\(BC=\sqrt{6^2+8^2}=10\)
\(\overrightarrow{AM}.\overrightarrow{BC}=\left(\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}\right).\overrightarrow{BC}=\overrightarrow{AB}.\overrightarrow{BC}+\dfrac{2}{3}BC^2=-\overrightarrow{BA}.\overrightarrow{BC}+\dfrac{2}{3}BC^2\)
\(=-BA.BC.cosB+\dfrac{2}{3}BC^2=-BA.BC.\dfrac{AB}{BC}+\dfrac{2}{3}BC^2\)
\(=-AB^2+\dfrac{2}{3}BC^2=-36+\dfrac{2}{3}.100=...\)
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