Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 11 / c / i / Solution

Use only the dot product to minimize the squared Euclidean distance from the origin:Completing the square givesThe square is minimized at $t_0=-({\mathbf{a}}_2\cdot {\mathbf{n}}_2)/\Vert {\mathbf{n}}_2{\Vert }^2$, yieldingGeometrically this subtracts the squared length of the orthogonal projection parallel to the line. At the minimizing point, $({\mathbf{a}}_2+t_0{\mathbf{n}}_2)\cdot {\mathbf{n}}_2=0$, as required for perpendicular distance.