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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: 📰 Alice Bell 6343713 📰 Dub Edition Midnight Club This Track Has Everyone Speakingsee The Hype Go Viral 2220425 📰 Windows 11 Retail License 9267679 📰 Golf Courses In Dallas Metroplex 4043733 📰 Blackline Stock 8727297 📰 Top Trainers Hidden Weapon Why Gardevoc Dominates Every Arena 3383439 📰 Amplified 4924087 📰 Alarabiya Arabic 2536981 📰 Cast Of Widow Clicquot 2015459 📰 This Simple Line Could Save Your Portfoliosee The Sp 500 Moving Average Surge Now 5185285 📰 We Are Charlie Krik 8678126 📰 Funfetti Cookies That Taste Like Nine Colored Joy Which You Need Today 6734688 📰 The Residence Review 5723880 📰 Thomas Girardi 5317657 📰 Inside The Final Fantasy Secret Lair Greatest Secrets Barlument Players Never Saw 8793280