Define the following metric on $P(\mathbb{R}^2)$: For $\bar{x},\bar{y}\in P(\mathbb{R}^2)$ $(x=(x_1,x_2),y=(y_1,y_2)\in \mathbb{R}^2$), define
$$d(\bar{x},\bar{y})=\frac{|x_2 y_1-x_1 y_2|}{\|x\|\cdot \|y\|}$$
It is clear that $d(\bar{x},\bar{x})=0 \Longleftrightarrow x=0$, and $d(\bar{x},\bar{y})=d(\bar{y},\bar{x})$. Also, it is easy to see that it is well defined.
The only thing I am struggling with is the triangular inequality, I'm not sure how to approach it (the usual methods don't seem to work here). How do you prove it?
The numerator is the magnitude of the cross product between the vectors, and the cross product is $\|x\|\|y\|\sin\theta$ where $\theta$ is the angle between them. (Reference)
So the distance $d$ is simply the sine of the angle between two vectors. The angles are all between $0$ and $\pi$, so the sine is nonnegative. We can also box them in $[0, \pi/2]$, by picking the smaller of two complementary angles.
So, let $\alpha$ be the angle between $\bar x$ and $\bar y$, and $\beta$ the angle between $\bar y$ and $\bar z$; make sure that $\alpha,\beta\in [0, \pi/2]$. Then $$ |\sin(\alpha \pm \beta)| = |\sin\alpha \cos\beta \pm \cos\alpha \sin\beta| \le \sin \alpha + \sin\beta $$ Since the angle between $\bar x$ and $\bar z$ is given by either $\alpha+\beta$ or $\alpha-\beta$ or $\beta-\alpha$, the triangle inequality follows.