Independence of parallelepiped volume on certain vector elements

Question

I am trying to calculate the volume of a parallelepiped defined by the vectors $\mathbf{\vec{a} = [a_x, 0, 0], \vec{b} = [b_x, b_y, 0], \vec{c} = [c_x, c_y, c_z]}$. However, when I apply the triple product to calculate the volume, $\mathbf{V = (\vec{a}~x~\vec{b}).\vec{c}}$, I get the result $\mathbf{V = a_xb_yc_z}$, which only includes the diagonal elements of the matrix $\mathbf{[a~b~c]}$. I feel like this is wrong and the off diagonal elements should play a role somehow. But I've checked my triple product dozens of times and can't find anything wrong with it.

Is it just that a parallelepiped expressed in this kind of a basis is unique, making the volume as easy to calculate as that of a rectangular prism?

Answer 1

The area of the parallelogram defined by $\vec{\mathbf a}$ and $\vec{\mathbf b}$ is $\|\vec{\mathbf a}\|$ times the height of the parallelogram, which is the component of $\vec{\mathbf b}$ that’s orthogonal to $\vec{\mathbf a}$. The latter lies along the $x$-axis, and the former lies in the $x$-$y$ plane, so the height is just the $y$-coordinate of $\vec{\mathbf b}$. The area of the parallelepiped is the area of this base times the height, which is the component of $\vec{\mathbf c}$ that’s orthogonal to both $\vec{\mathbf a}$ and $\vec{\mathbf v}$, i.e., that’s orthogonal to the $x$-$y$ plane, but that’s just the $z$-coordinate of $\vec{\mathbf c}$. $\|\vec{\mathbf a}\|$ is of course just the absolute value of its $x$-coordinate, so for the volume we get $a_xb_yc_z$, which are exactly the diagonal elements of $\begin{bmatrix}\vec{\mathbf a}&\vec{\mathbf b}&\vec{\mathbf c}\end{bmatrix}$.

Answer 2

The volume is indeed as easy to calculate as for a rectangular prism with $a$ and $b$ in the given forms (i.e. two and one zeros respectively).

Note that the determinant of the matrix formed by the three vectors gives the (signed) volume; here that matrix is triangular, and triangular matrices have their determinant given by the product of the diagonal entries only, which explains the surprising result.

Answer 3

The result is correct. Think about it this way: imagine that $\mathbf{c}$ is in the plane of $\mathbf{a}$ and $\mathbf{b}$ (meaning $c_z=0$). Then the area is zero because all three vectors are coplanar. So we need finite $c_z$ to make the result finite.

Also, think about the area of the parallelogram in the $(x,y)$ plane. Its just $a_x b_y$.

Putting these two results together make it easy to understand why the volume is $a_x b_y c_z$.