If θ is an acute angle and the vector $\vec a = (\sin θ)\hat i + (\cos θ)\hat j$ is perpendicular to the vector $\vec b=\hat i-\sqrt{3}\hat j$ then θ is equal to

$\frac{\pi}{3}$

The correct answer is Option (2) → $\frac{\pi}{3}$

Given $\vec a=(\sin\theta,\cos\theta)$ and $\vec b=(1,-\sqrt{3})$.

Since vectors are perpendicular:

$\vec a\cdot \vec b=0$

$\sin\theta\cdot 1+\cos\theta(-\sqrt{3})=0$

$\sin\theta-\sqrt{3}\cos\theta=0$

$\tan\theta=\sqrt{3}$

Since $\theta$ is acute:

$\theta=\frac{\pi}{3}$

Final answer: $\frac{\pi}{3}$