What is the geometric meaning of this vector equality? $\vec{BC}\cdot\vec{AD}+\vec{CA}\cdot\vec{BD}+\vec{AB}\cdot\vec{CD}=0$

Question

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.

Answer 1

Let $O$ be the orthocenter $O$ of $\triangle ABC$. Then \begin{align} &\overrightarrow{AB}\cdot\overrightarrow{CD} \ +\ \overrightarrow{BC}\cdot\overrightarrow{AD} \ +\ \overrightarrow{CA}\cdot\overrightarrow{BD}\\ =\ &\left(\overrightarrow{AB}\cdot\overrightarrow{CO} \ +\ \overrightarrow{BC}\cdot\overrightarrow{AO} \ +\ \overrightarrow{CA}\cdot\overrightarrow{BO}\right) + \left(\overrightarrow{AB}\cdot\overrightarrow{OD} \ +\ \overrightarrow{BC}\cdot\overrightarrow{OD} \ +\ \overrightarrow{CA}\cdot\overrightarrow{OD}\right)\\ =\ &\left(\overrightarrow{AB}\cdot\overrightarrow{CO} \ +\ \overrightarrow{BC}\cdot\overrightarrow{AO} \ +\ \overrightarrow{CA}\cdot\overrightarrow{BO}\right) + \left(\overrightarrow{AB}\ +\ \overrightarrow{BC}\ +\ \overrightarrow{CA}\right)\cdot\overrightarrow{OD}\tag{$\dagger$}\\ =\ &0+0=0.\\ \end{align} The first bracket on line $(\dagger)$ is zero because every side of $\triangle ABC$ is perpendicular to the altitude dropped from the opposite vertex. The second bracket is zero because it is the sum of directed edges of a closed circuit.

In short, the identity is basically a cyclic sum of expressions of the form "side dot altitude" on $\mathbb R^2$, but another cyclic sum of the form "side dot $\overrightarrow{OD}$" has been added to conceal the significance of the orthocenter and make the identity present in $\mathbb R^3$.

Answer 2

Here is another proof, maybe it will be of use: change $D$ by adding any vector $v$ to it. The sum changes by $\left(\overrightarrow{AB}\ +\ \overrightarrow{BC}\ +\ \overrightarrow{CA}\right)\cdot v=0$. So this is an expression independent of $D$. Similarly it is independent of $A$, $B$ and $C$, so is constant. Clearly this constant is $0$.

(In fact one could just move $D$ to $A$ and get zero right away. One of the proposed solutions moves $D$ to orthocenter $O$, but that is not really necessary.)

EDIT: To see independence from $A$ massage the formula by swapping direction of arrows so that $A$ is last:

$$\overrightarrow{BC} \cdot \overrightarrow{AD}\ +\ \overrightarrow{CA} \cdot \overrightarrow{BD}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CD}\ = \overrightarrow{CB} \cdot \overrightarrow{DA}\ +\ \overrightarrow{BD} \cdot \overrightarrow{CA} \ +\ \overrightarrow{DC} \cdot \overrightarrow{BA} $$

Now adding $v$ to $A$ changes the sum by $ (\overrightarrow{CB} + \overrightarrow{BD} + \overrightarrow{DC} )\cdot v=0$.

Same works for $B$ and $C$.

Answer 3

I am not sure if this is the "geometric" interpretation you hope, but here is a way to see why the strong "symetry" of the expression implies that it must be $0$.

Let's denote $\phi : (\mathbb{R}^3)^4 \rightarrow \mathbb{R}$ the application defined for all $A,B,C,D \in \mathbb{R}^3$ by $$\phi(A,B,C,D) = \overrightarrow{BC} \cdot \overrightarrow{AD}\ +\ \overrightarrow{CA} \cdot \overrightarrow{BD}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CD}$$

You can see that $\phi$ is a $4-$linear form on $\mathbb{R}^3$. Moreover, you have easily $$\phi(B,A,C,D) = \overrightarrow{AC} \cdot \overrightarrow{BD}\ +\ \overrightarrow{CB} \cdot \overrightarrow{AD}\ +\ \overrightarrow{BA} \cdot \overrightarrow{CD} = -\phi(A,B,C,D)$$

and this generalizes by saying that for every permutation $\sigma$ of the set $(A,B,C,D)$, one has $$\phi(\sigma(A),\sigma(B),\sigma(C),\sigma(D)) = \varepsilon(\sigma) \phi(A,B,C,D)$$

So $\phi$ is a $4-$linear antisymetric form on $\mathbb{R}^3$. And because $4 > 3$, the only antisymetric form on $\mathbb{R}^3$ is the null form, so $\phi \equiv 0$.

