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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!

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thanks, @Henry T. Horton for tex-ifying my question! –  tiswas Oct 30 '12 at 16:38
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Instead of projecting into 3-space, you could hit the vectors with a rigid transformation. Alternatively, simply compute the lengths of the edges ---the distance formula works in all dimensions-- and throw them into the Cayley-Menger determinant. (See the appropriate section of the Wikipedia "Tetrahedron" article: en.wikipedia.org/wiki/Tetrahedron ) –  Blue Oct 30 '12 at 16:44
    
Thanks, but I think Christian's approach looks a little faster. –  tiswas Oct 30 '12 at 23:03
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1 Answer 1

up vote 2 down vote accepted

If $n$-dimensional space ${\mathbb R}^n$ (in your case $n=4$) is equipped with, say, the standard euclidean scalar product (or metric) then this metric induces automatically a natural measure (length, surface area, $\ldots$) for lower-dimensional objects embedded in ${\mathbb R}^n$. In your case the situation can be handled by linear algebra:

Given $d\geq1$ vectors ${\bf a}_i\in {\mathbb R}^n$, they span a $d$-dimensional parallelotope $P$. To obtain its $d$-dimensional "volume" set up the $(d\times d)$-matrix $$g=\bigl[g_{ik}\bigr]\ ,\quad g_{ik}:={\bf a}_i\cdot{\bf a}_k\qquad(1\leq i\leq d,\ 1\leq k\leq d)$$ of the scalar products between the ${\bf a}_i$, and compute its determinant $$G({\bf a}_1,\ldots,{\bf a}_d):=\det(g)\ ,$$ called the Gram determinant of the ${\bf a}_i$. Then the $d$-dimensional volume of the parallelotope $P$ is given by $${\rm vol}_d(P)=\sqrt{G({\bf a}_1,\ldots,{\bf a}_d)}\ .$$ An example: The length of the segment $[{\bf 0},{\bf a}]\subset{\mathbb R}^n$, i.e., the $1$-dimensional volume of the parallelotope $P:=\{t{\bf a}\in{\mathbb R}^n\ | 0\leq t\leq 1\}$, is given by ${\rm vol}_1(P)=\sqrt{{\bf a}\cdot{\bf a}}$.

As you want only the $3$-dimensional volume of the simplex spanned by the three given vectors you have to divide the volume of the parallelepiped by $3!=6$.

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thanks - looks like a more elegant and quicker way than projecting! –  tiswas Oct 30 '12 at 17:57
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