Given a vector $x \in \mathbb{R}^{n}_{0+}$ such that $x = \sum^{k}_{i=1} \alpha_{i}v_{i}$, the vectors $(v_{1},...,v_{k}) \in \mathbb{R}^{n}_{0+}$ are an independent set, $k < n$, and $\alpha_{i} > 0$, it can be seen by simple combinatorial argument that for at least one vector $v_{i}$, $\frac{||x - \alpha_{i}v_{i}||_{2}}{||x||_{2}} \geq \frac{1}{k}$ (assume the opposite, then we have that the maximum value of $\frac{||x - \alpha_{i}v_{i}||_{2}}{||x||_{2}} < \frac{1}{k}$ but this implies that $||x - \sum^{k}_{i=1} \alpha_{i}v_{i}||_{2} > 0$ which is a contradiction).

Say we have an additional vector $y \in \mathbb{R}^{n}_{0+}$ such that $(v_{1},...,v_{k},y)$ is an independent set. What can be said about $\frac{||x - \beta y||_{2}}{||x||_{2}}$ where $\beta$ is the result of minimizing $||x-y\beta||_{2}$ subject to $\beta \geq 0$? If $(v_{1},...,v_{k},y)$ is orthogonal then $\frac{||x - \beta y||_{2}}{||x||_{2}} = 1$, but bounds in the absence of orthogonality don't seem as obvious. If we impose the additional requirement that $x - y\beta \geq 0$ element-wise are there stronger conclusions that can be drawn?

P.S. Apologies if the title I chose for this question is not perfectly representative of what I actually asked, I was unable to determine if there is a common technical term for this sort of question.

EDIT: Removed silly equivocation about the case when $(v_{1},...,v_{k},y)$ is orthogonal.

EDIT 2: Further research revealed a flaw in the reasoning presented here. I'm in the process of reformulating the question and examining if I can come to any conclusions myself in the meantime. I may repost at a later date if I can't make further progress, but as written here this question is ill-posed.