Can anyone think of a way of working out the volume of a triangle based pyramid embedded in 4D? The points are defined (from a 'central' point) by three vectors:
$a_1, a_2 \text{ and } a_3$
all of which are 4 dimensional. If it were simply in 3D space, I would simply do:
$\frac{(a_1 \times a_2) \cdot a_3}{6}$
but cross products aren't defined in 4D (as there is no one vector that is perpendicular to two vectors in 4 dimensions).
One way that this could be done is to call $V_w$ the volume of the pyramid if we ignore the $w$ components (e.g. project onto the $xyz$ plane), $V_x$ the volume if we project onto the $wyz$ plane, etc.
The volume would then be given by volume Pythagoras:
$V^2 = V_w^2 + V_x^2 + V_y^2 + V_z^2$
but this is a bit arduous, as it requires working out four volumes! there must be a better way!
Thanks in advance!