In $\mathbb{R}^2$, the distance from a point $(u,v)$ to the line $ax+by+c=0$ is given by $d_\mathbb{R}=\frac{au+bv+c}{\sqrt{a^2+b^2}}$.
Suppose instead we have a line in the projective plane, i.e. the set of homogeneous points $\{(x,y,z) \mid ax+by+cz=0\}$ for some real $a,b,c$ not all zero. What is the distance from the point $(u,v,w) \in \mathbb{R}P^2$ to this line? Extending the above to something like $d_{\mathbb{R}P}=\frac{au+bv+cw}{\sqrt{a^2+b^2+c^2}}$ is not well-defined since $u,v,w$ can be scaled arbitrarily.
Is there even a good sense of distance in $\mathbb{R}P^2$?
Thanks.
Edit: Here's an example of what I'm looking for in general. Find the distance from the homogenous point $(4\lambda,5\lambda,6\lambda)$ to the projective line $L$ defined by $x+2y+3z=0$.
If $a^2 + b^2 + c^2 = 1$ and $u^2 + v^2 + w^2 = 1$, the requested distance is $$ \tfrac{\pi}{2} - \arccos|au + bv + cw| = \arcsin|au + bv + cw|. $$
The "usual" metric on the projective plane is the Riemannian quotient of the round unit sphere by the antipodal map. To find the distance from a projective line $\ell$ to a point $p$, interpret the homogeneous equation of $\ell$ as the equation of a plane in $\mathbf{R}^3$, and the normalized homogeneous coordinates of $p$ as a point on the unit sphere.
The formula above comes from viewing (the absolute value of) the dot product $au + bv + cw$ as the cosine of the angle between the point $(u, v, w)$ and a unit normal $(a, b, c)$ to the great circle representing $\ell$. The complementary angle is the distance from $p$ to $\ell$.