GSEB Class 12 (HSC) • 2023 • 4 Marks

Vectors & 3D Geometry: Shortest Distance Between Skew Lines

Official examination question with verified M1/A1 mark scheme and step-by-step mathematical reasoning.

Problem Statement

Find the shortest distance between the lines: $\vec{r}_1 = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - 3\hat{j} + 2\hat{k})$ $\vec{r}_2 = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu(2\hat{i} + 3\hat{j} + \hat{k})$

Verified Solution & Marking Scheme

Compute a₂ - a₁
$\vec{a}_2 - \vec{a}_1 = 3\hat{i} + 3\hat{j} + 3\hat{k}$
Compute b₁ × b₂
$\vec{b}_1 \times \vec{b}_2 = -9\hat{i} + 3\hat{j} + 9\hat{k}, \quad |\vec{b}_1 \times \vec{b}_2| = \sqrt{81 + 9 + 81} = 3\sqrt{19}$
Apply Distance Formula
$d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right| = \frac{|-27 + 9 + 27|}{3\sqrt{19}} = \frac{9}{3\sqrt{19}} = \frac{3\sqrt{19}}{19}$
Practice this question with AI Socratic guidance on MonoMath →