$(\vec a - \vec b)$ perpendicular $(6\vec a + \vec b)$ what is angle between $\vec a$ and $\vec b$

Question

another info : $|\vec a| = 2$ , $|\vec b| = 3$

$(\vec a - \vec b)$ perpendicular $(6\vec a + \vec b)$

$6{\vec a} ^2 -5\vec a\vec b - {\vec b }^2 =0$

$12 -5\vec a\vec b - 3 =0$

$5(6\cos \theta) = 9$

But the answer is $\frac{\pi}{3}$. Where am i wrong?

Answer 1

$\vec{v}\cdot \vec{v}=\|v\|^2$. You used $\|v\|$ instead.