Solve $\mathbf{\vec{a}} \, \cdot \mathbf{\vec{x_{i+1}}} \gt \mathbf{\vec{a}} \, \cdot \mathbf{\vec{x_{i}}} $ for $\mathbf{\vec{a}}$ I have a set $X = \{\mathbf{\vec{x_1}},\mathbf{\vec{x_2}},\mathbf{\vec{x_3}},\ldots,\mathbf{\vec{x_k}}\}$ of vectors, all of which are $n$-dimensional.
I want to find an $n$-dimensional vector $\mathbf{\vec{a}}$, such that $\mathbf{\vec{a}} \, \cdot \mathbf{\vec{x_{i+1}}} \gt \mathbf{\vec{a}} \, \cdot \mathbf{\vec{x_{i}}} $ for $i = 1,2,3,\ldots, k$.
My question is: to be able to find such a vector, does the equality $n = k$ have to hold?
And, how does one go about solving this problem?

Background:
Basically, I was thinking of coming up with a single metric that measures athleticism in basketball. Each vector $\mathbf{\vec{x_i}}$ represents physical data about a player (i.e. wingspan, height, vertical jump, etc.). 
However, I do not know the weights of each of these elements (i.e. their contribution to athleticism), but I do know the ranking of the players (i.e. who is more athletic than who). So, I was wondering if there is a way to determine the weights ($\mathbf{\vec{a}}$) using the ranking.

[I write superscripts to index the vectors.]
If the vectors $ \mathbf{\vec{x^{i}}} $ are linearly independent, then for  $k\le n$ you can find such a vector $\mathbf{\vec{a}} $. You can do even  more: you can fix the overlaps $\mathbf{\vec{a}} \, \cdot \mathbf{\vec{x^{i}}} $. So let 
$\mathbf{\vec{a}} \, \cdot \mathbf{\vec{x^{i}}} = c^i\
$ and fix numbers $c^k > c^{k-1} > \cdots > c^1$. Then, writing the conditions in components (subscripts), you have
$$
\begin{pmatrix}x^1_1 & x^1_2 & \cdots & \cdots & x^1_n \\ 
x^2_1 & x^2_2 & \cdots &\cdots & x^2_n \\ 
&  & \cdots & & \\ 
x^k_1 & x^k_2 & \cdots & \cdots &x^k_n \\ 
\end{pmatrix}
\cdot  \begin{pmatrix}a_1 \\ 
a_2 \\ 
\vdots \\ 
\vdots \\ 
a_n \\ 
\end{pmatrix}  = \begin{pmatrix}c_1 \\ 
c_2 \\ 
\vdots \\ 
c_k \\ 
\end{pmatrix}
$$ 
or for short, $
\mathbf X \cdot \mathbf{\vec{a}} = \mathbf c
$. Note the different dimensionalities.
Now this system is underdetermined, and a particular solution  for $\mathbf{\vec{a}} $, which also is rather robust to measurement errors in individual values of given or new vectors $ \mathbf{\vec{x}} $, is given by the Pseudoinverse (or Moore–Penrose inverse) which is 
$$
\mathbf{\vec{a}} = \mathbf X^T (\mathbf X \mathbf X^T)^{-1}  \mathbf c
$$
For $k > n$ the system is overdetermined. Then feasibility in general is not given, one has to inspect. 
One can ask for the probability that such an overdetermined system 
can be solved not with fixed values $c^k > c^{k-1} > \cdots > c^1$, but such that at least one such set of values will exist. This probability is not well determined  for $n < k < 2 n$ but it is known that it drops sharply for $k > 2 n$. The treatment of those probabilities is a matter of linear separability / cf. the function counting theorem (Thomas Cover,  1965) and of later investigations  in the storage capacities of perceptrons (Elisabeth Gardner, 1988). 
