Từ đề bài ta có:
\(\left\{{}\begin{matrix}\left(\overrightarrow{a}+3\overrightarrow{b}\right)\left(7\overrightarrow{a}-5\overrightarrow{b}\right)=0\\\left(\overrightarrow{a}-4\overrightarrow{b}\right)\left(7\overrightarrow{a}-2\overrightarrow{b}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7\overrightarrow{a}^2-15\overrightarrow{b}^2+16\overrightarrow{a}.\overrightarrow{b}=0\left(1\right)\\7\overrightarrow{a}^2+8\overrightarrow{b}^2-30\overrightarrow{a}.\overrightarrow{b}=0\left(2\right)\end{matrix}\right.\)
Lần lượt lấy \(\left(1\right)-\left(2\right)\) và \(8.\left(1\right)+15.\left(2\right)\) ta được:
\(\left\{{}\begin{matrix}-23\overrightarrow{b}^2+46\overrightarrow{a}.\overrightarrow{b}=0\\161\overrightarrow{a}^2-322\overrightarrow{a}.\overrightarrow{b}=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{b}^2=2\overrightarrow{a}.\overrightarrow{b}\\\overrightarrow{a}^2=2\overrightarrow{a}.\overrightarrow{b}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{a}.\overrightarrow{b}=\dfrac{\overrightarrow{a}^2}{2}=\dfrac{\left|\overrightarrow{a}\right|^2}{2}\\\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|\end{matrix}\right.\)
\(\Rightarrow cos\left(\overrightarrow{a};\overrightarrow{b}\right)=\dfrac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|}=\dfrac{\dfrac{\left|\overrightarrow{a}\right|^2}{2}}{\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|}=\dfrac{1}{2}\)
\(\Rightarrow\left(\overrightarrow{a};\overrightarrow{b}\right)=60^0\)
