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Bài 2. 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) \(\overrightarrow{AC}\) + \(\overrightarrow{BA’}\) + \(\overrightarrow{DB}\) + \(\overrightarrow{C’D}\) = \(\overrightarrow{0}\).
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a) \(\overrightarrow{AB}\) + \(\overrightarrow{B’C’}\) + \(\overrightarrow{DD’}\) = \(\overrightarrow{AB}\) + \(\overrightarrow{BC}\) + \(\overrightarrow{CC’}\) = \(\overrightarrow{AC’}\);
b) \(\overrightarrow{BD}\) – \(\overrightarrow{D’D}\) – \(\overrightarrow{B’D’}\) = \(\overrightarrow{BD}\) + \(\overrightarrow{DD’}\) + \(\overrightarrow{D’B’}\) = \(\overrightarrow{BB’}\);
c) \(\overrightarrow{AC}\) + \(\overrightarrow{BA’}\) + \(\overrightarrow{DB}\) + \(\overrightarrow{C’D}\) = \(\overrightarrow{AC}\) + \(\overrightarrow{CD’}\) + \(\overrightarrow{D’B’}\) + \(\overrightarrow{B’A}\) = \(\overrightarrow{0}\).