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

February 22, 2026

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

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& 0 & 1 \\
& 1 & 2 \\
\end{vmatrix} = \mathbf{i}(0 \cdot 2 - 1 \cdot 1) - \mathbf{j}(1 \cdot 2 - 1 \cdot 0) + \mathbf{k}(1 \cdot 1 - 0 \cdot 0) = \langle -1, -2, 1 \rangle
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}

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