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

Solution 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 $\mathbb R^3$.

Solution 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 $C$.

Solution 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 $\phi \equiv 0$.

Solution 4:

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

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

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

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

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

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

Solution 5:

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 $0$.