Let $A \subseteq \mathbb{R}^n$ be the convex hull of finitely many points. Assume the original point is not in $A$. Let $a$ be the vector in $A$ with the minimum length (in $\ell_2$-norm). Is it true that all the points $v$ in $A$ are on the one side of the tangent plane at $a$, i.e., $$(v-a)\cdot a\ge 0?$$
And I remember there is a theorem saying that if the object function is linear, then the optimum should be obtained at some vertex of $A$. But here, $a$ can be on some face but not vertex of $A$. Is it because the object $\min\limits_{v\in A}\|v\|$ is quadratic? Any similar result for quadratic object function?
The answer is YES.
Let $S^{n-1}(a)$ be the sphere of dimension $n-1$ centered at the origin of radius $a$. As $A$ is convex and $a$ is a point of $A$ with minimum distance from the origin, then $$ S^{n-1}(a)\cap A=\{a\}, $$ otherwise, if there exists $a'\in S^{n-1}(a)\cap A$ with $a'\neq a$, by convexity the whole segment joining $a$ and $a'$ would be contained in $S^{n-1}(a)\cap A$ contradditting that $a$ has minimum distance from the origin.
It is easy to see that the cutting hyperplane $$ \pi:=\{x\in \mathbb{R}^n : (x-a)\cdot a=0\} $$ is exactly the hyperplane tangent to the sphere $S^{n-1}(a)$ at $a$.
Suppose now that there exists $v\in A$ such that $(v-a)\cdot a<0$, then by the convexity of $A$ the whole segment joining $v$ and $a$ is contained in A. But this is a contradiction as would exist a point $b$ in this segment contained in the interior of the sphere $S^{n-1}(a)$ and the distance of $b$ from the origin would be strictly smaller than the distance of $a$ from the origin.