Answer 4

This equation is true for any $4$ points in $\mathbb{R}^n$, for $n\ge1$. Since any $4$ points in $\mathbb{R}^n$, for $n\ge3$, live in a $3$-dimensional hyper-plane, we get full generality from $\mathbb{R}^3$. However, the result is just as easy to prove in $\mathbb{R}^n$, so we will.

This equation is true in each coordinate; the dot product then simply sums the zeros in the coordinates. The equation in each coordinate is simply a statement about vanishing triple products in $\mathbb{R}^3$: $$ \begin{align} &\color{#090}{(C-B)}\cdot\color{#00F}{(D-A)}+\color{#090}{(A-C)}\cdot\color{#00F}{(D-B)}+\color{#090}{(B-A)}\cdot\color{#00F}{(D-C)}\\[3pt] %&=\sum_{k=1}^n\begin{bmatrix}a_k&b_k&c_k\end{bmatrix}\begin{bmatrix}0&1&-1\\-1&0&1\\1&-1&0\end{bmatrix}\begin{bmatrix}d_k-a_k\\d_k-b_k\\d_k-c_k\end{bmatrix}\\ &=\sum_{k=1}^n[\color{#090}{(c_k-b_k)}\color{#00F}{(d_k-a_k)}+\color{#090}{(a_k-c_k)}\color{#00F}{(d_k-b_k)}+\color{#090}{(b_k-a_k)}\color{#00F}{(d_k-c_k)}]\tag1\\ &=\sum_{k=1}^n\color{#090}{\begin{bmatrix}1\\1\\1\end{bmatrix}\times\begin{bmatrix}a_k\\b_k\\c_k\end{bmatrix}}\cdot\color{#00F}{\begin{bmatrix}d_k-a_k\\d_k-b_k\\d_k-c_k\end{bmatrix}}\tag2\\ &=\sum_{k=1}^n\color{#090}{\vec u_k\times\vec v_k}\cdot\color{#00F}{(d_k\vec u_k-\vec v_k)}\tag3\\[9pt] &=0\tag4 \end{align} $$ Each triple product $\vec u_k\times\vec v_k\cdot(d_k\vec u_k-\vec v_k)=0$ because it represents the volume of the parallelepiped generated by $\vec u_k$, $\vec v_k$, and $d_k\vec u_k-\vec v_k$. Since these three vectors lie in the plane generated by $\vec u_k$ and $\vec v_k$, the parallelepiped is degenerate and has a volume of $0$.

Answer 5

Here is a geometrical interpretation

beeing $H$ the projection of $D$ onto the plane containing $A$, $B$ and $C$ such that

• $\overrightarrow{AD}=\overrightarrow{AH}+\overrightarrow{HD} $
• $\overrightarrow{BD}=\overrightarrow{BH}+\overrightarrow{HD} $
• $\overrightarrow{CD}=\overrightarrow{CH}+\overrightarrow{HD} $

and since $\overrightarrow{HD}$ is orthogonal to the plane containing $A$, $B$ and $C$, the given identity is equivalent to

$$\overrightarrow{BC} \cdot \overrightarrow{AH}\ +\ \overrightarrow{CA} \cdot \overrightarrow{BH}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CH}\ = 0$$

which is trivially true indeed by $\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AC}$ we obtain

$$(\overrightarrow{BA}+\overrightarrow{AC}) \cdot \overrightarrow{AH}\ +\ \overrightarrow{CA} \cdot \overrightarrow{BH}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CH}\ = $$

$$=\overrightarrow{BA}\cdot \overrightarrow{AH}+\overrightarrow{AC}\cdot \overrightarrow{AH}+\ \overrightarrow{CA} \cdot \overrightarrow{BH}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CH}\ = $$

$$=\overrightarrow{AB}\cdot \overrightarrow{HA}+\overrightarrow{CA}\cdot \overrightarrow{HA}+\ \overrightarrow{CA} \cdot \overrightarrow{BH}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CH}\ = $$

$$=\overrightarrow{AB}\cdot (\overrightarrow{CH}+\overrightarrow{HA})+\overrightarrow{CA}\cdot (\overrightarrow{BH}+\overrightarrow{HA})=$$

$$=\overrightarrow{AB}\cdot\overrightarrow{CA}+\overrightarrow{CA}\cdot\overrightarrow{BA}=0$$

