Question

Find the shortest distance between the following pairs of lines: \( \vec{r} = \hat{i} - 4\hat{j} + 5\hat{k} + \mu(5\hat{i} + 9\hat{j} + \hat{k}) \) and \( \vec{r} = 2\hat{i} + 8\hat{j} - 6\hat{k} + \lambda(3\hat{i} - 2\hat{j} + \hat{k}) \).

Hint:

When calculating \( \vec{b}_1 \times \vec{b}_2 \), double-check the signs of the components, especially the \( \hat{j} \) component, which has a negative sign in front of its minor.

Answer 1

Step 1: Understanding the Concept:
The shortest distance between two skew lines \( \vec{r} = \vec{a}_1 + \mu\vec{b}_1 \) and \( \vec{r} = \vec{a}_2 + \lambda\vec{b}_2 \) is given by the formula:
\[ d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|} \] Step 2: Detailed Explanation:
Identify parameters:
\( \vec{a}_1 = (1, -4, 5) \), \( \vec{b}_1 = (5, 9, 1) \).
\( \vec{a}_2 = (2, 8, -6) \), \( \vec{b}_2 = (3, -2, 1) \).
1) \( \vec{a}_2 - \vec{a}_1 = (1, 12, -11) \).
2) \( \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 9 & 1 \\ 3 & -2 & 1 \end{vmatrix} = \hat{i}(9+2) - \hat{j}(5-3) + \hat{k}(-10-27) = (11, -2, -37) \).
3) \( | \vec{b}_1 \times \vec{b}_2 | = \sqrt{11^2 + (-2)^2 + (-37)^2} = \sqrt{121 + 4 + 1369} = \sqrt{1494} \).
4) \( (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 1(11) + 12(-2) + (-11)(-37) = 11 - 24 + 407 = 394 \).
Distance \( d = \frac{394}{\sqrt{1494}} \).
Step 3: Final Answer:
The shortest distance is \( \frac{394}{\sqrt{1494}} \) units.