In 3D, we can find the area of the parallelogram spanned by two vectors by using the cross product:
$$Area = {\vert\vec a \times \vec b\vert}$$
In 2D, we can perform a similar operation using a 2D loose analog of the cross product that fills this purpose:
$$Area = {\vert\vec a_x \times \vec b_y - \vec a_y \times \vec b_x \vert}$$
How can the same be done in 4D? I know that we can just multiply the lengths and the sine of the angle between the vectors, but is there a way to do this without using trigonometry? Are there loose analogs of the cross product in 4D that fill this purpose? Also, is there a general solution that applies to 5D, 6D, etc?
I have found some other questions on Math SE asking about the areas of parallelograms between two 3D vectors, but I haven't been able to find any information about higher dimensions.
Write the two vectors $\vec a$ and $\vec b$ in $\mathbb{R}^4$ as column vectors and combine them in a $4 \times 2$ matrix $A = (\vec a, \vec b)$. Then the area of the parallelogram spanned by $\vec a$ and $\vec b$ is $$\sqrt{\det(A^T A)}$$
More generally, if we have $k$ vectors $v_1, \dots , v_k$ in $\mathbb{R}^n$, then the volume of the parallelepiped spanned by the vectors is $\sqrt{\det(A^T A)}$, where $A$ is the $n \times k$ matrix $(v_1, \dots ,v_k)$.
Source: All the Mathematics You Missed [But Need to Know for Graduate School] by Thomas A. Garrity.