What is the 3-volume of the 3-parallelepiped defined by $\left\{\vec{v_1},\vec{v_2},\vec{v_3}\right\}$? We have $\left\{\vec{v_1},\vec{v_2},\vec{v_3}\right\}=\left\{\begin{bmatrix}1\\0\\0\\0\end{bmatrix},\begin{bmatrix}1\\1\\1\\1\end{bmatrix},\begin{bmatrix}1\\2\\3\\4\end{bmatrix}\right\}$
QR-factorization gives $\text{det }(A^TA)=\text{det }(R^TR)=\text{det }(R^2)$ such that $\sqrt{\text{det }(A^TA)}$ yields the volume of our desired parallelepiped. Letting $A=\begin{pmatrix}1&1&1\\0&1&2\\0&1&3\\0&1&4\end{pmatrix}$,
$\sqrt{\text{det }(A^TA)}=\sqrt{\begin{pmatrix}1&0&0&0\\1&1&1&1\\1&2&3&4\end{pmatrix}\begin{pmatrix}1&1&1\\0&1&2\\0&1&3\\0&1&4\end{pmatrix}}=\sqrt{\text{det }\begin{pmatrix}1&1&1\\1&4&10\\1&10&30\end{pmatrix}}$
Expanding down the first column of $\begin{pmatrix}1&1&1\\1&4&10\\1&10&30\end{pmatrix}$ using Laplace yields 
$\text{det }\begin{pmatrix}1&1&1\\1&4&10\\1&10&30\end{pmatrix}=6$, such that $\sqrt{\text{det }(A^TA)}=\sqrt{6}$, which is the volume of the desired parallelepiped.

Is there a quicker and more energy-efficient algorithm to solve this? I know that for $3\times3$ matrices we can use $\text{det }A=\vec{u}\cdot(\vec{v}\times\vec{w})$, given $\left\{\vec{u},\vec{v},\vec{w}\right\}$ span the columns of $A$. Is there an analog here?
 A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}$Since volume of a parallelipiped spanned by a set of vectors is invariant under the operation of adding a scalar multiple of one vector to another, it suffices to compute the volume of the parallelipiped spanned by
$$
\left[\begin{array}{@{}c@{}}
    1 \\ 0 \\ 0 \\ 0 \\
  \end{array}\right],\qquad
\left[\begin{array}{@{}c@{}}
    0 \\ 1 \\ 1 \\ 1 \\
  \end{array}\right],\qquad
\left[\begin{array}{@{}c@{}}
    0 \\ 0 \\ 1 \\ 2 \\
  \end{array}\right].
\tag{1}
$$
The last two vectors (each lying in the coordinate $3$-space $\{x_{1} = 0\}$ orthogonal to the first vector) span a parallelogram $P$ whose area is the magnitude of the cross product
$$
(1, 1, 1) \times (0, 1, 2) = (1, -2, 1),
$$
namely $\sqrt{6}$. Since the first vector has unit length and is orthogonal to $P$, the three vectors in (1) span a prism of volume $\sqrt{6}$.
Naturally, there's "good fortune" in this example: After "column reduction", the first column was a standard basis vector orthogonal to the remaining columns.

Alternatively, there's a "cross product" for $(n - 1)$ vectors in $\Reals^{n}$ whose definition generalizes the usual cross product in $\Reals^{3}$: Assemble your vectors into an $(n - 1) \times n$ matrix, prepend the standard basis vectors into the first row (obtaining a square matrix) and take the determinant. The magnitude of the resulting vector is the volume of the parallelipiped of the original set of vectors. (See below.) Here you'd have
$$
\left\lvert\begin{array}{@{}cccc@{}}
\Basis_{1} & \Basis_{2} & \Basis_{3} & \Basis_{4} \\
1 & 0 & 0 & 0 \\
1 & 1 & 1 & 1 \\
1 & 2 & 3 & 4 \\
\end{array}\right\rvert
  = \left\lvert\begin{array}{@{}cccc@{}}
  \Basis_{1} & \Basis_{2} & \Basis_{3} & \Basis_{4} \\
  1 & 0 & 0 & 0 \\
  0 & 1 & 1 & 1 \\
  0 & 0 & 1 & 2 \\
  \end{array}\right\rvert
  = \Basis_{2} - 2\Basis_{3} + \Basis_{4},\qquad
\|\Basis_{2} - 2\Basis_{3} + \Basis_{4}\| = \sqrt{6}.
$$
(I leave to you the assessment of whether this is more energy-efficient for your purposes, though again I think there's a favorable comparison in this example.)
This "generalized cross product" has the advertised property for reasons analogous to the situation with the ordinary cross product:


*

*If $(v_{j})_{j=1}^{n-1}$ and $w$ are vectors in $\Reals^{n}$ and $v_{1} \times \dots \times v_{n-1}$ denotes the "cross product" described above, then
$$
\left\lvert\begin{array}{@{}c@{}}
w \\ v_{1} \\ \vdots \\ v_{n-1} \\
  \end{array}\right\rvert
  = (v_{1} \times \dots \times v_{n-1}) \cdot w.
\tag{2}
$$

*By (2), the cross product $v_{1} \times \dots \times v_{n-1}$ is orthogonal to each $v_{j}$, and therefore orthogonal to the parallelipiped spanned by the $(v_{j})$.

*The preceding item and the fact that the determinant in (2) is the $n$-dimensional volume of the parallelipiped spanned by the $(v_{j})$ and $w$ implies that the $(n - 1)$-dimensional volume of the parallelipiped spanned by the $(v_{j})$ is the magnitude of their cross product.
