I tried to solve this by directly multiplies a-b and 6a+b equal zero, and substitute some of it with the |a| and |b| to get a.b. but somehow I get different answer from the answer sheet that is 60 degree
It is still not complete but after that I get some bizzare answer so I didn't write it
Given $(\overrightarrow{a}-\overrightarrow{b})\perp(6\overrightarrow{a}+\overrightarrow{b})$, we know that \begin{align} (\overrightarrow{a}-\overrightarrow{b})\cdot(6\overrightarrow{a}+\overrightarrow{b})&=0\\ 6\overrightarrow{a}\cdot\overrightarrow{a}-5\overrightarrow{a}\cdot\overrightarrow{b}-\overrightarrow{b}\cdot\overrightarrow{b}&=0\\ 6|\overrightarrow{a}|^2-5\overrightarrow{a}\cdot\overrightarrow{b}-|\overrightarrow{b}|^2&=0\\ 6(2)^2-5\overrightarrow{a}\cdot\overrightarrow{b}-(3)^2&=0\\ -5\overrightarrow{a}\cdot\overrightarrow{b}&=-15\\ \overrightarrow{a}\cdot\overrightarrow{b}&=3 \end{align}
From the definition of the dot product: \begin{align} \cos\theta&=\dfrac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}\\ \cos\theta&=\dfrac{3}{(2)(3)}\\ \cos\theta&=\dfrac12\\ \theta&=60^{\circ} \end{align}
Anything from here that needs further explanation?