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📰 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}} $. 📰 Question: Find the vector $ \mathbf{v} $ such that $ \mathbf{v} \times \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} -4 \\ 6 \\ 2 \end{pmatrix} $. 📰 Solution: Let $ \mathbf{v} = \begin{pmatrix} a \\ b \\ c \end{pmatrix} $. The cross product is: 📰 How To Pair Airtag 📰 Macbook Pages 📰 How To Get Medicaid 📰 How Zfs Encryption Can Transform Your Data Privacy Overnight 6414408 📰 Match Three Games Free 📰 Sweetgreen Stock Price 📰 Vlc Player Mac Os X Download 📰 Fidelity Home Loan 6543664 📰 Shiny Stone Soul Silver 📰 Unlock Dubais Rich Flavor With The Ultimate Pistachio Dark Chocolate Bar 1579832 📰 Count Of Unique Values In Excel 📰 Mtas Traintime App 📰 What Is Cast N Chill 📰 Perfume Atelier 📰 The Forest Two