Cho hai vectơ \(\overrightarrow a  = \left( {1;\sqrt 3 } \right),\overrightarrow b  = \left( { - 2\sqrt 3 ;6} \right)\).

Ta có 

\(\begin{array}{l}
\left| {\overrightarrow a } \right| = \sqrt {{1^2} + {{\left( {\sqrt 3 } \right)}^2}}  = 2\\
\left| {\overrightarrow b } \right| = \sqrt {{{\left( { - 2\sqrt 3 } \right)}^2} + {6^2}}  = 4\sqrt 3 \\
\overrightarrow a .\overrightarrow b  = 1.\left( { - 2\sqrt 3 } \right) + \sqrt 3 .6 = 4\sqrt 3 \\
{\rm{cos}}\left( {\overrightarrow a ,\overrightarrow b } \right) = \frac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|}} = \frac{{4\sqrt 3 }}{{2.4\sqrt 3 }} = \frac{1}{2} \Rightarrow \left( {\overrightarrow a ,\overrightarrow b } \right) = {60^ \circ }
\end{array}\)