I've got this question on a problem sheet all about metric spaces. My idea was to consider points on the unit sphere, so that $d(L_1, L_2)$ would equal the sine of the angle between the two vectors. How would I use this to show that this is a metric? Or if there's a better way, could someone point me in the right direction?
Thanks
As you already noticed, $d(L_1,L_2)=\sin\theta$ where $\theta\in [0,\pi]$ is the angle between the lines. Now, it easy to note that $\sin\theta\geq 0~\forall~\theta\in [0,\pi]$ with equality iff $\theta=0$ iff $L_1=L_2$. Also, $d$ is obviously symmetric. What's left is to show the triangle inequality holds.
Suppose $d(L_1,L_3)=\sin a$ and $d(L_1,L_2)=\sin b$ and $d(L_2,L_3)=\sin c$. We need to show that $\sin a\leq\sin b+\sin c$
Now, there are two cases: the line $L_2$ can fall between $L_1$ and $L_3$ or outside.
For the first case, $a=b+c$ and so the triangle inequality follows simply by expanding $\sin(b+c)$ and using that $|\cos x|\leq 1~\forall~x\in\Bbb R$
For the second case, we have either $c=b+a$ or $b=a+c$ for which the proof follows similarly.
Hence, we conclude that $d$ is a metric.
