The 3DPGA cheat sheet states the norm of a euclidean line to be \lVert \ell \rVert = \sqrt{\ell^2} = \sqrt{d^2+e^2+f^2}.
When I use \ell = a{\bf e_{01}} + b{\bf e_{02}} + c{\bf e_{03}} + d{\bf e_{12}} + e{\bf e_{31}} + f{\bf e_{23}} and compute \ell^2 I get \ell^2 = -d^2-e^2-f^2 + 2(af + be + cd){\bf e_{0123}} which as the Section 8.1.3 of the SIGGRAPH lecture note mentions is a dual number. Even though I haven’t read it anywhere (?) I assume that all lines are simple bivectors (since they can be seen as the intersection of two planes and the product of two planes always results in a simple bivector), and similarly the Plücker quadric relation equals the pseudoscalar part being zero. Nevertheless, the norm then should be \sqrt{-\ell^2} or \sqrt{|\ell^2|}, right?
Bonus question: If not all bivectors are lines, what do non-simple bivectors represent then in 3D PGA?
Yes, the norm of a line should be \sqrt{l \tilde l} or \sqrt{\lvert l^2 \rvert}. (and yes, all lines are simple bivectors). The simple bivectors encode all translations and rotations.
On the bonus question, all euclidean non-simple bivectors are screw axi. (see secion 8.1.3 and 8.1.4 of the course notes).
I will add (for anyone’s convenience) that when you need to divide by the dual number a+b \epsilon to get the normalised line, you need to use the formula (a+b \epsilon)^{-1}=\frac{1}{a}-\frac{b}{a^2}\epsilon. That’s a simple little algebra exercise.
Thanks for pointing that out, @bellinterlab. I’ve added that to the discussion of dual numbers in the SIGGRAPH course notes (will appear in the next update). That can also be written as \dfrac{\bar{z}}{z \bar{z}} where z = a+b\epsilon and \bar{z} = a-b\epsilon is the conjugate of z. Note that this implies that pure dual numbers have no inverses; that’s equivalent to the fact that the metric is degenerate.
More formulas on the dual numbers and their noneucludean analogs can be found in Sec. 7.6 of my thesis.