Giải Câu 2 Bài Vecto trong không gian

Câu 2: Trang 91 - SGK Hình học 11

Cho hình hộp . Chứng minh rằng:

a) + \(\overrightarrow{B'C'}\) + \(\overrightarrow{DD'}\) = \(\overrightarrow{AC'}\);

b) - \(\overrightarrow{D'D}\) - \(\overrightarrow{B'D'}\) = \(\overrightarrow{BB'}\);

c) + \(\overrightarrow{BA'}\) + \(\overrightarrow{DB}\) + \(\overrightarrow{C'D}\) = \(\overrightarrow{0}\).

Bài làm:

a) Ta có: , \(\overrightarrow{DD'}\ =\overrightarrow{CC'}\)

=> + \(\overrightarrow{B'C'}\) + \(\overrightarrow{DD'}\) = + \(\overrightarrow{BC}\) + \(\overrightarrow{CC'}\) = \(\overrightarrow{AC'}\)

b) Ta có: , \(\overrightarrow{B'D'}\ =-\overrightarrow{D'B'}\)

=> - \(\overrightarrow{D'D}\) - \(\overrightarrow{B'D'}\) = + \(\overrightarrow{DD'}\) + \(\overrightarrow{D'B'}\) = \(\overrightarrow{BB'}\)

c) Ta có: , \(\overrightarrow{DB}\ =\overrightarrow{D'B'}\), \(\overrightarrow{C'D}\ =\overrightarrow{B'A}\)

=> + \(\overrightarrow{BA'}\) + \(\overrightarrow{DB}\) + \(\overrightarrow{C'D}\) = + \(\overrightarrow{CD'}\) + \(\overrightarrow{D'B'}\) + \(\overrightarrow{B'A}\) = \(\overrightarrow{0}\).