Câu 2: Trang 91 - SGK Hình học 11
Cho hình hộp \(ABCD.A'B'C'D'\). Chứng minh rằng:
a) \(\overrightarrow{AB}\) + \(\overrightarrow{B'C'}\) + \(\overrightarrow{DD'}\) = \(\overrightarrow{AC'}\);
b) \(\overrightarrow{BD}\) - \(\overrightarrow{D'D}\) - \(\overrightarrow{B'D'}\) = \(\overrightarrow{BB'}\);
c) \(\overrightarrow{AC}\) + \(\overrightarrow{BA'}\) + \(\overrightarrow{DB}\) + \(\overrightarrow{C'D}\) = \(\overrightarrow{0}\).
Bài Làm:
a) Ta có: \(\overrightarrow{B'C'}\ =\overrightarrow{BC}\), \(\overrightarrow{DD'}\ =\overrightarrow{CC'}\)
=> \(\overrightarrow{AB}\) + \(\overrightarrow{B'C'}\) + \(\overrightarrow{DD'}\) = \(\overrightarrow{AB}\) + \(\overrightarrow{BC}\) + \(\overrightarrow{CC'}\) = \(\overrightarrow{AC'}\)
b) Ta có: \(\overrightarrow{DD'}\ =-\overrightarrow{D'D}\), \(\overrightarrow{B'D'}\ =-\overrightarrow{D'B'}\)
=> \(\overrightarrow{BD}\) - \(\overrightarrow{D'D}\) - \(\overrightarrow{B'D'}\) = \(\overrightarrow{BD}\) + \(\overrightarrow{DD'}\) + \(\overrightarrow{D'B'}\) = \(\overrightarrow{BB'}\)
c) Ta có: \(\overrightarrow{BA'}\ =\overrightarrow{CD'}\), \(\overrightarrow{DB}\ =\overrightarrow{D'B'}\), \(\overrightarrow{C'D}\ =\overrightarrow{B'A}\)
=> \(\overrightarrow{AC}\) + \(\overrightarrow{BA'}\) + \(\overrightarrow{DB}\) + \(\overrightarrow{C'D}\) = \(\overrightarrow{AC}\) + \(\overrightarrow{CD'}\) + \(\overrightarrow{D'B'}\) + \(\overrightarrow{B'A}\) = \(\overrightarrow{0}\).