What is the shortest distance between the lines $\vec{r} = (1 + t)\hat{i} + (3 - 2t)\hat{j} + 2t\hat{k}$ and $\vec{r} = (2s - 1)\hat{i} + 2s\hat{j} + (s + 2)\hat{k}$?

Explanation

1. 01

   Split each into a point and a direction: $\vec{r} = (\hat{i} + 3\hat{j}) + t(\hat{i} - 2\hat{j} + 2\hat{k})$ and $\vec{r} = (-\hat{i} + 2\hat{k}) + s(2\hat{i} + 2\hat{j} + \hat{k})$.

2. 02

   $\vec{b}_1 \times \vec{b}_2 = -6\hat{i} + 3\hat{j} + 6\hat{k}$, of length 9, and $\vec{a}_2 - \vec{a}_1 = -2\hat{i} - 3\hat{j} + 2\hat{k}$.

3. 03

   $d = \frac{\lvert 12 - 9 + 12 \rvert}{9} = \frac{15}{9} = \frac{5}{3}$.