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Home » Set \( f'(x) = 0 \) to find critical points: \( 3x(x - 2) = 0 \). - Anadea

Set \( f'(x) = 0 \) to find critical points: \( 3x(x - 2) = 0 \). - Anadea

📅 May 15, 2026 👤 scraface

May 15, 2026

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📌 \|\mathbf{w} \times \mathbf{u}\| = \sqrt{(-1)^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}

📌 Thus, the maximum value of \( |\mathbf{v} \cdot (\mathbf{w} \times \mathbf{u})| \) is \( \sqrt{6} \), achieved when \( \mathbf{v} \) is in the direction of \( \mathbf{w} \times \mathbf{u} \). Since the dot product achieves its maximum absolute value when \( \mathbf{v} \) is aligned with \( \mathbf{w} \times \mathbf{u} \), and \( \|\mathbf{v}\| = 1 \), the maximum is \( \sqrt{6} \).

📌 But the original prompt said "find the largest possible value of", so we compute:

📌 \max_{\|\mathbf{v}\|=1} \left| \mathbf{v} \cdot (\mathbf{w} \times \mathbf{u}) \right| = \|\mathbf{w} \times \mathbf{u}\| = \sqrt{6}

📌 \boxed{\sqrt{6}}

📌 But to match the format and intent—clickbait style with SEO optimization and deep mathematical insight:

📌 Question: What is the maximum value of \( \left| \mathbf{v} \cdot (\mathbf{w} \times \mathbf{u}) \right| \) for a unit vector \( \mathbf{v} \), given \( \mathbf{w} = \langle 1, 0, 1 \rangle \) and \( \mathbf{u} = \langle 0, 1, 2 \rangle \)?

📌 Solution: The expression \( \left| \mathbf{v} \cdot (\mathbf{w} \times \mathbf{u}) \right| \) depends on the magnitude of the binormal vector. Compute \( \mathbf{w} \times \mathbf{u} = \left\langle -1, -2, 1 \right\rangle \), so

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