Let the manifold $S$ in $\mathbb R^n$ be defined by $g(x)=0$. If $p$ is a point not on $S$, and $q$ is the point of $S$ which is closest to $p$, show that the line from $p$ to $q$ is perpendicular to $S$ at $q$, Hint: Minimize $f(x)=|x-p|^2$
First off I don't understand why that is the function to minimize but since the method is simple I went on:
$\nabla f(x)=\lambda*\nabla g(x)$
$2(x-p)= \lambda*g_x$
For $\lambda$ different than 0:
$g_x=\frac{2(x-p)}\lambda$
From here I'm not sure what to do. I could integrate $g_x$ and find g(x), solve $g(x)=0$ for $\lambda$ then substitute. Is that the correct way of solving the problem?
"$q$ is the point of $S$ which is closest to $p$" means that $$ q = \mbox{arg}\min_{x \in S} |x - p|, $$ i.e., $$ |q - p| \leq |x - p| \qquad \forall x \in S. $$ Since the function $z\mapsto z^2$ is monotonic on $z \geq 0$, this is equivalent to $$ |q - p|^2 \leq |x - p|^2 \qquad \forall x \in S, $$ which may be solved by $$ q = \mbox{arg}\min_{x \in S} |x - p|^2. $$ That's why they minimize $f(x) = |x-p|^2$.
What you said should work. Note that your equation should really be $$ 2(x_i - p_i) = \lambda g_{x_i}, \quad 1 \leq i \leq n. $$ (Otherwise, the LHS and RHS do not live in the same space, really.)