We want to find a vector $\mathbf{v} \in \mathbb{R}^2$, while knowing vectors $\mathbf{e}, \mathbf{v}_0, \mathbf{v}_1$ all in $\mathbb{R}^2$. Please check the configuration of the vectors on the image below.
The best I could find about this problem is the following:
Let $\varepsilon \in [0,1]$, $$\mathbf{v} + \varepsilon \mathbf{e} = \mathbf{v}_0 \;\; (1)$$ $$\mathbf{v} - (1-\varepsilon) \mathbf{e} = \mathbf{v}_1 \;\; (2)$$
Now, multiply (1) with $(1-\varepsilon)$ and (2) with $\varepsilon$, then sum both equations:
$$\mathbf{v} = (1-\varepsilon)\mathbf{v}_0 + \varepsilon \mathbf{v}_1$$
But now I am stuck with not knowing $\varepsilon$.
Another thing I noticed, is that: $$\mathbf{v}_{\bot} = \mathbf{v}_{0\bot} =\mathbf{v}_{1\bot}$$
Where $\mathbf{v}_\bot$ is the orthogonal projection of $\mathbf{v}$ into $\mathbf{e}$
$\vec{e}=\|e\|\vec{e_1}$ where $\|\vec{e_1}\|=1$, from figure $\vec{v_0}-\vec{v_1}=\vec{e}$, $\vec{v_1}+k\vec{e_1}=-\vec{v}$ and $\vec{v_0}-(\|e\|-k)\vec{e_1}=-\vec{v}$; so $\vec{v}$ is not unique, it depends on $k$ constant length.