Question

The shortest distance between the lines \( \bar{r} = (3\bar{i} - 5\bar{j} + 2\bar{k}) + t(4\bar{i} + 3\bar{j} - \bar{k}) \) and \( \bar{r} = (\bar{i} + 2\bar{j} - 4\bar{k}) + s(6\bar{i} + 3\bar{j} - 2\bar{k}) \) is

• A. 7
• B. 8
• C. 9
• D. 12

Hint:

Always double-check signs when subtracting vectors and calculating determinants, as these are the most common sources of error in shortest distance problems.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

The problem asks for the shortest distance between two skew lines given in vector form. The shortest distance is measured along the line perpendicular to both direction vectors.
Step 2: Key Formula or Approach:

For lines \( \bar{r} = \bar{a}_1 + t\bar{b}_1 \) and \( \bar{r} = \bar{a}_2 + s\bar{b}_2 \), the shortest distance \( d \) is: \[ d = \left| \frac{(\bar{a}_2 - \bar{a}_1) \cdot (\bar{b}_1 \times \bar{b}_2)}{|\bar{b}_1 \times \bar{b}_2|} \right| \]
Step 3: Detailed Explanation:

Identify the components: \[ \bar{a}_1 = 3\bar{i} - 5\bar{j} + 2\bar{k}, \quad \bar{b}_1 = 4\bar{i} + 3\bar{j} - \bar{k} \] \[ \bar{a}_2 = \bar{i} + 2\bar{j} - 4\bar{k}, \quad \bar{b}_2 = 6\bar{i} + 3\bar{j} - 2\bar{k} \] Calculate \( \bar{a}_2 - \bar{a}_1 \): \[ \bar{a}_2 - \bar{a}_1 = (1-3)\bar{i} + (2-(-5))\bar{j} + (-4-2)\bar{k} = -2\bar{i} + 7\bar{j} - 6\bar{k} \] Calculate the cross product \( \bar{b}_1 \times \bar{b}_2 \): \[ \bar{b}_1 \times \bar{b}_2 = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k}\\4 & 3 & -1 \\ 6 & 3 & -2 \end{vmatrix} \] \[ = \bar{i}(-6 - (-3)) - \bar{j}(-8 - (-6)) + \bar{k}(12 - 18) \] \[ = \bar{i}(-3) - \bar{j}(-2) + \bar{k}(-6) = -3\bar{i} + 2\bar{j} - 6\bar{k} \] Calculate the magnitude \( |\bar{b}_1 \times \bar{b}_2| \): \[ \sqrt{(-3)^2 + (2)^2 + (-6)^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \] Calculate the dot product \( (\bar{a}_2 - \bar{a}_1) \cdot (\bar{b}_1 \times \bar{b}_2) \): \[ (-2)(-3) + (7)(2) + (-6)(-6) = 6 + 14 + 36 = 56 \] Calculate the distance: \[ d = \frac{|56|}{7} = 8 \]
Step 4: Final Answer:

The shortest distance is 8.