* \(\left|\overrightarrow{AB}+\overrightarrow{AC}\right|=\sqrt{\left(\overrightarrow{AB}+\overrightarrow{AC}\right)^2}=\sqrt{\left(\overrightarrow{AB}\right)^2+2\overrightarrow{AB}.\overrightarrow{AC}+\left(\overrightarrow{AC}\right)^2}\)
Với:
\(\left(\overrightarrow{AB}\right)^2=\left(\overrightarrow{AC}\right)^2=AB^2=a^2\)
\(\overrightarrow{AB}.\overrightarrow{AC}=AB.AC.\cos60^0=\dfrac{a^2}{2}\)
\(\Rightarrow\left|\overrightarrow{AB}+\overrightarrow{AC}\right|=\sqrt{a^2+2.\dfrac{a^2}{2}+a^2}=\sqrt{3}a\)
* \(\left|\overrightarrow{AB}-\overrightarrow{AC}\right|=\left|\overrightarrow{CB}\right|=CB=a\)
* \(\left|\overrightarrow{GB}+\overrightarrow{GC}\right|=\left|-\overrightarrow{GA}\right|=\left|\overrightarrow{AG}\right|=AG\)
Kẻ AM vuông góc với BC ( M thuộc BC)
Theo Pythagoras:\(AM^2=AB^2-\left(\dfrac{BC}{2}\right)^2=a^2-\dfrac{a^2}{4}=\dfrac{3a^2}{4}\)
M là đường cao trong tam giác đều nên cũng là đường trung tuyến trong tam giác đều\(\Rightarrow AG=\dfrac{2}{3}AM=\dfrac{2}{3}\cdot\dfrac{\sqrt{3}a}{2}=\dfrac{\sqrt{3}}{3}a\)
\(\Rightarrow\left|\overrightarrow{GB}+\overrightarrow{GC}\right|=\dfrac{\sqrt{3}}{3}a\)
