Đáp án:
\[\cos \left( {\overrightarrow a - \overrightarrow b ;\,\,\overrightarrow a + \overrightarrow b } \right) = - \frac{{\sqrt 2 }}{3}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left| {\overrightarrow a + \overrightarrow b } \right| = 6\\
\Leftrightarrow {\left| {\overrightarrow a + \overrightarrow b } \right|^2} = 36\\
\Leftrightarrow {\overrightarrow a ^2} + 2.\overrightarrow a .\overrightarrow b + {\overrightarrow b ^2} = 36\\
\Leftrightarrow {3^2} + 2\overrightarrow a .\overrightarrow b + {5^2} = 36\\
\Leftrightarrow \overrightarrow a .\overrightarrow b = 1\\
{\left| {\overrightarrow a - \overrightarrow b } \right|^2} = {\overrightarrow a ^2} - 2.\overrightarrow a .\overrightarrow b + {\overrightarrow b ^2} = {3^2} - 2.1 + {5^2} = 32 \Rightarrow \left| {\overrightarrow a - \overrightarrow b } \right| = 4\sqrt 2 \\
\cos \left( {\overrightarrow a - \overrightarrow b ;\,\,\overrightarrow a + \overrightarrow b } \right) = \frac{{\left( {\overrightarrow a - \overrightarrow b } \right)\left( {\overrightarrow a + \overrightarrow b } \right)}}{{\left| {\overrightarrow a - \overrightarrow b } \right|.\left| {\overrightarrow a + \overrightarrow b } \right|}} = \frac{{{{\overrightarrow a }^2} - {{\overrightarrow b }^2}}}{{6.4\sqrt 2 }} = \frac{{{3^2} - {5^2}}}{{24\sqrt 2 }} = - \frac{{\sqrt 2 }}{3}
\end{array}\)
