A flat connected to the origin point $o$ is represented by $X=o \wedge \textbf{A}_k \wedge \infty$, where $\textbf{A}_k$ is a k-blade of directions in the Euclidean subspace, and $\infty$ is the point at infinity. I want to calculate $$||X||^2 = X \widetilde{X} = (o\wedge \textbf{A}_k \wedge \infty)(\infty \wedge \widetilde{\textbf{A}_k} \wedge o)$$
I am not sure how to simplify the expression. The answer is supposed to be $X\widetilde X = -||\textbf{A}_k||^2$.
It is scalar, so $X\widetilde X=\langle (p\wedge \textbf{A}_k \wedge \infty)(\infty \wedge \widetilde{\textbf{A}_k} \wedge p) \rangle$.
In this CGA, we have $\infty \cdot \infty=0$, $o\cdot o=0$, $o\cdot \infty = -1$.
If there was a left contraction between $X$ and $\widetilde X$, then the solution would follow, but how would that be justified?
Despite my comments, I can answer this directly.
Note that $A$ in the Euclidean subspace is orthogonal to both $o$ and $\infty$. (Some would be more comfortable writing these vectors as $e_o$ and $e_\infty$.) Therefore $A\wedge o\wedge\infty=A\,(o\wedge\infty)$.
$$X\,X^\sim=(A\wedge o\wedge\infty)(\infty\wedge o\wedge A^\sim)$$
$$=A\,(o\wedge\infty)(\infty\wedge o)\,A^\sim$$
$$=A\,(-1)\,A^\sim$$
$$=-\lVert A\rVert^2$$