Problem Statement
Find the shortest distance between the two skew lines given by:\n\n$$\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - 3\hat{j} + 2\hat{k})$$\n\n$$\vec{r} = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu(2\hat{i} + 3\hat{j} + \hat{k})$$
Verified Solution & Marking Scheme
Identify vectors a₁, a₂, b₁, b₂
$$\vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}, \quad \vec{b}_1 = \hat{i} - 3\hat{j} + 2\hat{k}$$\n$$\vec{a}_2 = 4\hat{i} + 5\hat{j} + 6\hat{k}, \quad \vec{b}_2 = 2\hat{i} + 3\hat{j} + \hat{k}$$
Compute (a₂ - a₁)
$$\vec{a}_2 - \vec{a}_1 = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}$$
Compute cross product (b₁ × b₂)
$$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} = \hat{i}(-3 - 6) - \hat{j}(1 - 4) + \hat{k}(3 - (-6)) = -9\hat{i} + 3\hat{j} + 9\hat{k}$$\n\n$$|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-9)^2 + 3^2 + 9^2} = \sqrt{81 + 9 + 81} = \sqrt{171} = 3\sqrt{19}$$
Calculate Shortest Distance formula
$$d = \left| \frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|} \right| = \left| \frac{(-9)(3) + 3(3) + 9(3)}{3\sqrt{19}} \right| = \left| \frac{-27 + 9 + 27}{3\sqrt{19}} \right| = \frac{9}{3\sqrt{19}} = \frac{3}{\sqrt{19}}$$