Questions about the conformal model of GA The vectors $a$,$b$, and $e$ are null vectors in the GA generated by $R^{n+1,1}$ such that $a.e=b.e=1$ where $e$ is the vector representing a point at infinity. We are told that $a$ and $b$ (and indeed all null vectors) can represent points of an n-dimensional Euclidean space, with $a.b$ being proportional to the square of the Euclidean distance between $a$ and $b$: $(a-b)^2=-2a.b$.
The points $x$ which lie on the line $A$ that passes through $a$ and $b$ and extends to infinity must satisfy $x\wedge a \wedge b \wedge e=0$.
Hestenes (p.3) then makes a couple of claims that I don't quite follow.


*

*The tangent vector $n$ for the line $A$ is $n\equiv (a\wedge b).e = a-b$.
Why is this true?

*The length of line $A$ is given by $A^2=(a\wedge b \wedge e)^2=[(a\wedge b).e]^2=n^2$. Why is the equality in the middle true?
 A: You can use the BAC-CAB rule to evaluate the first formula.  For any three vectors $u, v, w$:
$$(u \wedge v) \cdot w = u (v \cdot w) - v (u \cdot w)$$
Since $a \cdot e = b \cdot e = 1$, this quickly evaluates to $a - b$.
For the second part, use grade projection.  In particular, remember that $a, b,e$ are all null, so you get
$$\langle abeeba \rangle_0 = 0$$
On the one hand, but using associativity, you can write the product in terms of vector and trivector parts:
$$\langle abeeba \rangle_0 = -(\langle abe \rangle_3 )^2 + (\langle abe \rangle_1)^2$$
The minus sign follows from the reverse of a 3-vector being its negative.
It's easy to get that $\langle abe\rangle_3 = a \wedge b \wedge e$, but the vector part requires a little more thought:
$$\langle abe \rangle_1 = (a \wedge b) \cdot e + (a \cdot b) e = a - b + (a \cdot b) e = n + (a \cdot b) e$$
Now, square it to get
$$n^2 + 2 (n \cdot e)(a \cdot b) + (a \cdot b)e^2$$
But $n = a - b$ and $(a -b) \cdot e = 0$.  Similarly, $e^2 = 0$, so the result follows.
