The shortest distance between the following lines:
\(\vec{r} = (\hat{\imath} + \hat{\jmath} - \hat{k}) + s(2\hat{\imath} + \hat{\jmath} + \hat{k})\)
\(\vec{r} = (\hat{\imath} + \hat{\jmath} + 2\hat{k}) + t(4\hat{\imath} + 2\hat{\jmath} + 2\hat{k})\), where s and t are scalars, is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\sqrt{\displaystyle\frac{15}{2}}\)
The direction \(4\hat{\imath} + 2\hat{\jmath} + 2\hat{k} = 2(2\hat{\imath} + \hat{\jmath} + \hat{k})\), so the lines are parallel. For parallel lines \(\vec{r} = \vec{a_1} + s\vec{b}\) and \(\vec{r} = \vec{a_2} + t\vec{b}\), the shortest distance is
\( \displaystyle d = \frac{|\vec{b} \times (\vec{a_2} - \vec{a_1})|}{|\vec{b}|} \)
Here:
\(\vec{b} = 2\hat{\imath} + \hat{\jmath} + \hat{k},\quad \vec{a_2} - \vec{a_1} = 3\hat{k}\)
\( \displaystyle \vec{b} \times 3\hat{k} = \begin{vmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\2&1&1\\0&0&3\end{vmatrix} = 3\hat{\imath} - 6\hat{\jmath} \)
\(|\vec{b} \times 3\hat{k}| = \sqrt{9 + 36} = \sqrt{45}\)
\(|\vec{b}| = \sqrt{4 + 1 + 1} = \sqrt{6}\)
So:
\( \displaystyle d = \frac{\sqrt{45}}{\sqrt{6}} = \sqrt{\frac{45}{6}} = \sqrt{\frac{15}{2}} \)