Solution: The magnitude of the cross product is $\|\mathbf{p} \times \mathbf{q}\| = \|\mathbf{p}\| \|\mathbf{q}\| \sin\theta$. Since $\mathbf{p}$ and $\mathbf{q}$ are unit vectors, $\|\mathbf{p} \times \mathbf{q}\| = \sin\theta = \frac{1}{2}$. Thus, $\theta = \arcsin\left(\frac{1}{2}\right)$, which gives $\theta = \frac{\pi}{6}$ or $\frac{5\pi}{6}$. However, the angle between vectors is taken as the acute angle, so $\theta = \boxed{\dfrac{\pi}{6}}$.

Thu, 30 Apr 2026 23:20:05 +0000

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