Cho vecto a = (1;2), vecto b = (-2;3) . Góc giữa hai vectơ

Đáp án đúng là: D

Với $\overrightarrow{a}=(1;2), \overrightarrow{b}=(-2;3)$ ta có:

+) $$\begin{gathered} 3\overrightarrow{a}=(3.1;3.2)=(3;6), 2\overrightarrow{b}=(2.(-2);2.3)=(-4;6) \\ Suy ra \overrightarrow{u}=3\overrightarrow{a}+2\overrightarrow{b}=(3-4;6+6)=(-1;12) \\ +) \overrightarrow{a}=(3;4), 5\overrightarrow{b}=(5.(-2);5.3)=(-10;15) \\ Suy ra \overrightarrow{v}=\overrightarrow{a}-5\overrightarrow{b}=(3-(-10);4-15)=(13;-11) \\ Ta có \cos (\overrightarrow{u},\overrightarrow{v})=\frac{\overrightarrow{u}.\overrightarrow{v}}{\left|\overrightarrow{u}\right|.\left|\overrightarrow{v}\right|} \\ =\frac{-1.13+12.(-11)}{\sqrt{{(-1)}^2+{12}^2}.\sqrt{{13}^2+{(-11)}^2}} \\ =\frac{-145}{\sqrt{145}.\sqrt{290}}=-\frac{1}{\sqrt{2}} \\ Suy ra (\overrightarrow{u},\overrightarrow{v})=135° \end{gathered}$$