Cho hình vuông \(ABCD\) cạnh \(a\). Tính $$\left|\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}\right|.$$
| \(\left|\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}\right|=2a\sqrt{2}\) | |
| \(\left|\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}\right|=3a\) | |
| \(\left|\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}\right|=2a+a\sqrt{2}\) | |
| \(\left|\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}\right|=3a\sqrt{2}\) |
Chọn phương án A.

Theo quy tắc hình bình hành ta có $$\overrightarrow{AB}+\overrightarrow{AD}=\overrightarrow{AC}.$$
Do đó, \(\left|\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}\right|=\left|2\overrightarrow{AC}\right|=2AC=2a\sqrt{2}\).