Câu hỏi:

Ta có $\vec{u} \perp \vec{v}$ nên $\vec{u} . \vec{v} = 0$.
Suy ra $(\frac{2}{5}\vec{a} - 3\vec{b}).(\vec{a} + \vec{b}) = 0$
$\Leftrightarrow \frac{2}{5}\vec{a}^2 + \frac{2}{5}\vec{a}.\vec{b} - 3\vec{a}.\vec{b} - 3\vec{b}^2 = 0$
$\Leftrightarrow \frac{2}{5}|\vec{a}|^2 - \frac{13}{5}\vec{a}.\vec{b} - 3|\vec{b}|^2 = 0$
$\Leftrightarrow \frac{2}{5} - \frac{13}{5}|\vec{a}||\vec{b}|cos\alpha - 3 = 0$
$\Leftrightarrow \frac{2}{5} - \frac{13}{5}cos\alpha - 3 = 0$
$\Leftrightarrow \frac{13}{5}cos\alpha = \frac{2}{5} - 3 = -\frac{13}{5}$
$\Leftrightarrow cos\alpha = -1$
$\Leftrightarrow \alpha = 180^0$.
Vậy góc giữa hai vecto $\vec{a}$ và $\vec{b}$ là $180^0$.