Let $\vec a$ and $\vec b$ be two unit vectors. If the vectors $\vec c = \vec a + 2\vec b$ and $\vec d=5\vec a-4\vec b$ are perpendicular to each other, then the angle between $\vec a$ and $\vec b$, is

$\frac{π}{3}$

Let θ be the angle between vectors $\vec a$ and $\vec b$. Then,

$\vec a.\vec b =|\vec a||\vec b|\cos θ = \cos θ$    $[∵|\vec a|=|\vec b|=1]$

Since $\vec c =\vec a + 2\vec b$ and $\vec d =5\vec a-4\vec b$ are perpendicular to each other. Therefore,

$\vec c.\vec d=0$

$⇒(\vec a + 2\vec b).(5\vec a-4\vec b)=0$

$⇒5(\vec a.\vec a)+6(\vec a.\vec b)-8(\vec b.\vec b)=0$

$⇒5|\vec a|^2+6\cos θ -8|\vec b|^2=0$

$⇒5+6\cos θ -8=0$   $[∵|\vec a|=|\vec b|=1]$

$⇒\cos θ=\frac{1}{2}$

$⇒θ=\frac{π}{3}$