Show that the lines \(\bar{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k})\) and \(\bar{r} = (4\hat{i} - 3\hat{j} + 2\hat{k}) + \mu(\hat{i} - 2\hat{j} + 2\hat{k})\) intersect each other.

Answer: The lines intersect at the point \((3, -1, 0)\).

A general point on the first line is \((1 + 2\lambda,\ 1 - 2\lambda,\ -1 + \lambda)\) and on the second is \((4 + \mu,\ -3 - 2\mu,\ 2 + 2\mu)\).

If the lines meet, for some \(\lambda, \mu\):

• \(1 + 2\lambda = 4 + \mu \Rightarrow 2\lambda - \mu = 3\) ... (1)
• \(1 - 2\lambda = -3 - 2\mu \Rightarrow 2\lambda - 2\mu = 4 \Rightarrow \lambda - \mu = 2\) ... (2)
• \(-1 + \lambda = 2 + 2\mu \Rightarrow \lambda - 2\mu = 3\) ... (3)

Subtracting (2) from (1): \(\lambda = 1\), then \(\mu = -1\). Check in (3): \(1 - 2(-1) = 3\), true.

Since the three equations are consistent, the lines intersect. Point of intersection: \(\lambda = 1\) gives \((3, -1, 0)\).

(Alternatively, \(\bar{a}_2 - \bar{a}_1 = 3\hat{i} - 4\hat{j} + 3\hat{k}\) and \(\begin{vmatrix} 3 & -4 & 3 \\ 2 & -2 & 1 \\ 1 & -2 & 2 \end{vmatrix} = 3(-2) + 4(3) + 3(-2) = 0\), so the shortest distance is zero and the non-parallel lines intersect.)