\(\begin{array}{l}a)\overrightarrow {AB} .\overrightarrow {CA} = \left| {\overrightarrow {AB} } \right|.\left| {\overrightarrow {CA} } \right|.\cos \left( {\overrightarrow {AB} ,\overrightarrow {CA} } \right)\\ = a.a\cos {120^0} = \dfrac{{{a^2}}}{2}\\b)\overrightarrow {GA} .\overrightarrow {GB} = \left| {\overrightarrow {GA} } \right|.\left| {\overrightarrow {GB} } \right|\cos \left( {\overrightarrow {GA} ,\overrightarrow {GB} } \right)\end{array}\)
Gọi \(M\) là trung điểm BC thì \(AM = AB\sin {60^0} = \dfrac{{a\sqrt 3 }}{2}\) \( \Rightarrow AG = \dfrac{2}{3}AM = \dfrac{{a\sqrt 3 }}{3}\)
\( \Rightarrow BG = \dfrac{{a\sqrt 3 }}{3}\).
\( \Rightarrow \overrightarrow {GA} .\overrightarrow {GB} = \dfrac{{a\sqrt 3 }}{3}.\dfrac{{a\sqrt 3 }}{3}.\cos {120^0} = \dfrac{{{a^2}}}{6}\)
