 Solution

ID: 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:

D (t)^{2} = (a_{2} + t n_{2}) \cdot (a_{2} + t n_{2}) = ∥ a_{2} ∥^{2} + 2 t (a_{2} \cdot n_{2}) + t^{2} ∥ n_{2} ∥^{2} .

D (t)^{2} = ∥ n_{2} ∥^{2} (t + \frac{a _{2} \cdot n _{2}}{∥ n _{2} ∥ ^{2}})^{2} + ∥ a_{2} ∥^{2} - \frac{( a _{2} \cdot n _{2} ) ^{2}}{∥ n _{2} ∥ ^{2}} .

The square is minimized at

t_{0} = - (a_{2} \cdot n_{2}) /∥ n_{2} ∥^{2}

, yielding

d_{0} = a_{2} \cdot a_{2} - \frac{( a _{2} \cdot n _{2} ) ^{2}}{n _{2} \cdot n _{2}} .

Geometrically this subtracts the squared length of the orthogonal projection parallel to the line. At the minimizing point,

(a_{2} + t_{0} n_{2}) \cdot n_{2} = 0

 New to topics? Read the docs here!