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Home » However, in such problems, we compute: - Aurero

However, in such problems, we compute: - Aurero

📅 March 14, 2026 👤 scraface

Mar 14, 2026

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📌 5Question: Let \( \mathbf{v} \) be a vector in \( \mathbb{R}^3 \) such that \( \|\mathbf{v}\| = 1 \) and \( \mathbf{v} \cdot (\mathbf{w} \times \mathbf{u}) = \frac{1}{2} \), where \( \mathbf{w} = \langle 1, 0, 1 \rangle \), \( \mathbf{u} = \langle 0, 1, 2 \rangle \). Find the maximum possible value of \( \|\mathbf{v}\| \) under the constraint—wait, correction: \( \|\mathbf{v}\| \) is fixed at 1, so instead reinterpret: find the maximum of \( \|\mathbf{v}\|^2 \) given the dot product condition, but since \( \|\mathbf{v}\| = 1 \), we instead seek consistent interpretation.

📌 Wait—reformulate properly.

📌 Corrected interpretation: Find the maximum value of \( k \) such that \( \mathbf{v} \cdot (\mathbf{w} \times \mathbf{u}) = \frac{1}{2} \) is possible for a unit vector \( \mathbf{v} \), or equivalently find the maximum efficiency of such a dot product under normalization. But since \( \|\mathbf{v}\| \) is constrained to 1, the equation defines a constraint; perhaps instead ask: find the maximum possible value of \( \left| \mathbf{v} \cdot (\mathbf{w} \times \mathbf{u}) \right| \) over all unit vectors \( \mathbf{v} \), which is always 1 via Cauchy-Schwarz. But that’s trivial.

📌 Alternate meaningful version: Given fixed vectors \( \mathbf{w} = \langle 1, 0, 1 \rangle \), \( \mathbf{u} = \langle 0, 1, 2 \rangle \), and a unit vector \( \mathbf{v} \), find the maximum value of \( \left| \mathbf{v} \cdot (\mathbf{w} \times \mathbf{u}) \right| \), which measures projection onto the binormal vector.

📌 And since \( \|\mathbf{v}\| = 1 \), the maximum of the absolute value is \( \|\mathbf{w} \times \mathbf{u}\| \).

📌 \mathbf{w} \times \mathbf{u} = \begin{vmatrix}

📌 & 0 & 1 \\

📌 & 1 & 2 \\

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