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Home » 36^12 ≈ 3.56 × 10^21 → 3× = 1.07 × 10^22 - Baxtercollege

36^12 ≈ 3.56 × 10^21 → 3× = 1.07 × 10^22 - Baxtercollege

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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📌 \mathbf{u} \cdot \mathbf{v} = 1,\quad \mathbf{v} \cdot \mathbf{w} = 6.

📌 Assume all vectors lie in a 3D subspace. Express \(\mathbf{u}\) and \(\mathbf{w}\) in terms of an orthonormal basis adapted to the constraints. Use geometric insight: the maximum \(\mathbf{u} \cdot \mathbf{w}\) occurs when \(\mathbf{u}\) and \(\mathbf{w}\) are aligned as much as possible, consistent with intermediate alignments to \(\mathbf{v}\).

📌 Let us apply the inequality:

📌 but refine using vector decomposition. Decompose \(\mathbf{w}\) into components parallel and perpendicular to \(\mathbf{v}\):

📌 \mathbf{w} = \mathbf{w}_\parallel + \mathbf{w}_\perp, \quad \mathbf{w}_\parallel = \frac{\mathbf{v} \cdot \mathbf{w}}{\|\mathbf{v}\|^2} \mathbf{v} = \frac{6}{9} \mathbf{v} = \frac{2}{3} \mathbf{v}.

📌 \|\mathbf{w}_\perp\|^2 = \|\mathbf{w}\|^2 - \|\mathbf{w}_\parallel\|^2 = 16 - \left(\frac{2}{3}\right)^2 \cdot 9 = 16 - 4 = 12.

📌 Now, \(\mathbf{u} \cdot \mathbf{w} = \mathbf{u} \cdot \mathbf{w}_\parallel + \mathbf{u} \cdot \mathbf{w}_\perp = \mathbf{u} \cdot \left(\frac{2}{3} \mathbf{v}\right) + \mathbf{u} \cdot \mathbf{w}_\perp = \frac{2}{3} (\mathbf{u} \cdot \mathbf{v}) + \mathbf{u} \cdot \mathbf{w}_\perp = \frac{2}{3}(1) + \mathbf{u} \cdot \mathbf{w}_\perp = \frac{2}{3} + \mathbf{u} \cdot \mathbf{w}_\perp.

📌 Since \(\|\mathbf{u} - \mathbf{u} \cdot \mathbf{v} \mathbf{v}\| \geq 0\), the projection of \(\mathbf{u}\) onto the plane perpendicular to \(\mathbf{v}\) satisfies:

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