§2. Tổng và hiệu của hai vectơ

\(\left|\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{DC}\right|=\left|\overrightarrow{AB}+\overrightarrow{AC}\right|=\left|3\cdot\overrightarrow{AD}\right|=3a\)

a: \(\left|\overrightarrow{AB}-\overrightarrow{AC}\right|=\left|\overrightarrow{CB}\right|=10a\)

b: \(\left|\overrightarrow{AB}+\overrightarrow{AC}\right|=\dfrac{BC}{2}=5a\)

b: \(\left|\overrightarrow{BA}+\overrightarrow{BC}\right|=\left|\dfrac{\overrightarrow{AC}}{2}\right|=\dfrac{5}{2}a\)

a: \(\overrightarrow{MN}+\overrightarrow{NP}+\overrightarrow{PQ}\)

\(=\overrightarrow{MP}+\overrightarrow{PQ}\)

\(=\overrightarrow{MQ}\)

a: \(\overrightarrow{CA}+\overrightarrow{AB}+\overrightarrow{BC}\)

\(=\overrightarrow{CB}+\overrightarrow{BC}\)

\(=\overrightarrow{0}\)

b: \(\overrightarrow{AM}+\overrightarrow{AP}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)=\dfrac{1}{2}\cdot2\cdot\overrightarrow{AN}=\overrightarrow{AN}\)

\(\overrightarrow{BM}+\overrightarrow{CN}+\overrightarrow{AP}\)

\(=\dfrac{1}{2}\left(\overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{AP}\right)\)

\(=\overrightarrow{0}\)