Mojtaba has a winning strategy.
Let him consider a large vector $\vec{V}$ to the left direction and $2017$ small vectors facing towards right. Let these vectors be $\vec{V} = (-V, 0)$, $\vec{v}_1 = (x_1, y_1)$, $\vec{v}_2 = (x_2, y_2)$, $\dots$, $\vec{v}_{2017} = (x_{2017}, y_{2017})$. So that we have
$$x_1 + \dots + x_{2017} = V, \quad \forall\ 1 \le i \le 2017 : x_i > 0$$
and
$$y_1 + \dots + y_{2017} = 0.$$
On his first move, Mojtaba chooses the large vector $\vec{V}$ and places it on the plane.

Now both players must select a vector from the remaining vectors, which all are from left to right. Therefore none of these small vectors intersect each other, and the only possible way to have a pair of vectors intersecting each other is to have a small vector $\vec{u}_i$ that cuts $\vec{V}$ at some point other than the two ending points of $\vec{V}$. Assume that $\vec{u}_i = (x_i^*, y_i^*)$ is the first vector to intersect with $\vec{V}$ ($\vec{u}_1 = (x_1^*, y_1^*)$, ..., $\vec{u}_{2017} = (x_{2017}^*, y_{2017}^*)$) is a permutation of the vectors, sorted by the time they're chosen). There's two cases, we either have $y_j^* > 0, \forall j < i$ and $y_1^* + y_2^* + \cdots + y_i^* < 0$, or $y_j^* < 0, \forall j < i$ and $y_1^* + y_2^* + \cdots + y_i^* > 0$. So we can translate the problem as following.
There are $2017$ real numbers $y_1, \dots, y_{2017}$ with $y_1 + \dots + y_{2017} = 0$. Each player, starting from Hooman chooses a number $y_k$ and writes it on the plane. Assume that numbers are arranged by $y_1^*, y_2^*, \dots, y_{2017}^*$, in order of the time they're chosen. Without loss of generality assume that $y_1^* \ge 0$ (we will define $y_i$'s so that there's no loss of generality, see the following). Mojtaba wins if for all $i \le 2017$ we have
$$S_i = y_1^* + \cdots + y_i^* \ge 0.$$
Also add this assumption that Mojtaba drew the initial $2018$ vectors such that $\forall 1 \le 2i + 1, 2j \le 2017 : y_{2i+1} = -y_{2j} = 1$ except for $2i + 1 = 1397$ where $|y_{1397}| = 0$. Now the winning strategy is quite simple. In each move, Mojtaba only needs to choose a positive (or if $y_1^* < 0$, a negative) $y_i^*$, if such $y_i^*$ exists. Otherwise, he chooses an arbitrary remaining number. Now we have
$$S_2 = y_1^* + y_2^* \geq 1$$
We also always have $S_i \ge 0$ unless all non-negative $y_i^*$'s are chosen, and Mojtaba is forced to choose from the remaining negative numbers. Now if there's an index $i \le 2017$ such that $S_i < 0$, since $S_{2017} = 0$, there must be an index $j > i$ such that $y_j^* > 0$. But this is impossible, because $S_i < 0$ means Mojtaba is out of positive numbers to choose from, so $y_j^*$ cannot exist which is a contradiction. Thus, there's no such $i$ and the claim that Mojtaba has a winning strategy is proved. ■