No, this is just wrong! The volume of the parallelepiped would be all outer products, [itex]e_1 \wedge e_2 \wedge e_3[/itex].
[itex](e_1 \wedge e_2)[/itex] is a
bivector. So you're asking: what is the inner product of a vector with a bivector? That has a clear geometrical interpretation. For a vector [itex]a[/itex] and bivector [itex]B[/itex], [itex]a \cdot B[/itex] does the following:
- Project [itex]a[/itex] onto the plane defined by [itex]B[/itex]
- Rotate 90 degrees in the "sense" of [itex]B[/itex]
- Dilate by the magnitude of [itex]B[/itex]
Note that this uses all three defining characteristics of the bivector [itex]B[/itex]:
- Attitude (basically the angle the plane makes in space)
- Orientation (clockwise vs. counterclockwise)
- Magnitude (i.e. area)
With the inner product used by Hestenes et al, you also have
[tex]a \cdot B = - B \cdot a[/tex]
which let's you answer your question.
By the way, in 3D, your construction is equivalent to the "double cross product" (not the "vector triple product"):
[tex]
(e_1 \wedge e_2) \cdot e_3 = - (e_1 \times e_2) \times e_3[/tex]
Note how the GA version (described above) is much more intuitive and easy to visualize -- the VA version (double cross product) will give you carpal tunnel from all those applications of the right-hand rule!