Problem Statement
Find the shortest distance between the two lines:
$L_1: \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda (2\hat{i} + 3\hat{j} + 4\hat{k})$
$L_2: \vec{r} = (2\hat{i} + 4\hat{j} + 5\hat{k}) + \mu (3\hat{i} + 4\hat{j} + 5\hat{k})$
Verified Solution & Marking Scheme
Extract points and direction vectors
$\vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}, \quad \vec{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}$
$\vec{a}_2 = 2\hat{i} + 4\hat{j} + 5\hat{k}, \quad \vec{b}_2 = 3\hat{i} + 4\hat{j} + 5\hat{k}$
$\vec{a}_2 - \vec{a}_1 = \hat{i} + 2\hat{j} + 2\hat{k}$
Compute cross product b₁ × b₂
$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = \hat{i}(15 - 16) - \hat{j}(10 - 12) + \hat{k}(8 - 9) = -\hat{i} + 2\hat{j} - \hat{k}$
$|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$
Evaluate scalar triple product and shortest distance d
$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (1)(-1) + (2)(2) + (2)(-1) = -1 + 4 - 2 = 1$
$d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right| = \frac{1}{\sqrt{6}} = \frac{\sqrt{6}}{6}\text{ units}$