Vtyjkyuk

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 $$langleomega-v^*,v-v^*rangle = 0?$$

If this is correct, what is the intuition behind this? (writing the equation using the scalar product of two vectors?)

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 $$beginalign langleomega-v^*,v-v^*rangle &= langleomega,v-v^*rangle - langle v^*,v-v^*rangle \ &= a_1x_1+cdots+a_nx_n+dendalign$$ 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:

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.