 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1/11/c/ii/solution

The nonzero direction vectors now have linear dependence, so the lines are parallel or coincident. Subtract from

d = a_{2} - a_{1}

v_{1}

; the remaining component is the shortest joining displacement. The distance between parallel lines is

d (L_{1}, L_{2}) = d - \frac{d \cdot v _{1}}{v _{1} \cdot v _{1}} v_{1} = \frac{∥ d \times v _{1} ∥}{∥ v _{1} ∥} .

The second expression follows from the relation between the cross product and the perpendicular component. It includes distance zero for coincident lines.

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