Question

In a regular hexagon ABCDEF, $\overrightarrow{AC}+\overrightarrow{AF}-\overrightarrow{AB}=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.$

Answer 1

Given:
ABCDEF is a regular hexagon

$$\begin{gathered} \overrightarrow{AC}+\overrightarrow{AF}-\overrightarrow{AB}=\overrightarrow{AC}+\overrightarrow{CD}-\overrightarrow{AB}\left(\because \overrightarrow{AF}=\overrightarrow{CD}\right) \\ =\left(\overrightarrow{AC}+\overrightarrow{CD}\right)-\overrightarrow{AB} \\ =\left(\overrightarrow{AD}\right)-\overrightarrow{AB} \\ =\left(\overrightarrow{AB}+\overrightarrow{BD}\right)-\overrightarrow{AB} \\ =\overrightarrow{AB}+\overrightarrow{BD}-\overrightarrow{AB} \\ =\overrightarrow{BD} \end{gathered}$$


Hence, $\overrightarrow{AC}+\overrightarrow{AF}-\overrightarrow{AB}=\underline{\overrightarrow{BD}}.$