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\(\overrightarrow{NP}=\overrightarrow{NC}+\overrightarrow{CP}\)
\(=\dfrac{2}{3}\overrightarrow{BC}+\dfrac{1}{3}\overrightarrow{CA}\)
\(=-\dfrac{2}{3}\overrightarrow{CB}+\dfrac{1}{3}\overrightarrow{CA}\)
\(\overrightarrow{PM}=\overrightarrow{PA}+\overrightarrow{AM}\)
\(=\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{3}\overrightarrow{AB}\)
\(=\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{3}\left(\overrightarrow{AC}+\overrightarrow{CB}\right)\)
\(=\dfrac{1}{3}\overrightarrow{CA}+\dfrac{1}{3}\overrightarrow{CB}\)
\(\overrightarrow{AB}+\overrightarrow{NA}+\overrightarrow{BM}+\overrightarrow{BQ}=\overrightarrow{NB}+\overrightarrow{BM}+\overrightarrow{BQ}=\overrightarrow{NM}+\overrightarrow{BQ}\)
\(\overrightarrow{NM}=\overrightarrow{BQ}=\overrightarrow{QC}\)
\(\Rightarrow\overrightarrow{NM}+\overrightarrow{BQ}=\overrightarrow{QC}+\overrightarrow{BQ}=\overrightarrow{BC}\)
a: \(\overrightarrow{CN}=\dfrac{1}{2}\overrightarrow{CA}+\dfrac{1}{2}\overrightarrow{CB}\)
\(=\dfrac{1}{2}\overrightarrow{CB}+\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{2}\overrightarrow{CB}\)
\(=\dfrac{1}{2}\overrightarrow{u}-\overrightarrow{v}\)
a: \(\overrightarrow{CA}+\overrightarrow{AB}+\overrightarrow{BC}\)
\(=\overrightarrow{CB}+\overrightarrow{BC}\)
\(=\overrightarrow{0}\)
b: \(\overrightarrow{AM}+\overrightarrow{AP}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)=\dfrac{1}{2}\cdot2\cdot\overrightarrow{AN}=\overrightarrow{AN}\)