Trong mặt phẳng tọa độ Oxy, hãy tính góc giữa hai vecto a và vecto b

* Hướng dẫn giải

a) Ta có: $\overrightarrow{a}.\overrightarrow{b}=\left(-3\right).2+1.6=0\Rightarrow \left(\overrightarrow{a},\overrightarrow{b}\right)={90}^0.$

b)  Ta có: $\overrightarrow{a}.\overrightarrow{b}=3.2+1.4=10$

$\left|\overrightarrow{a}\right|=\sqrt{3^2+1^2}=\sqrt{10},\left|\overrightarrow{b}\right|=\sqrt{2^2+4^2}=2\sqrt{5}$

$\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|.c\text{os}\left(\overrightarrow{a},\overrightarrow{b}\right)\Rightarrow c\text{os}\left(\overrightarrow{a},\overrightarrow{b}\right)=\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|}=\frac{10}{\sqrt{10}.2\sqrt{5}}=\frac{1}{\sqrt{2}}\Rightarrow \left(\overrightarrow{a},\overrightarrow{b}\right)={45}^0.$

c) Ta có: $\overrightarrow{a}.\overrightarrow{b}=\left(-\sqrt{2}\right).2+1.\left(-\sqrt{2}\right)=-3\sqrt{2}$

$\left|\overrightarrow{a}\right|=\sqrt{{\left(-\sqrt{2}\right)}^2+1^2}=\sqrt{3},\left|\overrightarrow{b}\right|=\sqrt{2^2+{\left(-\sqrt{2}\right)}^2}=\sqrt{6}$

$\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|.c\text{os}\left(\overrightarrow{a},\overrightarrow{b}\right)\Rightarrow c\text{os}\left(\overrightarrow{a},\overrightarrow{b}\right)=\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|}=\frac{-3\sqrt{2}}{\sqrt{3}.\sqrt{6}}=-1\Rightarrow \left(\overrightarrow{a},\overrightarrow{b}\right)={180}^0.$