Q.19 Find the shortest distance between the lines \(\bar{r} = (4\hat{i} - \hat{j}) + \lambda(\hat{i} + 2\hat{j} - 3\hat{k})\) and \(\bar{r} = (\hat{i} - \hat{j} - 2\hat{k}) + \mu(\hat{i} + 4\hat{j} - 5\hat{k})\)
Best answer
Answer: \(\frac{5}{\sqrt{3}}\) units
\(\bar{a}_1 = 4\hat{i} - \hat{j}\), \(\bar{b}_1 = \hat{i} + 2\hat{j} - 3\hat{k}\); \(\bar{a}_2 = \hat{i} - \hat{j} - 2\hat{k}\), \(\bar{b}_2 = \hat{i} + 4\hat{j} - 5\hat{k}\).
\(\bar{a}_2 - \bar{a}_1 = -3\hat{i} + 0\hat{j} - 2\hat{k}\).
\(\bar{b}_1 \times \bar{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 1 & 4 & -5 \end{vmatrix} = \hat{i}(-10 + 12) - \hat{j}(-5 + 3) + \hat{k}(4 - 2) = 2\hat{i} + 2\hat{j} + 2\hat{k}\).
\(|\bar{b}_1 \times \bar{b}_2| = \sqrt{4 + 4 + 4} = 2\sqrt{3}\).
\((\bar{a}_2 - \bar{a}_1) \cdot (\bar{b}_1 \times \bar{b}_2) = -6 + 0 - 4 = -10\).
Shortest distance \(d = \frac{|(\bar{a}_2 - \bar{a}_1) \cdot (\bar{b}_1 \times \bar{b}_2)|}{|\bar{b}_1 \times \bar{b}_2|} = \frac{10}{2\sqrt{3}} = \frac{5}{\sqrt{3}}\) units.
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