Cho hai vectơ a và b đều khác vecto 0. Biết: (vecto a,b)=60 độ, vecto a.b=4 và

* Hướng dẫn giải

Hướng dẫn giải:

Đáp án đúng là: D.

$\overrightarrow{a}.\overrightarrow{b}=4\iff \left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|.cos\left(\overrightarrow{a},\overrightarrow{b}\right)=4\iff \left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|.cos60°=4$

$\iff \left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|.\frac{1}{2}=4\iff \left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|=8$.

Ta có: $\left\{\begin{array}{l}\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|=8\\ \left|\overrightarrow{a}\right|+\left|\overrightarrow{b}\right|=6\end{array}\right.\iff \left\{\begin{array}{l}\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|=8\\ \left|\overrightarrow{b}\right|=6-\left|\overrightarrow{a}\right|\end{array}\right.\iff \left\{\begin{array}{l}\left|\overrightarrow{a}\right|.\left(6-\left|\overrightarrow{a}\right|\right)=8\left(1\right)\\ \left|\overrightarrow{b}\right|=6-\left|\overrightarrow{a}\right|\end{array}\right.$

Xét (1) ta có : $6\left|\overrightarrow{a}\right|-{\left|\overrightarrow{a}\right|}^2=8\iff -{\left|\overrightarrow{a}\right|}^2+6\left|\overrightarrow{a}\right|-8=0\iff \left[\begin{array}{l}\left|\overrightarrow{a}\right|=4\\ \left|\overrightarrow{a}\right|=2\end{array}\right.$

Vì $\left|\overrightarrow{a}\right|>3$ nên ta có: $\left|\overrightarrow{a}\right|=4$.