Consider a point, $v$ outside a closed convex set $C$ and $\omega \in C$. Let $v^*$ be the orthogonol projection of $v$ onto $C$. Can I write the equation of the hyperplane goes through the projection which separates convex set and the point $v$ as $$\langle\omega-v^*,v-v^*\rangle = 0?$$
If this is correct, what is the intuition behind this? (writing the equation using the scalar product of two vectors?)
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There could be many hyperplanes separating a point from a general convex set, so it does not make sense to write "the equation of the hyperplane", but yes, if you take a hyperplane passing through the closest point in a convex set to some point $p$ outside the set, which is orthogonal to the line connecting $p$ to its closest point, then that hyperplane will separate the convex set from the point $p$, as you can easily prove.
This is just the point-normal form of implicit equation for a hyperplane, but expressed in a way that you might not have seen before. It’s easily converted into a more familiar form: if $\omega = (x_1,\dots,x_n)$, then $$\begin{align} \langle\omega-v^*,v-v^*\rangle &= \langle\omega,v-v^*\rangle - \langle v^*,v-v^*\rangle \\ &= a_1x_1+\cdots+a_nx_n+d\end{align}$$ where the $a_i$ are some fixed set of constants and $d=-\langle v^*,v-v^*\rangle$, also a constant. This plane clearly passes through $v^*$, since $\langle v^*-v^*,v-v^*\rangle = \langle0,v-v^*\rangle = 0$.
Geometrically, this is the set of all points $\omega$ such that $\omega-v^*$ is orthogonal to the fixed vector $v-v^*$. Now, $\omega-v^*$ is just the displacement vector that goes from the fixed point $v^*$ to $\omega$, so this equation describes the set of all points that can be reached by moving purely in a direction orthogonal to $v-v^*$. This example in $\mathbb R^2$ should give you the idea: