Question

The shortest distance between the lines \(\frac{x-3}{-1} = \frac{y-2}{4} = \frac{z-1}{2}\) and \(\frac{x-1}{2} = \frac{y-1}{1} = \frac{z-2}{5}\) is:

• A. \(\sqrt{6}\)
• B. \(6\sqrt{6}\)
• C. \(\sqrt{5}\)
• D. \(5\sqrt{5}\)

Hint:

When calculating \(\vec{b}_1 \times \vec{b}_2\), you can factor out common terms (like 9 here) to simplify the magnitude and dot product calculations, as long as you account for it in the final division.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The shortest distance between two skew lines is the projection of the vector joining a point on each line onto the common perpendicular vector (cross product of the direction vectors).

Step 2: Key Formula or Approach:
S.D. = \(\left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|\) Lines: \(L_1: \vec{a}_1 = (3, 2, 1), \vec{b}_1 = (-1, 4, 2)\) \(L_2: \vec{a}_2 = (1, 1, 2), \vec{b}_2 = (2, 1, 5)\)

Step 3: Detailed Explanation:
1. \(\vec{a}_2 - \vec{a}_1 = (1-3, 1-2, 2-1) = (-2, -1, 1)\). 2. \(\vec{b}_1 \times \vec{b}_2\): \[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-1 & 4 & 2
2 & 1 & 5 \end{vmatrix} = \hat{i}(20-2) - \hat{j}(-5-4) + \hat{k}(-1-8) = 18\hat{i} + 9\hat{j} - 9\hat{k} \] 3. Magnitude \(|\vec{b}_1 \times \vec{b}_2| = \sqrt{18^2 + 9^2 + (-9)^2} = \sqrt{324 + 81 + 81} = \sqrt{486} = 9\sqrt{6}\). 4. Dot product: \((\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-2)(18) + (-1)(9) + (1)(-9) = -36 - 9 - 9 = -54\). 5. S.D. = \(\left| \frac{-54}{9\sqrt{6}} \right| = \frac{6}{\sqrt{6}} = \sqrt{6}\).

Step 4: Final Answer:
The shortest distance between the lines is \(\sqrt{6}\).