Answer 6

We take (assume) vector $AD$ given is directed as shown, that is, essentially it is given as the resulting sum of vectors $ AB,BC,CD $, i.e.,

$$ \overrightarrow{AB} +\overrightarrow{BC}+ \overrightarrow{CD} = \overrightarrow{AD} \tag 1 $$

We verify this proposition with direct dot products of position vectors taken with $(x,y) $ components. Position vectors are shown without overhead arrows.

$$ (ab,bc,cd,ad)= [(p,q),(r,s),(u,v),(p+r+u,q+s+v)]\;\tag1$$

$$ ac= (p+r,q+s),bd=(r+u),(s+v) \tag 2 $$

then dot vector product sums of opposite sides

$$ab.cd + bc.ad =\tag 3$$

$$ p r + r^2 + q s + s^2 + p u + r u + q v + s v \tag 4$$

and dot vector product sums of diagonals

$$(ac.bd)=(ab + bc).(bc + cd)=$$

$$ p r + r^2 + q s + s^2 + p u + r u + q v + s v \tag 5 $$

Since (4) and (5) are same, the assumed sign convention in (1) is validated and then we have the vector sign convention:

$$\overrightarrow{BC} \cdot \overrightarrow{AD}\ -\ \overrightarrow{CA} \cdot \overrightarrow{BD}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CD}=0 \tag 6 $$

The central diagonal products has different signs for $ca$ and $ac$ so these relations are identical and the middle term is left as it is for consistency. Moreover the diagonal product term is given negative in Ptolemy Inequality the Wiki reference, retaining sign as it is more appropriate.

In search of a geometrical significance

1. The given relation is a beautiful 3-space vectorisation /generalisation of the Ptolemy theorem which deals with scalars arising from dot products.

2. This post has prompted me to define New Oval shapes here in the plane allied to the Circle... that circumscribe non-cyclic quadrilaterals having the new non-zero constant as property $e$.

This approach resulted in Ptolemy theorem generalization in the plane (where the New Ovals circumscribe plane quadrilaterals whose sides product sum and diagonals product sum bears a constant ratio $e.$

3. It has been verified by the OP and me that the given scalar result from dot products are valid in $\mathbb R^3 $ also. Hece it must be concluded the given vector dot products relation is valid for a skew quadrilateral sitting inside a sphere (vertices lie on the sphere). This is a clear possible geometrical interpretation.

$$\overrightarrow{BC} \cdot \overrightarrow{AD}\ -\ \overrightarrow{CA} \cdot \overrightarrow{BD}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CD}\ =0 \tag 7 $$

The above property has been verified in Mathematica vector computation for a zero sum taking four arbitrary points $(A,B,C,D)$ on a unit sphere computed and sketched thus:

This and what follows is strictly not part of answer, but mentioned for the sake of continuity of the subject.

It is very exciting to imagine that we could even validate in $\mathbb R^3 $ * non-spherical Ovaloid surfaces circumscribing skew quadrilaterals* obeying the modified ( by me) Ptolemy inequality ...using the relation that includes $e$:

$$\overrightarrow{BC} \cdot \overrightarrow{AD}\ -\ \overrightarrow{CA} \cdot \overrightarrow{BD}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CD}\ = e \tag 8 $$

5. Further on it is not difficult to demonstrate that the Ovaloid surface circumscribes a skew quadrilateral and the given scalar product including non-zero right-hand side $e$.

6. Finally imho the relation is having a physical significance in Mechanics rather than the asked geometrical interpretation.

Force equilibrium is easily established by zero vector sum. When right hand side vanishes the moment equilibrium exists making for * full static equilibrium of forces and moments*.

$$ \sum F_i=0; \sum M_i=0 ;\tag9$$

When it does n't, i.e., with RHS =$e$ there is a constant unbalanced moment in dynamic equilibrium.

$$ \sum F_i=0;\sum M_i= e ;\tag{10}$$

To establish it with finality within time available before bounty is tough for me, so this may be considered fwiw in conceptual relation to Newtonian Mechanics.

Thanks for the indulgence.. Regards

Answer 7

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).