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.
\(\left|\overrightarrow{AB}+\overrightarrow{BC}\right|=\left|\overrightarrow{AC}\right|=AC=a\).