Cho hai vectơ \(\vec{a}, \vec{b}\) thỏa mãn: \(|\vec{a}|=4 ;|\vec{b}|=3 ; \vec{a} \cdot \vec{b}=10\) . Xét hai vectơ \(\bar{y}=\vec{a}-\vec{b}; \quad \vec{x}=\vec{a}-2 \vec{b}\) ....

* Hướng dẫn giải

\(\begin{array}{l} \text { Ta có } \vec{x} \cdot \vec{y}=(\vec{a}-2 \vec{b})(\vec{a}-\vec{b})=(\vec{a})^{2}+2(\vec{b})^{2}-3 \vec{a} \cdot \vec{b}=4 \\ |\vec{x}|=\sqrt{(\vec{x})^{2}}=\sqrt{(\vec{a}-2 \vec{b})^{2}}=\sqrt{(\vec{a})^{2}+4(\vec{b})^{2}-4 \vec{a} \cdot \vec{b}}=2 \sqrt{3} \\ |\vec{y}|=\sqrt{(\vec{y})^{2}}=\sqrt{(\vec{a}-\vec{b})^{2}}=\sqrt{(\vec{a})^{2}+(\vec{b})^{2}-2 \vec{a} \cdot \vec{b}}=\sqrt{5} \\ \cos \alpha=\frac{\vec{x} \cdot \vec{y}}{|x| \cdot|\vec{y}|}=\frac{4}{2 \sqrt{3} \cdot \sqrt{5}}=\frac{2}{\sqrt{15}} \end{array}\)