Đáp án:
$\left( {\overrightarrow a ,\overrightarrow b } \right) = {60^0}$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
\overrightarrow p = \overrightarrow a + 2\overrightarrow b ;\overrightarrow q = 5\overrightarrow a - 4\overrightarrow b \\
\Rightarrow \overrightarrow p .\overrightarrow q = \left( {\overrightarrow a + 2\overrightarrow b } \right)\left( {5\overrightarrow a - 4\overrightarrow b } \right)\\
\Leftrightarrow 0 = 5{\left( {\overrightarrow a } \right)^2} + 6\overrightarrow a .\overrightarrow b - 8{\left( {\overrightarrow b } \right)^2}\\
\Leftrightarrow 0 = 5{\left| {\overrightarrow a } \right|^2} + 6\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|.\cos \left( {\overrightarrow a ,\overrightarrow b } \right) - 8{\left| {\overrightarrow b } \right|^2}\\
\Leftrightarrow 0 = - 3{\left| {\overrightarrow a } \right|^2} + 6{\left| {\overrightarrow a } \right|^2}.\cos \left( {\overrightarrow a ,\overrightarrow b } \right)\\
\Leftrightarrow \cos \left( {\overrightarrow a ,\overrightarrow b } \right) = \dfrac{{3{{\left| {\overrightarrow a } \right|}^2}}}{{6{{\left| {\overrightarrow a } \right|}^2}}}\\
\Leftrightarrow \cos \left( {\overrightarrow a ,\overrightarrow b } \right) = \dfrac{1}{2}\\
\Leftrightarrow \left( {\overrightarrow a ,\overrightarrow b } \right) = {60^0}
\end{array}$
Vậy $\left( {\overrightarrow a ,\overrightarrow b } \right) = {60^0}$
