Such a metric does exist. Simply let $d(L_1,L_2)$ equal the angle between $L_1$ and $L_2$ plus the distance between the closest point in $L_1$ to $0$ and the closest point in $L_2$ to $0$.

This is a metric because it is a sum of two pseudometrics and the distance between distinct lines is $0$. We easily verify that the two bullet points are satisfied, because parallel lines have their closest point to $0$ in the plane which is perpendicular to the lines and passes through $0$.

It dodges Vidit's proof because it is not, as Robert notes, invariant under the group of Euclidean motions.

Such a metric does exist. Simply let $d(L_1,L_2)$ equal the angle between $L_1$ and $L_2$ plus the distance between $P_1$ and $Q_1$, where $P_1$ is the closest point in $L_1$ to $0$ and $Q_2$ is the closest point in $L_2$ to $0$.

This is a metric because it is a sum of two pseudometrics and the distance between distinct lines is $0$. We easily verify that the two bullet points are satisfied, because parallel lines have their closest point to $0$ in the plane which is perpendicular to the lines and passes through $0$.

It dodges Vidit's proof because it is not, as Robert notes, invariant under the group of Euclidean motions.

Such a metric does exist. Simply let $d(L_1,L_2)$ equal the angle between $L_1$ and $L_2$ plus the distance between the closest point in $L_1$ to $0$ and the closest point in $L_2$ to $0$.

Such a metric does exist. Simply let $d(L_1,L_2)$ equal the angle between $L_1$ and $L_2$ plus the distance between the closest point in $L_1$ to $0$ and the closest point in $L_2$ to $0$.

This is a metric because it is a sum of two pseudometrics and the distance between distinct lines is $0$. We easily verify that the two bullet points are satisfied, because parallel lines have their closest point to $0$ in the plane which is perpendicular to the lines and passes through $0$.

It dodges Vidit's proof because it is not, as Robert notes, invariant under the group of Euclidean motions.