Gradient formula of volume of tetrahedron involving the scalar triple product Let $a,b,c,d \in \mathbb{R}^3$ be the vertices of a tetrahedron (I’m unsure whether or not the order of the vertices is important for what follows). The volume of the tetrahedron is
$$
    \begin{align}
    \operatorname{vol}(a,b,c,d)
        &= \frac{1}{6}\big((b-a) \times (c - a)\big) \cdot (d - a) \\
        &= \frac{1}{6}\det
           \left|\begin{matrix}
               (b-a) & (c-a) & (d-a)
           \end{matrix}\right|
    \end{align}
$$
The gradient can be represented nicely by
$$
    \nabla\operatorname{vol}(a,b,c,d) = \frac{1}{6}\left(\begin{matrix}
        (d - b) \times (c - b) \\
        (c - a) \times (d - a) \\
        (d - a) \times (b - a) \\
        (b - a) \times (c - a) \\
    \end{matrix}\right).
$$
I have verified this formula by hand, i.e. I have manually calculated $\nabla\operatorname{vol}$ componentwise and compared the result to the componentwise expansion of the given formula — but I’m not satisfied. I prefer derivation to “left-hand side equals right-hand side” verification.
Can anyone give an insightful explanation of the gradient formula?
 A: This comes down to a straightforward calculation. We have for $a,b,x \in \mathbb{R}^3$
$$\begin{align}
    \nabla_x(a \times b)\cdot x
    &= \nabla_x \big( (a_2b_3 - a_3b_2)x_1 + (a_3b_1 - a_1b_3)x_2 + (a_1b_2 - a_2b_1)x_3 \big) \\
    &= a \times b
\end{align}$$
and
$$\begin{align}
    \nabla_x(a \times x) \cdot b
    &= \nabla_x \big( (a_2x_3 - a_3x_2)b_1 + (a_3x_1 - a_1x_3)b_2 + (a_1x_2 - a_2x_1)b_3 \big) \\
    &= b \times a
\end{align}$$
and
$$\begin{align}
    \nabla_x (x \times a) \cdot b 
        &= \nabla_x \big( (x_2a_3 - x_3a_2)b_1 + (x_3a_1 - x_1a_3)b_2 + (x_1a_2 - x_2a_1)b_3 \big) \\
        &= a \times b.
\end{align}$$
From this, the three gradients $\nabla_b\operatorname*{vol}$, $\nabla_c\operatorname*{vol}$ and $\nabla_d\operatorname*{vol}$ immediately follow. For $\nabla_a\operatorname*{vol}$ we expand $\operatorname*{vol}(a,b,c,d)$ and get
$$\begin{align*}
    6\cdot\operatorname*{vol}(a,b,c,d)
        &= \big( (b-a) \times (c-a) \big) \cdot (d-a) \\
        &= \big( b \times c - b \times a - a \times c + a \times a \big) \cdot (d - a) \\
        &= (b \times c) \cdot d - (b \times a) \cdot d - (a \times c) \cdot d - (b \times c) \cdot a.
\end{align*}$$
Then use the formulae given above and obtain
$$\begin{align*}
    6\cdot\nabla_a\operatorname*{vol}
        &= b \times d - c \times d - b \times c && \\
        &= b \times d - c \times d - b \times c - b \times b && \text{(add $0$)} \\
        &= b \times d - c \times d + c \times b - b \times b && \text{(antisymmetry)} \\
        &= (b - c) \times (d - b) && \text{(factor out)} \\
        &= - (c - b) \times (d - b) && \\
        &= (d - b) \times (c - b) && \text{(antisymmetry)}.
\end{align*}$$
