《2025年一本知识大盘点高中数学全一册人教版》

例 4 (1) 如图，已知四棱锥 $ P - ABCD $ 的各棱长均为 2，则 $ \overrightarrow{AP} · \overrightarrow{BC} = $(2)
A.$ 2\sqrt{3} $
B.$ \sqrt{3} $
C.1
D.2

答案：
(1)
$ \because $四棱锥 $ P - ABCD $ 的各棱长均为 $ 2 $，
$ \therefore \triangle ADP $ 为等边三角形，四边形 $ ABCD $ 为菱形，
$ \therefore AP = AD $，$ \angle PAD = 60^{\circ} $，$ AD // BC $，$ AD = BC $，
$ \therefore \overrightarrow{AD} = \overrightarrow{BC} $，
$ \therefore \overrightarrow{AP} · \overrightarrow{BC} $
$ = \overrightarrow{AP} · \overrightarrow{AD} $
$ = |\overrightarrow{AP}| |\overrightarrow{AD}| \cos \angle PAD $
$ = 2 × 2 × \cos 60^{\circ} $
$ = 2 $
答案：D

(2) 如图，在各棱长都为 2 的四面体 $ ABCD $ 中，$ \overrightarrow{CE} = \overrightarrow{ED} $，$ \overrightarrow{AF} = 2\overrightarrow{FD} $，则 $ \overrightarrow{BE} · \overrightarrow{CF} = $(A)


A.$ -\frac{1}{3} $
B.$ \frac{1}{3} $
C.$ -\frac{1}{2} $
D.$ \frac{1}{2} $

答案：
(2)
$ \because $各棱长都为 $ 2 $，
$ \therefore \overrightarrow{BA} $，$ \overrightarrow{BC} $ 的夹角，$ \overrightarrow{BD} $，$ \overrightarrow{BC} $ 的夹角，$ \overrightarrow{BD} $，$ \overrightarrow{BA} $ 的夹角均为 $ \frac{\pi}{3} $，
$ \because \overrightarrow{CE} = \overrightarrow{ED} $，$ \overrightarrow{AF} = 2\overrightarrow{FD} $，
$ \therefore \overrightarrow{BE} = \frac{1}{2}(\overrightarrow{BC} + \overrightarrow{BD}) $，$ \overrightarrow{AF} = \frac{2}{3}\overrightarrow{AD} $，
$ \therefore \overrightarrow{CF} $
$ = \overrightarrow{BF} - \overrightarrow{BC} $
$ = \overrightarrow{BA} + \overrightarrow{AF} - \overrightarrow{BC} $
$ = \overrightarrow{BA} + \frac{2}{3}\overrightarrow{AD} - \overrightarrow{BC} $
$ = \overrightarrow{BA} + \frac{2}{3}(\overrightarrow{BD} - \overrightarrow{BA}) - \overrightarrow{BC} $
$ = \frac{1}{3}\overrightarrow{BA} - \overrightarrow{BC} + \frac{2}{3}\overrightarrow{BD} $
$ \therefore \overrightarrow{BE} · \overrightarrow{CF} $
$ = \frac{1}{2}(\overrightarrow{BC} + \overrightarrow{BD}) · (\frac{1}{3}\overrightarrow{BA} - \overrightarrow{BC} + \frac{2}{3}\overrightarrow{BD}) $
$ = \frac{1}{6}\overrightarrow{BA} · \overrightarrow{BC} - \frac{1}{2}\overrightarrow{BC}^2 - \frac{1}{6}\overrightarrow{BC} · \overrightarrow{BD} + \frac{1}{6}\overrightarrow{BA} · \overrightarrow{BD} + \frac{1}{3}\overrightarrow{BD}^2 $
$ = \frac{1}{6}|\overrightarrow{BA}| |\overrightarrow{BC}| \cos \frac{\pi}{3} - \frac{1}{2}|\overrightarrow{BC}|^2 - \frac{1}{6}|\overrightarrow{BC}| |\overrightarrow{BD}| \cos \frac{\pi}{3} + \frac{1}{6}|\overrightarrow{BA}| |\overrightarrow{BD}| \cos \frac{\pi}{3} + \frac{1}{3}|\overrightarrow{BD}|^2 $
$ = \frac{1}{6} × 2 × 2 × \frac{1}{2} - \frac{1}{2} × 2^2 - \frac{1}{6} × 2 × 2 × \frac{1}{2} + \frac{1}{6} × 2 × 2 × \frac{1}{2} + \frac{1}{3} × 2^2 $
$ = -\frac{1}{3} $
答案：A