Bổ sung đề; AC=3cm
=>BC=5cm
a: \(\overrightarrow{AB}\cdot\overrightarrow{BC}=-\overrightarrow{BA}\cdot\overrightarrow{BC}=-BA\cdot BC\cdot cosABC=-4\cdot5\cdot\dfrac{AB}{BC}=-4\cdot5\cdot\dfrac{4}{5}=-16\)
b: Gọi Mlà trung điểm của BC
=>AM=BC/2=2,5cm
\(P=2\cdot\overrightarrow{AM}\cdot\overrightarrow{BC}\)
\(=2\cdot\overrightarrow{BC}\left(\overrightarrow{BM}-\overrightarrow{BA}\right)\)
\(=2\cdot\overrightarrow{BC}\cdot\overrightarrow{BM}-2\cdot\overrightarrow{BC}\cdot\overrightarrow{BA}\)
\(=2\cdot BC\cdot BM\cdot cos0^0-2\cdot BC\cdot BA\cdot cosB\)
\(=2\cdot5\cdot2.5-2\cdot5\cdot4\cdot\dfrac{AB}{BC}=25-40\cdot\dfrac{4}{5}=25-32=-7\)
Đúng 0
Bình luận (0)
