Cho \(\overrightarrow{u}=\overrightarrow{a}+3\overrightarrow{b}\) vuông góc với \(\overrightarrow{v}=7\overrightarrow{a}-5\overrightarrow{b}\) và \(\overrightarrow{x}=\overrightarrow{a}-4\overrightarrow{b}\)vuông góc với \(\overrightarrow{y}=7\overrightarrow{a}-2\overrightarrow{b}\). Khi đó góc giữa hai vectơ \(\overrightarrow{a}\) và \(\overrightarrow{b}\) bằng:
A. \(75^o\)
B. \(60^o\)
C. \(120^o\)
D. \(45^o\)
\(\overrightarrow{u}\overrightarrow{v}=0\Rightarrow\left(\overrightarrow{a}+3\overrightarrow{b}\right)\left(7\overrightarrow{a}-5\overrightarrow{b}\right)=7a^2+16\overrightarrow{a}\overrightarrow{b}-15b^2=0\left(1\right)\)
\(\overrightarrow{x}\overrightarrow{y}=0\Rightarrow\left(\overrightarrow{a}-4\overrightarrow{b}\right)\left(7\overrightarrow{a}-2\overrightarrow{b}\right)=7a^2-30\overrightarrow{a}\overrightarrow{b}+8b^2=0\left(2\right)\)
(1) và (2): \(\left\{{}\begin{matrix}7a^2+16\overrightarrow{a}\overrightarrow{b}-15b^2=0\\7a^2-30\overrightarrow{a}\overrightarrow{b}+8b^2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{a}\overrightarrow{b}=\frac{b^2}{2}\\a^2=b^2\Rightarrow\left|a\right|=\left|b\right|\end{matrix}\right.\)
\(\Rightarrow cos\left(\overrightarrow{a},\overrightarrow{b}\right)=\frac{\overrightarrow{a}\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{\frac{b^2}{2}}{\left|a\right|.\left|b\right|}=\frac{\frac{b^2}{2}}{b^2}=\frac{1}{2}\)
\(\Rightarrow\left(\overrightarrow{a};\overrightarrow{b}\right)=60^0\)
Cảm ơn bạn rất nhiều !