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Home » $ S(6,3) = 90 $ - Tacotoon

$ S(6,3) = 90 $ - Tacotoon

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 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:

📌 \|\mathbf{u}_\perp\|^2 = \|\mathbf{u}\|^2 - (\mathbf{u} \cdot \mathbf{v})^2 = 4 - 1 = 3.

📌 Let \(\mathbf{u}_\perp = \mathbf{u} - \frac{1}{9} \mathbf{v}\) (since \(\mathbf{v}\) has norm 3), so:

📌 \|\mathbf{u}_\perp\|^2 = \|\mathbf{u}\|^2 - (\mathbf{u} \cdot \mathbf{v})^2 / \|\mathbf{v}\|^2 \cdot

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