Trong mặt phẳng tọa độ cho hai vectơ a (1;m),b (√3;1). Góc giữa hai vectơ

* Hướng dẫn giải

Ta có: $\text{cos}\left(\overrightarrow{a};\overrightarrow{b}\right)=c{\text{os 60}}^0\iff \frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|}=\frac{1.\sqrt{3}+\text{}m.1}{\sqrt{1^2+m^2}.\sqrt{3+1^2}}=\frac{1}{2}$ 

$\begin{array}{l}\iff \frac{\sqrt{3}+\text{}m}{\sqrt{1+\text{}{m}^2}.2}=\frac{1}{2}\iff \frac{\sqrt{3}+\text{}m}{\sqrt{1+\text{}{m}^2}}=1\\ \iff \sqrt{3}+\text{}m=\sqrt{1+\text{}{m}^2}\iff 3+\text{}2\sqrt{3}m+\text{}{m}^2=1+\text{}{m}^2\\ \iff 2\sqrt{3}m=-2\iff m=\frac{-\sqrt{3}}{3}\end{array}$

Chọn  C.