What is the geometric meaning of this vector equality? $\vec{BC}\cdot\vec{AD}+\vec{CA}\cdot\vec{BD}+\vec{AB}\cdot\vec{CD}=0$ I was doing some exercises for linear algebra. One of them was to prove that for any four points $A, B, C, D \in \mathbb{R}^3$ the following equality holds:
$$\overrightarrow{BC} \cdot \overrightarrow{AD}\ +\ \overrightarrow{CA} \cdot \overrightarrow{BD}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CD}\ = 0$$
The proof is easy; you just make three vectors starting in $A$ and then see that all the terms cancel out.
My question is: what is the geometric interpretation of this equality? How can I visualize it or understand its deeper meaning? Does this equality have a name or where can I read more about it?
I'm asking this because it turns out that it is not just a random equality and is rather useful. For example, if we want to prove the existence of orthocenter, we can do it surprisingly easily and quickly using this equality.

It seems when looking at
\begin{align*}
\overrightarrow{BC} \cdot \overrightarrow{AD}\ +\ \overrightarrow{CA} \cdot \overrightarrow{BD}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CD}\ = 0\tag{1}
\end{align*}
in full generality, which means in this context considering any four points in $\mathbb{R}^3$ there is rather some (not so deep) combinatorial meaning than any deep geometrical meaning. Nevertheless there are beautiful geometrical visualisations when considering special cases instead.
Essentially one-dimensional: Note the identity (1) which is stated in $\mathbb{R}^3$ is essentially a relationship of four points in $\mathbb{R}$. When taking for instance the projection mapping $\pi_1(X)=X_1$ where $X=(X_1,X_2,X_3)$, the identity (1) boils down to

\begin{align*}
&\color{blue}{(C_1-B_1)(D_1-A_1)}+\color{red}{(A_1-C_1)(D_1-B_1)}+(B_1-A_1)(D_1-C_1)\\
&\qquad=\ \ \color{blue}{C_1D_1-B_1D_1-A_1C_1+A_1B_1}\\
&\qquad\quad\color{red}{-C_1D_1\qquad\qquad\qquad\quad\ -A_1B_1+A_1D_1+B_1C_1}\tag{2}\\
&\qquad\qquad\qquad\ +B_1D_1+A_1C_1\qquad\qquad-A_1D_1-B_1C_1\\
&\qquad=0
\end{align*}
We observe in (2) the terms in each dimension cancel away independently from terms of other dimensions. From this we might conclude that any essence of geometrical information should be already contained within each coordinate dimension separately.

Moreover the order of Points $A_1,B_1,C_1$ and $D_1$ is also irrelevant. We may assume $A_1\leq B_1\leq C_1\leq D_1$ or any other of the $4!=24$ permutations, since any change in the order may change the sign from $X_1Y_1$ but then $-X_1Y_1$ does also change the sign.
Beautiful special cases: As @darigrinberg indicated in the comment section we have for instance Ptolemy's theorem
which claims the identity (1) in the one-dimensional case using lengths $|\overline{BC}|,   |\overline{AD}|$, etc, for the special case that the $4$ points $A,B,C,D\in\mathbb{R}^2$ are cyclic quadrilateral.  A nice proof of this theorem is based upon circle inversion and triangle similarities as shown in the figure below.

The four points $A,B,C,D$ are elements of the blue circle with point $D$ being the center of the red circle. By circle inversion the points $A,B,C$ are mapped to a line giving $A',B',C'$. We so derive $|\overline{DA}||\overline{DA'}|=|\overline{DB}||\overline{DB'}|=|\overline{DC}||\overline{DC'}|$ and find this way similar triangles as the shaded pair and two more pairs from which the theorem follows. This proof is nicely shown here by Adam Hrankowsi which will soon be available in When Am I Ever Gonna Use This Stuff?.
Note the Pythagorean theorem and the Cosine rule follow from special cases of Ptolemy's theorem.
Combinatorial aspects: We look again at the vectorised identity (1) and consider the products
\begin{align*}
&\overrightarrow{\color{blue}{B}C} \cdot \overrightarrow{\color{blue}{A}D}\tag{3}\\
&\overrightarrow{C\color{blue}{A}} \cdot \overrightarrow{\color{blue}{B}D}\tag{4}\\
&\overrightarrow{\color{blue}{A}\color{blue}{B}} \cdot \overrightarrow{CD}\tag{5}\\
\end{align*}
The points $A$ and $B$ for example occur in three variations: $A$ and $B$ both at the left side of the vector in (3) which can be encoded as $(L,L)$. $A$ at the right side and $B$ at the left side in (4), encoded as $(R,L)$ and $A$ and $B$ both at the same side, encoded as $(0,0)$. This relationship holds for each of the six possible selections of two points.
Reducing this relationship to one coordinate by the projection map we observe: Whenever two points $X_1, Y_1$ are encoded as $(L,L)$ or  $(R,R)$ the product $X_1Y_1$ has different sign than the product resulting from a constellation $(L,R)$ or $(R,L)$. Checking each pair of points in (1) for this relationship we obtain
\begin{align*}
\begin{array}{cccccc}
(A_1,B_1)&(A_1,C_1)&(A_1,D_1)&(B_1,C_1)&(B_1,D_1)&(C_1,D_1)\\
\hline
(L,L)&(L,R)&(0,0)&(0,0)&(L,R)&(R,R)\tag{6}\\
(R,L)&(0,0)&(R,R)&(L,L)&(0,0)&(L,R)\\
(0,0)&(L,L)&(L,R)&(R,L)&(R,R)&(0,0)
\end{array}
\end{align*}
showing that each pair of points results in two products which cancel away, while $(0,0)$ means it does not produce a product at all.
Conclusion: In order to provide a geometric visualisation of any four points $A,B,C,D$ in $\mathbb{R}^3$ it should appropriately encode the information stated in table (6).
