Cho tam giác \(ABC\) đều, cạnh \(a\). Tính \(\left|\overrightarrow{AB}+\overrightarrow{BC}\right|\).
| \(\left|\overrightarrow{AB}+\overrightarrow{BC}\right|=2a\) | |
| \(\left|\overrightarrow{AB}+\overrightarrow{BC}\right|=a\sqrt{3}\) | |
| \(\left|\overrightarrow{AB}+\overrightarrow{BC}\right|=\dfrac{\sqrt{3}}{2}\) | |
| \(\left|\overrightarrow{AB}+\overrightarrow{BC}\right|=a\) |
Chọn phương án D.
\(\left|\overrightarrow{AB}+\overrightarrow{BC}\right|=\left|\overrightarrow{AC}\right|=AC=a\).