Question

Góc giữa hai vectơ \(\overrightarrow{a}và\overrightarrow{b}\) bằng 60 độ và \(a^2=b^2\).Đặt \(\overrightarrow{u}=\overrightarrow{a}-\overrightarrow{b}\) thì góc giữa hai vectơ \(\overrightarrow{a}và\overrightarrow{u}\) bằng:

Answer 1

\(\overrightarrow{u}^2=\left(\overrightarrow{a}-\overrightarrow{b}\right)^2=a^2+b^2-2\overrightarrow{a}.\overrightarrow{b}=2a^2-2a^2.cos60^0=a^2\)

\(\Rightarrow\left|\overrightarrow{u}\right|=a\)

\(\overrightarrow{a}.\overrightarrow{u}=\overrightarrow{a}\left(\overrightarrow{a}-\overrightarrow{b}\right)=a^2-\overrightarrow{a}.\overrightarrow{b}=a^2-a^2.cos60^0=\frac{a^2}{2}\)

\(\Rightarrow cos\left(\overrightarrow{a};\overrightarrow{u}\right)=\frac{\overrightarrow{a}.\overrightarrow{u}}{\left|\overrightarrow{a}\right|.\left|\overrightarrow{u}\right|}=\frac{a^2}{2a^2}=\frac{1}{2}\Rightarrow\left(\overrightarrow{a};\overrightarrow{u}\right)=60^0\)