Past exam of the mathematical methods I course of the University of Oxford / 2015 / 10 / a / Solution

For the remaining distance calculation with $b=c$, every point on the line is $a+tb$. The nearest point to $p$ is found by orthogonal projection onto $b$:Indeed, differentiating $|a+tb-p{|}^2$ sets its component along $b$ to zero. Its second derivative is $2|b{|}^2>0$, proving this is the minimum.