If the vectors $\vec{a}$ and $\vec{b}$ are such that $|\vec{a}| = 3, |\vec{b}| = \frac{2}{3}$ and $\vec{a} \times \vec{b}$ is a unit vector, then find the angle between $\vec{a}$ and $\vec{b}$.

$\frac{\pi}{6}$

The correct answer is Option (1) → $\frac{\pi}{6}$ ##

Given that, $|\vec{a}| = 3, |\vec{b}| = \frac{2}{3}$ and $\vec{a} \times \vec{b}$ is a unit vector.

Therefore, $|\vec{a} \times \vec{b}| = 1$

$ |\vec{a}||\vec{b}| \sin \theta = 1 $

$ 3 \cdot \frac{2}{3} \sin \theta = 1 $

$ \sin \theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{6} $

The angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$.