Discussion
The shortest distance between the lines \(\vec{r} = \hat{i} + \hat{j} + 3\hat{k} + \lambda(2\hat{i} + 2\hat{j} + \hat{k})\) and \(\vec{r} = (2\mu + 1)\hat{i} + (2\mu - 1)\hat{j} + (\mu + 1)\hat{k}\) , where \(\lambda\) and \(\mu\) are parameters, is
Answer: E
The shortest distance is 2.
The second line is \((\hat{i} - \hat{j} + \hat{k}) + \mu(2\hat{i} + 2\hat{j} + \hat{k})\), so both lines have direction \(\vec{d} = (2, 2, 1)\) and are parallel. Use a point on each: (1, 1, 3) and (1, -1, 1).
\[ \begin{aligned} \vec{w} &= (1, -1, 1) - (1, 1, 3) = (0, -2, -2) \\ \vec{w} \times \vec{d} &= (2, -4, 4),\ |\vec{w} \times \vec{d}| = 6 \\ d &= \frac{6}{|\vec{d}|} = \frac{6}{3} = 2 \end{aligned} \]
6 is the size of the cross product before dividing by \(|\vec{d}| = 3\). The skew-line formula does not apply here because the lines are parallel.
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