Solution: The dot product of two unit vectors is $\mathbf{u} \cdot \mathbf{v} = \cos\theta$. Given $\cos\theta = \frac{\sqrt{3}}{2}$, the angle $\theta$ satisfies $\theta = \arccos\left(\frac{\sqrt{3}}{2}\right)$. This corresponds to $\theta = 30^\circ$ or $\frac{\pi}{6}$ radians. However, since cosine is positive in both the first and fourth quadrants, but angles between vectors are typically taken in $[0, \pi]$, the solution is $\boxed{\dfrac{\pi}{6}}$.

Tue, 05 May 2026 22:46:11 +0000

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