Solution: Let $ \theta $ be the angle between $ \mathbf{a} $ and $ \mathbf{b} $, so $ \cos\theta = \frac{1}{2} \Rightarrow \theta = 60^\circ $. Let $ \phi $ be the angle between $ \mathbf{b} $ and $ \mathbf{c} $, so $ \cos\phi = \frac{\sqrt{3}}{2} \Rightarrow \phi = 30^\circ $. To maximize $ \mathbf{a} \cdot \mathbf{c} = \cos(\alpha) $, where $ \alpha $ is the angle between $ \mathbf{a} $ and $ \mathbf{c} $, arrange $ \mathbf{a}, \mathbf{b}, \mathbf{c} $ in a plane. The maximum occurs when $ \mathbf{a} $ and $ \mathbf{c} $ are aligned, but constrained by their angles relative to $ \mathbf{b} $. The minimum angle between $ \mathbf{a} $ and $ \mathbf{c} $ is $ 60^\circ - 30^\circ = 30^\circ $, so $ \cos(30^\circ) = \frac{\sqrt{3}}{2} $. However, if they are aligned, $ \alpha = 0^\circ $, but this requires $ \theta = \phi = 0^\circ $, which contradicts the given dot products. Instead, use the cosine law for angles: $ \cos\alpha \leq \cos(60^\circ - 30^\circ) = \cos(30^\circ) = \frac{\sqrt{3}}{2} $. Thus, the maximum is $ \boxed{\frac{\sqrt{3}}{2}} $.

Sun, 10 May 2026 08:38:35 +0000

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