Prove that for any points A,B,C,D in space…
Prove that for any points $A,B,C,D$ in space $\overrightarrow{BC}\cdot\overrightarrow{AD}+\overrightarrow{CA}\cdot\overrightarrow{BD}+\overrightarrow{AB}\cdot\overrightarrow{CD}=0$.
Explanation of Method
A scalar-product identity for any four points A, B, C, D.
Step 1: Position vectors
With $\vec a,\vec b,\vec c,\vec d$ the position vectors of the points, $\overrightarrow{XY}=\vec y-\vec x$.
Step 2: Expansion
\[ (\vec c-\vec b)\cdot(\vec d-\vec a)=\vec c\cdot\vec d-\vec a\cdot\vec c-\vec b\cdot\vec d+\vec a\cdot\vec b, \]
\[ (\vec a-\vec c)\cdot(\vec d-\vec b)=\vec a\cdot\vec d-\vec a\cdot\vec b-\vec c\cdot\vec d+\vec b\cdot\vec c, \]
\[ (\vec b-\vec a)\cdot(\vec d-\vec c)=\vec b\cdot\vec d-\vec b\cdot\vec c-\vec a\cdot\vec d+\vec a\cdot\vec c. \]
Step 3: Cancellation
Each of the six products appears once with $+$ and once with $-$:
| Product | Occurrences | Sum |
|---|---|---|
| $\vec a\cdot\vec b$ | $+$ (1st), $-$ (2nd) | $0$ |
| $\vec a\cdot\vec c$ | $-$ (1st), $+$ (3rd) | $0$ |
| $\vec a\cdot\vec d$ | $+$ (2nd), $-$ (3rd) | $0$ |
| $\vec b\cdot\vec c$ | $+$ (2nd), $-$ (3rd) | $0$ |
| $\vec b\cdot\vec d$ | $-$ (1st), $+$ (3rd) | $0$ |
| $\vec c\cdot\vec d$ | $+$ (1st), $-$ (2nd) | $0$ |
\[\overrightarrow{BC}\cdot\overrightarrow{AD}+\overrightarrow{CA}\cdot\overrightarrow{BD}+\overrightarrow{AB}\cdot\overrightarrow{CD}=0\qquad\blacksquare\]
- $BC,\ AD$
- $CA,\ BD$
- $AB,\ CD$
Step 4: Consequences
- If two pairs of opposite edges of a tetrahedron are perpendicular, the third pair is perpendicular too (two terms vanish, so the third does).
- With $D=H$ in the plane of $ABC$, where $AH\perp BC$ and $BH\perp CA$, the identity gives $CH\perp AB$. This is again the concurrency of the altitudes (Problem 62).
Result
\[\boxed{\,\overrightarrow{BC}\cdot\overrightarrow{AD}+\overrightarrow{CA}\cdot\overrightarrow{BD}+\overrightarrow{AB}\cdot\overrightarrow{CD}=0\qquad\blacksquare\,}\]