Q.20 Find the shortest distance between the lines, \(\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}\) and \(\frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}\).
Best answer
Answer: \(\frac{1}{\sqrt{6}}\) units
Here \(\bar{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}\), \(\bar{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}\), \(\bar{a}_2 = 2\hat{i} + 4\hat{j} + 5\hat{k}\), \(\bar{b}_2 = 3\hat{i} + 4\hat{j} + 5\hat{k}\).
\(\bar{a}_2 - \bar{a}_1 = \hat{i} + 2\hat{j} + 2\hat{k}\).
\(\bar{b}_1 \times \bar{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = \hat{i}(15 - 16) - \hat{j}(10 - 12) + \hat{k}(8 - 9) = -\hat{i} + 2\hat{j} - \hat{k}\), \(|\bar{b}_1 \times \bar{b}_2| = \sqrt{6}\).
\((\bar{a}_2 - \bar{a}_1)\cdot(\bar{b}_1 \times \bar{b}_2) = -1 + 4 - 2 = 1\).
\(d = \frac{|(\bar{a}_2 - \bar{a}_1)\cdot(\bar{b}_1 \times \bar{b}_2)|}{|\bar{b}_1 \times \bar{b}_2|} = \frac{1}{\sqrt{6}}\) units.
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