Complexity of finding vectors with optimal projection?
The problem is NP-Hard and the reduction is from partition. 

So, assume you are given an instance of partition $S=${$x_1,\ldots,x_n$}. Let $t = (\sum_{i=1}^n x_i) / 2$. The question is whether there is a subset of $S$ such that its sum is $t$.

To this end, let $v_0 = (0, 2t, t)$ be a special vector. For every number $x_i$, create the vector 

$v_i = (x_i, -x_i, 0)$,

for $i=1,\ldots, n$. Now, the claim is that there is a subset of the vectors with sum $\geq (t, t, t)$ (in absolute value) if and only if there is a subset of $S$ that add up to $t$.

So, consider a subset $X \subseteq S$, with $\alpha = \sum_{x\in X}$. We have that the corresponding sum of vectors (together with the special vector $v_0$) is

$(\alpha, 2t -\alpha, t)$. 

Clearly, this vector is maximized when $\alpha=t$, as required. 

Note, that you must include the special vector $v_0$ - otherwise the third coordinate would be zero in the sum of vectors.

QED

Note, that the problem is solvable in polynomial time if the numbers are polynomially small, doing dynamic programming (like the one used for solving subset sum if the numbers are small).

2d

If you set the special vector to be $(8t, 10t)$ then the above reduction would work verbatim in two dimensions. (The constant $8$ here is somewhat arbitrary...)