 4\hat{k} \)
• \( \vec{a}_2 = 3\hat{i} + 3\hat{j} - 5\hat{k} \)
• Shared Direction Vector \( \vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k} \)

Calculate the initial displacement vector connecting the two starting points: \[ \vec{a}_2 - \vec{a}_1 = (3-1)\hat{i} + (3-2)\hat{j} + (-5 - (-4))\hat{k} = 2\hat{i} + \hat{j} - \hat{k} \]

Step 2: Compute the cross product vector \( (\vec{a}_2 - \vec{a}_1) \times \vec{b} \).
Set up the matrix determinant layout using standard unit coordinates: \[ (\vec{a}_2 - \vec{a}_1) \times \vec{b} = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} 2 & 1 & -1 2 & 3 & 6 \end{matrix} \right| \] Expanding along the top row: \[ = \hat{i}(6 - (-3)) - \hat{j}(12 - (-2)) + \hat{k}(6 - 2) = 9\hat{i} - 14\hat{j} + 4\hat{k} \] Calculate its magnitude: \[ |(\vec{a}_2 - \vec{a}_1) \times \vec{b}| = \sqrt{9^2 + (-14)^2 + 4^2} = \sqrt{81 + 196 + 16} = \sqrt{293} \]

Step 3: Divide by the magnitude of the direction vector \( |\vec{b}| \).
Find the magnitude of the shared direction heading: \[ |\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \] Combine these values into the parallel distance formula: \[ d = \frac{\sqrt{293}}{7} \]