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📰 Solution: The cosine of the angle between vectors $\mathbf{a}$ and $\mathbf{b}$ is given by $\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}$. Compute the dot product: $\mathbf{a} \cdot \mathbf{b} = (1)(0) + (0)(1) + (1)(1) = 1$. The magnitudes are $\|\mathbf{a}\| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2}$ and $\|\mathbf{b}\| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$. Thus, $\cos\theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$. The final answer is $\boxed{\dfrac{1}{2}}$. 📰 Question: A science policy analyst models the efficiency of a renewable energy grid using complex numbers. If $z = \cos\theta + i\sin\theta$ satisfies $z^6 = -1$, find $\theta$ in radians. 📰 Solution: By De Moivre's Theorem, $z^6 = \cos(6\theta) + i\sin(6\theta) = -1 + 0i$. This implies $6\theta = \pi + 2\pi k$ for integer $k$. Solving for $\theta$ gives $\theta = \frac{\pi}{6} + \frac{\pi k}{3}$. The principal solution in $[0, 2\pi)$ is $\theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}$. The smallest positive $\theta$ is $\boxed{\dfrac{\pi}{6}}$.