Cho hai vectơ \(\overrighta...

D

- Hướng dẫn giải

Ta có:

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

Vậy chọn D.