Problem Statement
Find the shortest distance between the following two skew lines in vector form:
$$\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - 3\hat{j} + 2\hat{k})$$
$$\vec{r} = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu(2\hat{i} + 3\hat{j} + \hat{k})$$
Verified Solution & Marking Scheme
Identify position vectors and direction vectors
$$\vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}, \quad \vec{b}_1 = \hat{i} - 3\hat{j} + 2\hat{k}$$
$$\vec{a}_2 = 4\hat{i} + 5\hat{j} + 6\hat{k}, \quad \vec{b}_2 = 2\hat{i} + 3\hat{j} + \hat{k}$$
$$\vec{a}_2 - \vec{a}_1 = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}$$
Compute cross product b₁ × b₂
$$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} = \hat{i}(-3 - 6) - \hat{j}(1 - 4) + \hat{k}(3 - (-6)) = -9\hat{i} + 3\hat{j} + 9\hat{k}$$
$$|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-9)^2 + 3^2 + 9^2} = \sqrt{81 + 9 + 81} = \sqrt{171} = 3\sqrt{19}$$
Compute scalar dot product (b₁ × b₂) · (a₂ - a₁)
$$(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1) = (-9)(3) + (3)(3) + (9)(3) = -27 + 9 + 27 = 9$$
Apply shortest distance formula
$$d = \left| \frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|} \right| = \frac{|9|}{3\sqrt{19}} = \frac{3}{\sqrt{19}} = \frac{3\sqrt{19}}{19}\text{ units}$$