Condition under which a point belongs to the convex hull of some other points I can't understand a fact used in the proof of a theorem I am reading.  It can be stated as follows:

Let $x_0,\dots,x_n$ be points in some Euclidean space $\Bbb R^d$. Then a point $p\in\Bbb R^d$ belongs to the convex hull of $x_0,\dots,x_n$ if there does not exist a vector $v\in\Bbb R^d$ such that $\langle v,x_i-p\rangle>0$ for all $x_i$.

I know that a point $p$ is in the convex hull if, by definition, it can be written as
$$
p=\sum_{i=0}^n\lambda_i x_i
$$
for some $\lambda_i>0$ such that $\sum\limits_{i=0}^n\lambda_i=1$.  I tried taking the scalar product with $x_i-p$ and separating parts by linearity, but I didn't succeed in proving it.  How can I prove it?
Vectors $v$ that satisfy the statement are called separating vectors.  Why is this term used?  What's the intuition behind?  
 A: The intuition is that a point $p$ and a vector $v$ define a hyperplane
in $\mathbb R^d$ through $p$ perpendicular to $v$.
The points $q$ such that the inner product of the vector $q - p$ with $v$
is positive, $\langle v, q-p \rangle > 0$,
are all on one side of this hyperplane, the points $q$ such that
$\langle v, q-p \rangle < 0$ are all on the opposite side of this hyperplane,
and the points $q$ such that $\langle v, q-p \rangle = 0$
are on the hyperplane itself.
That is, for any point $p$, the vector $v$ defines a hyperplane through
$p$ that "separates" $\mathbb R^d$.
If there is a vector $v$ such that $\langle v,x_i-p\rangle>0$ for all $x_i$,
it means all the points in the set  $\{x_0,\ldots,x_n\}$
lie strictly on the same side of the hyperplane 
through $p$ perpendicular to $v$; no points are on the "other side"
or on the hyperplane itself.
If there is a hyperplane in $\mathbb R^d$ such that 
all of the points of the convex hull of
$\{x_0,\ldots,x_n\}$ are strictly on one side of that hyperplane,
what does that say about the rest of the points in the convex hull
of $\{x_0,\ldots,x_n\}$?
What does it say about a point $p$ that lies exactly on
that hyperplane?
