Bài tập 4.66. Cho bốn điểm A, B, C, D trong mặt phẳng. Chứng minh rằng
$\overrightarrow{AB} . \overrightarrow{CD} + \overrightarrow{BC} . \overrightarrow{AD} + \overrightarrow{CA} . \overrightarrow{BD} = 0$
Bài Làm:
Có $\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD}$
$\overrightarrow{CA} = \overrightarrow{BA} - \overrightarrow{BC}$
$\overrightarrow{BD} = \overrightarrow{BC} + \overrightarrow{CD}$
Có:
$\overrightarrow{AB} . \overrightarrow{CD} + \overrightarrow{BC} . \overrightarrow{AD} + \overrightarrow{CA} . \overrightarrow{BD}$
= $\overrightarrow{AB} . \overrightarrow{CD} + \overrightarrow{BC} (\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD}) + (\overrightarrow{BA} - \overrightarrow{BC}) (\overrightarrow{BC} + \overrightarrow{CD})$
$\overrightarrow{AB} . \overrightarrow{CD} + \overrightarrow{BC} . \overrightarrow{AD} + \overrightarrow{CA} . \overrightarrow{BD}$
= $\overrightarrow{AB} . \overrightarrow{CD} + \overrightarrow{BC} . \overrightarrow{AB} + \overrightarrow{BC}^{2} + \overrightarrow{BC} . \overrightarrow{CD} + \overrightarrow{BA} . \overrightarrow{BC} + \overrightarrow{BA} . \overrightarrow{CD} - \overrightarrow{BC}^{2} - \overrightarrow{BC} . \overrightarrow{CD}$
$\overrightarrow{AB} . \overrightarrow{CD} + \overrightarrow{BC} . \overrightarrow{AD} + \overrightarrow{CA} . \overrightarrow{BD}$
= $(\overrightarrow{AB} . \overrightarrow{CD} + \overrightarrow{BA} . \overrightarrow{CD}) + (\overrightarrow{BC} . \overrightarrow{AB}$
+ $\overrightarrow{BA} . \overrightarrow{BC}) +(\overrightarrow{BC}^{2} - \overrightarrow{BC}^{2}) + (\overrightarrow{BC} . \overrightarrow{CD} - \overrightarrow{BC} . \overrightarrow{CD}$)
= 0
$\Rightarrow \overrightarrow{AB} . \overrightarrow{CD} + \overrightarrow{BC} . \overrightarrow{AD} + \overrightarrow{CA} . \overrightarrow{BD}$ = 0