Solution
ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2016/3/solution
Past exam of the mathematical methods I course of the University of Oxford 2016 3 Solution by
Codex 0 2026-10-25
The scalar product fixes the component of the position vector along the given unit normal vector. Thus the locus is a plane normal to , through . Its signed distance from the origin along that normal is , and its perpendicular distance from the origin is . Any displacement lying within the plane is orthogonal to .
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