Question

Giúp e vs ạ e đg cần gấp e cảm ơn

Answer 1

a: Ta có: \(\overrightarrow{PA}+2\cdot\overrightarrow{PB}=\overrightarrow{0}\)

=>\(\overrightarrow{PA}=-2\cdot\overrightarrow{PB}\)

=>P nằm giữa A và B sao cho AP=2PB

AP+PB=AB

=>AB=2PB+PB=3BP

=>\(BP=\frac13BA;AP=\frac23AB\)

Ta có: \(5\cdot\overrightarrow{AQ}-2\cdot\overrightarrow{AC}=\overrightarrow{0}\)

=>\(5\cdot\overrightarrow{AQ}=2\cdot\overrightarrow{AC}\)

=>\(\overrightarrow{AQ}=\frac25\cdot\overrightarrow{AC}\)

Ta có: \(\overrightarrow{PQ}=\overrightarrow{PA}+\overrightarrow{AQ}\)

\(=-\frac23\cdot\overrightarrow{AB}+\frac25\cdot\overrightarrow{AC}=-2\left(\frac13\cdot\overrightarrow{AB}-\frac15\cdot\overrightarrow{AC}\right)\)

\(=-\frac{2}{15}\left(5\cdot\overrightarrow{AB}-3\cdot\overrightarrow{AC}\right)\) (1)
b: Xét ΔABC có AM là đường trung tuyến

nên \(\overrightarrow{AM}=\frac12\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)

=>\(\overrightarrow{AI}=\frac12\cdot\overrightarrow{AM}=\frac14\cdot\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)

\(\overrightarrow{PI}=\overrightarrow{PA}+\overrightarrow{AI}\)

\(=-\frac23\cdot\overrightarrow{AB}+\frac14\left(\overrightarrow{AB}+\overrightarrow{AC}\right)=-\frac23\cdot\overrightarrow{AB}+\frac14\cdot\overrightarrow{AB}+\frac14\cdot\overrightarrow{AC}\)

\(=\frac{-5}{12}\cdot\overrightarrow{AB}+\frac{3}{12}\cdot\overrightarrow{AC}=-\frac{1}{12}\left(5\cdot\overrightarrow{AB}-3\cdot\overrightarrow{AC}\right)\) (2)

Từ (1),(2) suy ra \(\frac{\overrightarrow{PI}}{\overrightarrow{PQ}}=\frac{-1}{12}:\frac{-2}{15}=\frac{1}{12}\cdot\frac{15}{2}=\frac{15}{24}=\frac58\)

=>P,I,Q thẳng hàng