Visualize the orbits of exoplanets

Nice article, thanks. So this inclination is a measure of an offset of the plane of the exoplanet's orbit from the line passing through Earth and the star? So the reason most of these planets have very low inclination is A) those are easiest to detect since they block the sun the most, in particular we couldn't detect a planet with 90 degree inclination at all, and B) there's a symmetry involved so there are more ways to have 0 inclination than one might first guess (eg: take the picture in the article are rotated in the screen, it keeps the same angle of inclination).

Part (B) is kind of fun, now that I'm thinking about it. It's saying that the average angle between two randomly chosen planes is greater than the average angle between a randomly chosen plane and a randomly chosen vector. To see this, we can fix a plane, the X-Y plane and always rotate the system to use that as the (first) randomly chosen plane. Then there's one plane of 0 degree separation from the XY plane, namely itself, but a whole circle's worth of planes perpendicular to it (any plane with it's normal vector lying on the XY plane). In the opposite manner, we see that there's a single line perpendicular to the XY-plane but a circle's worth of lines of inclination 0. This isn't the most intuitive thing in the world, and I second-guessed myself after I wrote (B) down.

(Actually, I guess it's a real projective line's worth, not a circle's worth, of lines of inclination 0 (directions in a plane choose a line but two directions choose the same line so we've doubled counted lines).)