Solution
ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-2/11/e/solution
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2024 Paper 2 11 e Solution by
Codex 0 2026-10-18
A necessary compatibility condition is , because a cross product with is perpendicular to it. If this condition fails, there is no solution.
If it holds, take the equation and use the vector triple product identity:Thus the perpendicular orthogonal projection is fixed and the parallel component is free:Conversely, under the compatibility condition, so every displayed vector solves the equation. The solution set is a straight line, including the line through the origin when .
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