Question

Find the shortest distance between lines \( \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} \) and \( \frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5} \).

Hint:

Use the vector formula for shortest distance; ensure accurate cross and dot product calculations.

Answer 1

Line 1: Point \( \mathbf{a}_1 = (1, 2, 3) \), direction vector \( \mathbf{b}_1 = (2, 3, 4) \).
Line 2: Point \( \mathbf{a}_2 = (2, 4, 5) \), direction vector \( \mathbf{b}_2 = (3, 4, 5) \).
Shortest distance formula: \[ d = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)|}{|\mathbf{b}_1 \times \mathbf{b}_2|}. \] \[ \mathbf{a}_2 - \mathbf{a}_1 = (2-1, 4-2, 5-3) = (1, 2, 2). \] \[ \mathbf{b}_1 \times \mathbf{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4\\ 3 & 4 & 5 \end{vmatrix} = \hat{i}(3 \cdot 5 - 4 \cdot 4) - \hat{j}(2 \cdot 5 - 4 \cdot 3) + \hat{k}(2 \cdot 4 - 3 \cdot 3) = (-1, 2, -1). \] \[ |\mathbf{b}_1 \times \mathbf{b}_2| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}. \] \[ (\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2) = (1, 2, 2) \cdot (-1, 2, -1) = -1 + 4 - 2 = 1. \] \[ d = \frac{|1|}{\sqrt{6}} = \frac{1}{\sqrt{6}}. \] Answer: \( \frac{1}{\sqrt{6}} \).