# The formula for the volumn of three vectors in $\mathbb{R}^4$

For two vectors $v_1,v_2 \in \mathbb{R}^3$,there is a formula for the area of parallelogram generated by $v_1,v_2$. That is given by the cross product $|v_1 \times v_2|$ or $|det(v_1,v_2,\frac{v_1\times v_2}{|v_1\times v_2|})|$.

I am wondering if such formula exists in $\mathbb{R}^4$.

More precisely, let $w_1,w_2,w_3\in \mathbb{R}^4$ be independent vectors in $\mathbb{R}^4$. Then what is the volumn of parallelepiped generated by $w_1,w_2,w_3$?

I suppose that if there is a vector $u_1\in \mathbb{R}^4$ such that $u_1 \perp w_i$ for $1\le i \le 3$ and $|u_1|=1$, then the volumn should be $|det(w_1,w_2,w_3,u_1)|$. But I couldn't find a formula for such $u_1$ using $w_1,w_2,w_3$. (Namely, I am trying to find some concept similar to the cross product in $\mathbb{R}^3$.)

Could you shed me a light on this?

Any comment would be highly appreciated.

• Also see for example this thread, or this. Dec 9 '16 at 5:37

The following works for $k$ vectors $u_1,u_2,\ldots,u_k\in\Bbb{R}^n$ ($k<n$). Let $M$ be the $k\times n$ matrix having those $k$ vectors as rows. Let $V$ be the (hyper)volume of the $k$-dimensional parallelotope with edges $u_1,u_2,\ldots$. Then $$V^2=\det MM^T.\qquad(*)$$
A way to see this is as follows. Any transformation of the type $u_i\mapsto u_i+cu_j$ leaves $V$ unchanged (think: a shearing map transforming a parallelopiped to another with same base and height). It also amount to premultiplying $M$ (and postmultiplying $M^T$) by an elementary matrix with determinant $1$ (leaving $\det MM^T$ unchanged). But, a sequence of such transformations performs Gram-Schmidt orthogonalization on the set $u_1,\ldots,u_k$. So if the formula $(*)$ holds for an orthogonal set of vectors, it holds for all vectors.
But, the entries of the matrix $MM^T$ are the inner products $(u_i,u_j)$. So when $u_1,u_2,\ldots,u_k$ are orthogonal, that matrix is diagonal with elements $||u_i||^2$. The claim follows.
• Thank you very much! May I ask one more? What is the precise definition of volumn of k-dimensional parallelotope gen'd by $\mu_1,\cdots,\mu_k$? I am wondering whether the following is right. Let $V$ be the subspace of $\mathbb{R}^n$ generated by $\mu_1,\cdots,\mu_k$. And choose an orthonormal basis ${w_{k+1},\cdots ,w_{n}}$ of $V^{\perp}$. Then I suppose the volumn of k-dimensional parallelotope gen'd by $\mu_1,\cdots,\mu_k$ is $det(\mu_1,\cdots,\mu_k,w_{k+1},\cdots ,w_{n})$. Right? Dec 9 '16 at 6:25