Giải bài 2 trang 17 – SGK Hình học lớp 10

Vì $K$ là trung điểm của $BC$ nên $\overrightarrow{AB} +\overrightarrow{AC} =2\overrightarrow{AK}$  (1)

    $M$ là trung điểm của $AC$ nên $\overrightarrow{BA} +\overrightarrow{BC} =2\overrightarrow{BM}$  (2)

Trừ từng vế của (1) cho (2), ta được:

$\stackrel{}{\mathrm{<span\; class="math-tex">\$\backslash overrightarrow\{AB\}\; +\backslash overrightarrow\{AC\}\; -\backslash overrightarrow\{BA\}-\backslash overrightarrow\{BC\}=2\backslash overrightarrow\{AK\}-2\backslash overrightarrow\{BM\}\$</span>}}\left(\stackrel{}{}\right)$

$\left(\stackrel{}{\mathrm{<span\; class="math-tex">\$\backslash Leftrightarrow\; \backslash left(\backslash overrightarrow\{AB\}-\backslash overrightarrow\{BA\}\backslash right)+\backslash left(\backslash overrightarrow\{AC\}-\backslash overrightarrow\{BC\}\backslash right)=2\backslash overrightarrow\{u\}-2\backslash overrightarrow\{v\}\$</span>}}\right)\stackrel{}{}$

$\Leftrightarrow \left(​​​​\overrightarrow{AB}+\overrightarrow{AB}\right)+​​​​\left(\overrightarrow{AC}​+\overrightarrow{CB}\right)=2\overrightarrow{u}-2\overrightarrow{v}$

$\Leftrightarrow 2\overrightarrow{AB}+​​\overrightarrow{AB}=2\overrightarrow{u}-2\overrightarrow{v}$

$\Leftrightarrow 3\overrightarrow{AB}=2\overrightarrow{u}-2\overrightarrow{v}\\ \Leftrightarrow \overrightarrow{AB}=\dfrac{2}{3}\overrightarrow{u}-\dfrac{2}{3}\overrightarrow{v}$

Từ (1), ta có:  $\overrightarrow{AC}=2\overrightarrow{AK} -\overrightarrow{AB} =2\overrightarrow{u} -\left(\dfrac{2}{3}\overrightarrow{u}-\dfrac{2}{3}\overrightarrow{v}\right)=\dfrac{4}{3}\overrightarrow{u}+\dfrac{2}{3}\overrightarrow{v}$

Suy ra:$\overrightarrow{CA}=-\dfrac{4}{3}\overrightarrow{u} -\dfrac{2}{3}\overrightarrow{v}$ .

Từ (2), ta có: $\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=\left(\dfrac{4}{3}\overrightarrow{u}+\dfrac{2}{3}\overrightarrow{v}\right)- \left(\dfrac{2}{3}\overrightarrow{u}-\dfrac{2}{3}\overrightarrow{v}\right)=\dfrac{2}{3}\overrightarrow{u}+\dfrac{4}{3}\overrightarrow{v}$.

Vậy $\overrightarrow{AB} =\dfrac{2}{3}\overrightarrow{u}-\dfrac{2}{3};\,\overrightarrow{CA} =-\dfrac{4}{3}\overrightarrow{u}-\dfrac{2}{3}\overrightarrow{v};\,\overrightarrow{BC} =\dfrac{2}{3}\overrightarrow{u}+\dfrac{4}{3}\overrightarrow{v}$.

Ghi nhớ:

Nếu AI là đường trung tuyến trong tam giác ABC (I thuộc BC) thì $\overrightarrow{AB} +\overrightarrow{AC} =2\overrightarrow{AI}$