Find the distance between parallel lines \(\frac{x}{2} = \frac{y}{-1} = \frac{z}{2}\) and \(\frac{x+1}{2} = \frac{y-1}{-1} = \frac{z+1}{2}\)

Answer: \(\frac{\sqrt{2}}{3}\) units

The lines pass through \(\bar{a}_1 = \bar{0}\) and \(\bar{a}_2 = -\hat{i} + \hat{j} - \hat{k}\), with common direction \(\bar{b} = 2\hat{i} - \hat{j} + 2\hat{k}\), \(|\bar{b}| = 3\).

\(\bar{a}_2 - \bar{a}_1 = -\hat{i} + \hat{j} - \hat{k}\).

\((\bar{a}_2 - \bar{a}_1)\times\bar{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -1 \\ 2 & -1 & 2 \end{vmatrix} = \hat{i}(2 - 1) - \hat{j}(-2 + 2) + \hat{k}(1 - 2) = \hat{i} - \hat{k}\), with magnitude \(\sqrt{2}\).

Distance \(d = \frac{|(\bar{a}_2 - \bar{a}_1)\times\bar{b}|}{|\bar{b}|} = \frac{\sqrt{2}}{3}\) units.