Metric for a set of skew lines in three dimensional space Suppose we have a set $M = \left\{\boldsymbol{x}\in \mathbb{R}^3_{++}: \boldsymbol{x} = \boldsymbol{\alpha}+\lambda \boldsymbol{\beta}, \lambda\in (0,t), \boldsymbol{\alpha},\boldsymbol{\beta} \in \mathbb{R}^3 \right\},$ where $t$ is a positive number and $\mathbb{R}^3_{++}$ implies that all coordinates in $\boldsymbol{x}$ is strictly positive. Also, for any $\boldsymbol{x'},\boldsymbol{x''}\in M$, $\boldsymbol{x'}$ are not parallel/orthogonal to $\boldsymbol{x''}$.
I am trying to find a valid metric for this set but failed. I have tried the shortest distance between the two lines (i.e., $d(\boldsymbol{x'},\boldsymbol{x''})=\left\|\frac{\boldsymbol{\beta'}\times\boldsymbol{\beta''}\cdot(\boldsymbol{\alpha'}-\boldsymbol{\alpha''})}{\|\boldsymbol{\beta'}\times \boldsymbol{\beta''}\|}\right\|$) and the volume of the tetrahedron bounded by the four endpoints of the two lines (i.e., $d(\boldsymbol{x'},\boldsymbol{x''})=\left\|\frac{\boldsymbol{\beta'}\times\boldsymbol{\beta''}\cdot(\boldsymbol{\alpha'}-\boldsymbol{\alpha''})}{6}\right\|$). But they all violate the triangle inequality condition. Can anyone help me with this? Thank you so much for your attention and participation.
Edit: I also tried the angle between the lines $$d(\boldsymbol{x'},\boldsymbol{x''}) = \left\|\frac{\alpha \cdot \beta}{||\alpha||*||\beta||} \right\|=\left\|cos(\theta)\right\|,$$ where $\theta$ is the angle between the two lines. But I failed to prove the triangle inequality as well.
Edit: I am looking for a metric that involves the $L^2-$norm.
 A: Define a function $d: M \times M  \to \mathbb R^+$ by $d(x, y) = sup_{\lambda \in (0, t)}(||(x - y)(\lambda)||) $ where $||.||$ is the $L_2$ norm in $R^3$.
Firstly, is this well defined ?
Certainly, $x - y$ is well defined for $\lambda \in (0, t)$ and $||(x - y)(\lambda)||$ is a positive ($\ge 0$) real value.
$||(x - y)(\lambda)|| \le ||x(\lambda)|| + ||y(\lambda)||$ and with $\alpha, \beta$ fixed for $x$, $t$ fixed, and $\lambda \in (0, t)$  then $||x(\lambda)||$  is bounded above as is $||y(\lambda)||$.
So, $||(x - y)(\lambda)|| $ is bounded above and therefore has a supremum.
Secondly, is $d$ a metric ?

*

*$d(x, y) \ge 0$ follows from the property of $||.|| \ge 0$

*$d(x,x) = sup(||0||) = 0$ and if $x \ne y$ then clearly, $d(x, y) \ne 0$

*$d(x, z) = sup_{\lambda \in (0, t)}(||(x - z)(\lambda)||) $
$= sup_{\lambda \in (0, t)}(||x(\lambda) - z(\lambda)||) $
$= sup_{\lambda \in (0, t)}(||x(\lambda) - y(\lambda)+ y(\lambda)- z(\lambda)||) $
$\le sup_{\lambda \in (0, t)}(||x(\lambda) - y(\lambda)|| + ||y(\lambda)- z(\lambda)||) $  by property of $||.||$
$\le sup_{\lambda \in (0, t)}(||x(\lambda) - y(\lambda)||) + sup_{\lambda \in (0, t)}(||y(\lambda)- z(\lambda)||) $ since $||.||$ is positive
I.e. $d(x, z) \le d(x, y) + d(y, z) $
So, $d$ is a metric.
Note, ||.|| doesn't have to be the $L_2$ norm, any norm on $\mathbb R^3$ results in a supremum metric.
This is just a specific case of a supremum metric on "bounded" functions. Your functions ($x \in M$) are bounded in terms of any norm on $\mathbb R^3$ since they are a linear combination of two vectors using a parameter $\lambda$ in a finite range $(0, t)$.
