Đáp án:
c.$\vec{GA}.\vec{GB}+\vec{GB}.\vec{GC}+\vec{GC}.\vec{GA}=-\dfrac{29}{6}$
d.$AD=\dfrac{3\sqrt[]{6}}{5}$
Giải thích các bước giải:
c.Ta có:
$\begin{split}(3\vec{GB})^2&=(\vec{BA}+\vec{BC})^2\\&=BA^2+BC^2+2BA.BC.cos(\widehat{\vec{BA},\vec{BC}}\\&=BA^2+BC^2+(BA^2+BC^2-CA^2)\\&=2(BA^2+BC^2)-AC^2\end{split}$
$\rightarrow BG^2=\dfrac{2(BA^2+BC^2)-AB^2}{9}=\dfrac{31}{9}$
Tương tự:
$\rightarrow \begin{cases}CG^2=\dfrac{2(CA^2+CB^2)-AB^2}{9}=\dfrac{46}{9}\\AG^2=\dfrac{2(AB^2+AC^)-BC^2}{9}=\dfrac{10}{9}\end{cases}$
Đặt $S=\vec{GA}.\vec{GB}+\vec{GB}.\vec{GC}+\vec{GC}.\vec{GA}$
$\vec{GA}+\vec{GB}+\vec{GC}=0$
$\rightarrow (\vec{GA}+\vec{GB}+\vec{GC})^2=0$
$\rightarrow GA^2+GB^2+GC^2+2\vec{GA}.\vec{GB}+2\vec{GB}.\vec{GC}+2\vec{GC}.\vec{GA}=0$
$\rightarrow GA^2+GB^2+GC^2+2S=0$
$\rightarrow S=-\dfrac{GA^2+GB^2+GC^2}{2}=-\dfrac{29}{6}$
d.Do AD là phân giác góc A nên ta có:
$\dfrac{DB}{DC}=\dfrac{AB}{AC}\rightarrow AC.\vec{DB}=-AB.\vec{DC}$
$\rightarrow 3(\vec{AB}-\vec{AD})=-2(\vec{AC}-\vec{AD})\rightarrow \vec{AD}=\dfrac{3}{5}.\vec{AB}+\dfrac{2}{5}.\vec{AC}$
$\rightarrow \vec{AD}^2=(\dfrac{3}{5}.\vec{AB}+\dfrac{2}{5}.\vec{AC})^2=\dfrac{9}{25}.AB^2+\dfrac{4}{25}.AC^2+\dfrac{12}{25}.\vec{AB}.\vec{AC}=\dfrac{54}{25}$
$\rightarrow AD=\dfrac{3\sqrt[]{6}}{5}$
