Question: If $ \mathbf{a}, \mathbf{b}, \mathbf{c} $ are unit vectors with $ \mathbf{a} \cdot \mathbf{b} = \frac{1}{2} $, $ \mathbf{b} \cdot \mathbf{c} = \frac{\sqrt{3}}{2} $, find the maximum value of $ \mathbf{a} \cdot \mathbf{c} $.

["Title: Maximize $ \mathbf{a} \cdot \mathbf{c} $: Geometry and Vector Calculation with Unit Vectors", "Meta Description:\nExplore the maximum value of $ \mathbf{a} \cdot \mathbf{c} $ given $ \mathbf{a}, \mathbf{b}, \mathbf{c} $ are unit vectors, $ \mathbf{a} \cdot \mathbf{b} = \frac{1}{2} $, and $ \mathbf{b} \cdot \mathbf{c} = \frac{\sqrt{3}}{2} $. Learn how angles and vector orientation influence this dot product.", "---", "## Understanding the Maximum of $ \mathbf{a} \cdot \mathbf{c} $", "In foundational vector algebra, the dot product $ \mathbf{u} \cdot \mathbf{v} $ between two unit vectors equals the cosine of the angle $ \ heta $ between them:", "$$\n\mathbf{u} \cdot \mathbf{v} = \cos \ heta\n$$", "Thus, maximizing $ \mathbf{a} \cdot \mathbf{c} $ corresponds to minimizing the angle $ \ heta_{ac} $ between vectors $ \mathbf{a} $ and $ \mathbf{c} $.", "Given:\n- $ \mathbf{a} \cdot \mathbf{b} = \frac{1}{2} \Rightarrow \cos \ heta_{ab} = \frac{1}{2} \Rightarrow \ heta_{ab} = 60^\circ $\n- $ \mathbf{b} \cdot \mathbf{c} = \frac{\sqrt{3}}{2} \Rightarrow \cos \ heta_{bc} = \frac{\sqrt{3}}{2} \Rightarrow \ heta_{bc} = 30^\circ $", "These angular constraints fix the relative orientations of unit vectors $ \mathbf{a}, \mathbf{b}, \mathbf{c} $ in space.", "---", "## Geometric Interpretation", "With $ \mathbf{b} $ as a fixed reference vector, vectors $ \mathbf{a} $ and $ \mathbf{c} $ lie on cones around $ \mathbf{b} $:\n- Vector $ \mathbf{a} $ lies on a cone of half-angle $ 60^\circ $ about $ \mathbf{b} $\n- Vector $ \mathbf{c} $ lies on a cone of half-angle $ 30^\circ $ about $ \mathbf{b} $", "The actual orientation of $ \mathbf{a} $ and $ \mathbf{c} $ around $ \mathbf{b} $ determines the minimal possible angle $ \ heta_{ac} $, and thus the maximum value $ \cos \ heta_{ac} = \mathbf{a} \cdot \mathbf{c} $.", "To maximize $ \mathbf{a} \cdot \mathbf{c} $, align $ \mathbf{a} $ and $ \mathbf{c} $ as closely as possible, subject to their angular separation around $ \mathbf{b} $.", "---", "## Optimal Configuration and Maximum Value", "Let’s fix $ \mathbf{b} $ along the positive $ z $-axis in a 3D coordinate system. Because angles depend only on relative orientation, this simplifies calculations.", "Then:\n- $ \mathbf{a} $ makes $ 60^\circ $ with $ \mathbf{b} $: it lies in a cone around the $ z $-axis, such that the azimuthal angle (around $ z $) is arbitrary, but optimized for minimal $ \ heta_{ac} $\n- $ \mathbf{c} $ makes $ 30^\circ $ with $ \mathbf{b} $", "To minimize $ \ heta_{ac} $, align $ \mathbf{a} $ and $ \mathbf{c} $ on the same side of $ \mathbf{b} $ and as close as possible — ideally, after rotating about $ \mathbf{b} $, they lie in the same plane.", "The smallest possible angle between $ \mathbf{a} $ and $ \mathbf{c} $ occurs when both $ \mathbf{a} $ and $ \mathbf{c} $ lie in the same vertical plane (say the $ xz $-plane), symmetric or symmetric-aligned with respect to $ \mathbf{b} $.", "Then:\n$$\n\ heta_{ac} = |\ heta_{ab} - \ heta_{bc}| = |60^\circ - 30^\circ| = 30^\circ\n$$", "This configuration minimizes $ \ heta_{ac} $, so maximizes the dot product:", "$$\n\mathbf{a} \cdot \mathbf{c} = \cos(30^\circ) = \frac{\sqrt{3}}{2}\n$$", "Could alignment be better by shifting $ \mathbf{c} $ to the opposite side? No — since $ \mathbf{a} $ is constrained to only $ 60^\circ $ from $ \mathbf{b} $, and $ \mathbf{c} $ to $ 30^\circ $, moving $ \mathbf{c} $ away increases $ \ heta_{ac} $. The minimum possible $ \ heta_{ac} $ is achieved when both $ \mathbf{a} $ and $ \mathbf{c} $ lie in the same rotational plane around $ \mathbf{b} $ and are aligned to minimize separation.", "Hence:", "$$\n\max(\mathbf{a} \cdot \mathbf{c}) = \cos(30^\circ) = \frac{\sqrt{3}}{2}\n$$", "---", "## Verifying Optimality via Vector Algebra", "Suppose $ \mathbf{a}, \mathbf{b}, \mathbf{c} $ are unit vectors. Express $ \mathbf{a} $ and $ \mathbf{c} $ in terms of components along and perpendicular to $ \mathbf{b} $:", "Let $ \hat{b} = \mathbf{b} $ (unit vector). Let $ \hat{a}\perp $, $ \hat{a}\parallel $ be components perpendicular and parallel to $ \mathbf{b} $, respectively. Similar for $ \mathbf{c} $.", "Then:\n$$\n\mathbf{a} = \mathbf{a}\parallel + \mathbf{a}\perp, \quad \mathbf{c} = \mathbf{c}\parallel + \mathbf{c}\perp\n$$", "Since $ \mathbf{a} \cdot \mathbf{b} = \mathbf{a}\parallel \cdot \hat{b} = |\mathbf{a}\parallel| = \cos 60^\circ = \frac{1}{2} $, so $ |\mathbf{a}\perp| = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \frac{\sqrt{3}}{2} $ — similarly, $ |\mathbf{c}\perp| = \sqrt{1 - \left(\frac{\sqrt{3}}{2}\right)^2} = \frac{1}{2} $.", "Now,\n$$\n\mathbf{a} \cdot \mathbf{c} = (\mathbf{a}\parallel + \mathbf{a}\perp) \cdot (\mathbf{c}\parallel + \mathbf{c}\perp)\n= \mathbf{a}\parallel \cdot \mathbf{c}\parallel + \mathbf{a}\perp \cdot \mathbf{c}\perp + \mathbf{a}\parallel \cdot \mathbf{c}\perp + \mathbf{a}\perp \cdot \mathbf{c}\parallel\n$$", "Dot products of perpendicular components are at most $ |\mathbf{a}\perp| \cdot |\mathbf{c}\perp| = \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = \frac{\sqrt{3}}{4} $. Since $ \mathbf{a}\parallel \parallel \mathbf{c}\parallel $, their dot product is $ |\mathbf{a}\parallel| |\mathbf{c}\parallel| \cos \phi $, maximized when $ \phi = 0 $.", "To maximize $ \mathbf{a} \cdot \mathbf{c} $, align all directions: $ \mathbf{a}\parallel = \mathbf{c}\parallel $, $ \mathbf{a}\perp = \mathbf{c}\perp $ (same perpendicular orientation), and $ \phi = 0 $ between parallels.", "Then:\n$$\n\mathbf{a} \cdot \mathbf{c} = |\mathbf{a}\parallel| |\mathbf{c}\parallel| + \mathbf{a}\perp \cdot \mathbf{c}\perp = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) = \frac{1}{4} + \frac{\sqrt{3}}{4} = \frac{1 + \sqrt{3}}{4}\n$$", "Wait — this contradicts our earlier result. Where is the error?", "Ah — no! This computation assumes $ \mathbf{a}\perp $ and $ \mathbf{c}\perp $ are both aligned with $ \mathbf{a}\parallel $ and $ \mathbf{c}\parallel $ in the same direction — but $ \mathbf{a}\perp \cdot \mathbf{c}\perp $ is maximum when $ \mathbf{a}\perp $ and $ \mathbf{c}\perp $ are parallel, which we assumed.", "But $ \frac{1 + \sqrt{3}}{4} \approx \frac{1 + 1.732}{4} = \frac{2.732}{4} = 0.683 $", "Compare to $ \frac{\sqrt{3}}{2} \approx 0.866 $ — clearly smaller.", "So what went wrong?", "The mistake: $ |\mathbf{a}\perp| = \sqrt{1 - (1/2)^2} = \sqrt{3}/2 \approx 0.866 $, yes.\n$ |\mathbf{c}\perp| = \sqrt{1 - ( \sqrt{3}/2 )^2 } = \sqrt{1 - 3/4} = \sqrt{1/4} = 1/2 $", "So $ \mathbf{a}\perp \cdot \mathbf{c}\perp \leq |\mathbf{a}\perp| |\mathbf{c}\perp| = (\sqrt{3}/2)(1/2) = \sqrt{3}/4 \approx 0.433 $", "And $ \mathbf{a}\parallel \cdot \mathbf{c}\parallel = |\mathbf{a}\parallel| |\mathbf{c}\parallel| = (1/2)(1/2) = 1/4 = 0.25 $", "Total maximum: $ 0.25 + 0.433 = 0.683 $, so $ \frac{1 + \sqrt{3}}{4} \approx 0.683 $", "But earlier geometric reasoning gave $ \cos(30^\circ) = \sqrt{3}/2 \approx 0.866 $ — contradiction.", "So where is the error in the geometric reasoning?", "Ah! The angles $ \ heta_{ab} $ and $ \ heta_{bc} $ are fix, but $ \mathbf{a} $ and $ \mathbf{c} $ need not both be in the same plane.", "But the minimal angle between $ \mathbf{a} $ and $ \mathbf{c} $ occurs when both make minimal deviation from $ \mathbf{b} $ and lie in the same side plane — but the dot product depends on their relative azimuthal angle in that plane.", "Let $ \phi $ be the azimuthal angle between $ \mathbf{a}\perp $ and $ \mathbf{c}\perp $ in the plane perpendicular to $ \mathbf{b} $. Then:", "$$\n\mathbf{a} \cdot \mathbf{c} = \cos(\ heta_{ab} - \ heta_{bc}) + \cos \phi \cdot \sin(\ heta_{ab}) \sin(\ heta_{bc}) + \sin(\ heta_{ab}) \cos(\ heta_{bc}) \cos \phi \quad \ ext{(standard dot product formula)}\n$$", "But simpler: use the identity for dot product in 3D:", "Let $ \mathbf{a} = (\sin \alpha, 0, \cos \alpha) $, $ \alpha = 60^\circ $\nLet $ \mathbf{c} = (\sin \beta \cos \phi, \sin \beta \sin \phi, \cos \beta) $, $ \beta = 30^\circ $", "Then:\n$$\n\mathbf{a} \cdot \mathbf{c} = \sin 60^\circ \cos \phi \cdot \sin 30^\circ + \cos 60^\circ \cdot \sin 30^\circ \cos \beta\n$$", "Wait — full dot product:", "$$\n\mathbf{a} \cdot \mathbf{c} = a_x c_x + a_y c_y + a_z c_z\n$$", "Set $ \mathbf{a} $ in $ xz $-plane: $ \mathbf{a} = (\sin 60^\circ, 0, \cos 60^\circ) = \left( \frac{\sqrt{3}}{2}, 0, \frac{1}{2} \right) $\nSet $ \mathbf{c} $ general in $ 30^\circ $ cone:\n$$\n\mathbf{c} = (\sin 30^\circ \cos \ heta, \sin 30^\circ \sin \ heta, \cos 30^\circ) = \left( \frac{1}{2} \cos \ heta, \frac{1}{2} \sin \ heta, \frac{\sqrt{3}}{2} \right)\n$$", "Then:", "$$\n\mathbf{a} \cdot \mathbf{c} = \left( \frac{\sqrt{3}}{2} \right)\left( \frac{1}{2} \cos \ heta \right) + (0) + \left( \frac{1}{2} \right)\left( \frac{\sqrt{3}}{2} \right) = \frac{\sqrt{3}}{4} \cos \ heta + \frac{\sqrt{3}}{4}\n= \frac{\sqrt{3}}{4} (1 + \cos \ heta)\n$$", "Maximum when $ \cos \ heta = 1 \Rightarrow \ heta = 0 $:", "$$\n\mathbf{a} \cdot \mathbf{c} = \frac{\sqrt{3}}{4} (1 + 1) = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}\n$$", "Ah! The earlier geometric intuition was correct — but only when the azimuthal angles align. The minimum $ \ heta_{ac} $ is $ |\ heta_{ab} - \ heta_{bc}| = 30^\circ $, and dot product is:", "$$\n\cos(\ heta_{ac}) = \cos(30^\circ) = \frac{\sqrt{3}}{2}\n$$", "And it is achievable when $ \ heta = 0 $ in the perpendicular plane.", "Thus, the maximum value is indeed $ \frac{\sqrt{3}}{2} $", "---", "## Final Conclusion", "Given:\n- $ \mathbf{a} \cdot \mathbf{b} = \frac{1}{2} \Rightarrow \ heta_{ab} = 60^\circ $\n- $ \mathbf{b} \cdot \mathbf{c} = \frac{\sqrt{3}}{2} \Rightarrow \ heta_{bc} = 30^\circ $", "These define angular constraints. The dot product $ \mathbf{a} \cdot \mathbf{c} = \cos \ heta_{ac} $, where $ \ heta_{ac} $ is the angle between $ \mathbf{a} $ and $ \mathbf{c} $.", "By aligning $ \mathbf{a} $ and $ \mathbf{c} $ in the same plane perpendicular to $ \mathbf{b} $ and making their relative azimuthal angle zero, the minimal $ \ heta_{ac} = |60^\circ - 30^\circ| = 30^\circ $. Thus:", "$$\n\max(\mathbf{a} \cdot \mathbf{c}) = \cos(30^\circ) = \frac{\sqrt{3}}{2}\n$$", "This maximum is attainable under the given conditions.", "---", "Related Topics:\n- Vector algebra\n- Dot product and angles\n- Optimization of dot products under geometric constraints\n- Use of spherical coordinates in vector geometry", "Keywords:\nunit vectors, dot product, $ \mathbf{a} \cdot \mathbf{b} $, $ \mathbf{b} \cdot \mathbf{c} $, maximum value, $ 60^\circ $, $ 30^\circ $, vector geometry, minimum angle between vectors", "---", "For further reading on vector optimization, see “Linear Algebra and Its Applications” by Gilbert Strang."]