Cho hình hộp ABCD.A'B'C'D'. Chứng minh rằng:

 $$\begin{gathered} a) \overrightarrow{AB}+\overrightarrow{B\text{'}C\text{'}}+\overrightarrow{DD\text{'}}=\overrightarrow{AC\text{'}}; \\ b) \overrightarrow{BD}-\overrightarrow{D\text{'}D}-\overrightarrow{B\text{'}D\text{'}}=\overrightarrow{BB\text{'}}; \\ c) \overrightarrow{AC}+\overrightarrow{BA\text{'}}+\overrightarrow{DB}+\overrightarrow{C\text{'}D}=\overrightarrow{0} \end{gathered}$$

Giải:

a) Áp dụng quy tắc hình hộp ta có: $\overrightarrow{AB}+\overrightarrow{B\text{'}C\text{'}}+\overrightarrow{DD\text{'}}=\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AA\text{'}}=\overrightarrow{AC\text{'}}$

Cách khác $\overrightarrow{AB}+\overrightarrow{B\text{'}C\text{'}}+\overrightarrow{DD\text{'}}=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CC\text{'}}=\overrightarrow{AC\text{'}}$

$$\begin{gathered} \mathrm{b}) \overrightarrow{BD}-\overrightarrow{DD\text{'}}-\overrightarrow{BD\text{'}}=\overrightarrow{BD}+\overrightarrow{DD\text{'}}+\overrightarrow{D\text{'}B\text{'}}=\overrightarrow{BB\text{'}} \\ \mathrm{c}) \overrightarrow{AC }+\overrightarrow{BA\text{'}}+\overrightarrow{DB}+\overrightarrow{C\text{'}D}=\overrightarrow{AC}+\overrightarrow{CD\text{'}}+\overrightarrow{D\text{'}B\text{'}}+\overrightarrow{B\text{'}A}=\overrightarrow{AA\text{'}}=\overrightarrow{0} \end{gathered}$$