Cho hình tứ diện \(ABCD\). Chứng minh rằng: \(\overrightarrow{AB}.\overrightarrow{CD}+\overrightarrow{AC}.\overrightarrow{DB}+\overrightarrow{AD}.\overrightarrow{BC}=0.\)

\(\overrightarrow{AB}.\overrightarrow{CD}=\overrightarrow{AB}.(\overrightarrow{AD}-\overrightarrow{AC})\)

\(\overrightarrow{AC}.\overrightarrow{DB}=\overrightarrow{AC}.(\overrightarrow{AB}-\overrightarrow{AD})\)

\(\overrightarrow{AD}.\overrightarrow{BC}=\overrightarrow{AD}.(\overrightarrow{AC}-\overrightarrow{AB}).\)

Cộng từng vế ba đẳng thức trên ta được:

\(\overrightarrow {AB} .\overrightarrow {CD}  + \overrightarrow {AC} .\overrightarrow {DB}  + \overrightarrow {AD} .\overrightarrow {BC} \)

\( = \overrightarrow {AB} \left( {\overrightarrow {AD}  - \overrightarrow {AC} } \right)\) \( + \overrightarrow {AC} .\left( {\overrightarrow {AB}  - \overrightarrow {AD} } \right)\) \( + \overrightarrow {AD} \left( {\overrightarrow {AC}  - \overrightarrow {AB} } \right)\)

\( = \overrightarrow {AB} .\overrightarrow {AD}  - \overrightarrow {AB} .\overrightarrow {AC} \) \( + \overrightarrow {AC} .\overrightarrow {AB}  - \overrightarrow {AC} .\overrightarrow {AD} \) \( + \overrightarrow {AD} .\overrightarrow {AC}  - \overrightarrow {AD} .\overrightarrow {AB} \)

\( = \overrightarrow {AB} .\overrightarrow {AD}  - \overrightarrow {AD} .\overrightarrow {AB} \) \( + \overrightarrow {AC} .\overrightarrow {AB}  - \overrightarrow {AB} .\overrightarrow {AC} \) \( + \overrightarrow {AD} .\overrightarrow {AC}  - \overrightarrow {AC} .\overrightarrow {AD} \)

\( =  0  +  0  +  0  = 0 \)