Question

Bài 12. Cho tam giác ABC. M là điểm thuộc cạnh BC sao cho 3BM = 7CM. Biểu diễn vecto AM qua vecto AB và AC .
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Answer 1

\(3BM=7CM=7\left(BC-BM\right)\Rightarrow10BM=7BC\)

\(\Rightarrow BM=\dfrac{7}{10}BC\Rightarrow\overrightarrow{BM}=\dfrac{7}{10}\overrightarrow{BC}\)

Ta có:

\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=\overrightarrow{AB}+\dfrac{7}{10}\overrightarrow{BC}=\overrightarrow{AB}+\dfrac{7}{10}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\overrightarrow{AB}-\dfrac{7}{10}\overrightarrow{AB}+\dfrac{7}{10}\overrightarrow{AC}\)

\(\Rightarrow\overrightarrow{AM}=\dfrac{3}{10}\overrightarrow{AB}+\dfrac{7}{10}\overrightarrow{AC}\)

Answer 2

Hình vẽ:

Answer 3

Answer 4

Lời giải:

Theo đề thì $\overrightarrow{3BM}=7\overrightarrow{MC}=-7\overrightarrow{CM}$

Lại có:

$\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}$

$\Rightarrow 3\overrightarrow{AM}=3\overrightarrow{AB}+3\overrightarrow{BM}=3\overrightarrow{AB}-7\overrightarrow{CM}(1)$

$\overrightarrow{AM}=\overrightarrow{AC}+\overrightarrow{CM}$

$\Rightarrow 7\overrightarrow{AM}=7\overrightarrow{AC}+7\overrightarrow{CM}(2)$

Từ $(1);(2)\Rightarrow 10\overrightarrow{AM}=3\overrightarrow{AB}+7\overrightarrow{AC}$

$\Rightarrow \overrightarrow{AM}=\frac{3}{10}\overrightarrow{AB}+\frac{7}{10}\overrightarrow{AC}$