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 \( \mathbf{u} = \langle 0, 1, 2 \rangle \)?

["Maximizing the Scalar Triple Product: Finding the Maximum Value of ( \left| \mathbf{v} \cdot (\mathbf{w} \ imes \mathbf{u}) \right| )", "In vector calculus, one powerful and elegant expression is the scalar triple product:\n[\n\left| \mathbf{v} \cdot (\mathbf{w} \ imes \mathbf{u}) \right|\n]\nThis quantity represents the volume of the parallelepiped formed by vectors ( \mathbf{v} ), ( \mathbf{w} ), and ( \mathbf{u} ). When ( \mathbf{v} ) is constrained to be a unit vector, maximizing this expression becomes a meaningful optimization problem.", "### Given Vectors\nWe are provided with fixed vectors:\n[\n\mathbf{w} = \langle 1, 0, 1 \rangle, \quad \mathbf{u} = \langle 0, 1, 2 \rangle\n]\nOur goal is to find the maximum value of ( \left| \mathbf{v} \cdot (\mathbf{w} \ imes \mathbf{u}) \right| ) for unit vectors ( \mathbf{v} = \langle v_1, v_2, v_3 \rangle ) such that ( v_1^2 + v_2^2 + v_3^2 = 1 ).", "---", "### Step 1: Compute ( \mathbf{w} \ imes \mathbf{u} )\nThe cross product ( \mathbf{w} \ imes \mathbf{u} ) is a constant vector perpendicular to both ( \mathbf{w} ) and ( \mathbf{u} ). Compute it directly:", "[\n\mathbf{w} \ imes \mathbf{u} = \n\begin{vmatrix}\n\mathbf{i} & \mathbf{j} & \mathbf{k} \\n1 & 0 & 1 \\n0 & 1 & 2 \\n\end{vmatrix}\n= \mathbf{i}(0 \cdot 2 - 1 \cdot 1) - \mathbf{j}(1 \cdot 2 - 1 \cdot 0) + \mathbf{k}(1 \cdot 1 - 0 \cdot 0)\n]\n[\n= \mathbf{i}(-1) - \mathbf{j}(2) + \mathbf{k}(1) = \langle -1, -2, 1 \rangle\n]", "---", "### Step 2: Use the Property of the Scalar Triple Product\nA key identity in vector algebra states:", "[\n\left| \mathbf{v} \cdot (\mathbf{w} \ imes \mathbf{u}) \right| = \left| \mathbf{v} \cdot \mathbf{n} \right|\n]\nfor any constant vector ( \mathbf{n} = \mathbf{w} \ imes \mathbf{u} ), when ( \mathbf{v} ) is a unit vector. This follows from the Cauchy-Schwarz inequality, where the dot product is maximized when ( \mathbf{v} ) points in the direction of ( \mathbf{n} ) (or opposite, but absolute value makes it the same).", "Thus,\n[\n\left| \mathbf{v} \cdot (\mathbf{w} \ imes \mathbf{u}) \right| \leq |\mathbf{w} \ imes \mathbf{u}|\n]\nand equality occurs when\n[\n\mathbf{v} \parallel \mathbf{w} \ imes \mathbf{u}\n]", "Therefore, the maximum value is simply the magnitude of ( \mathbf{w} \ imes \mathbf{u} ).", "---", "### Step 3: Compute ( | \mathbf{w} \ imes \mathbf{u} | )\nWe already found ( \mathbf{w} \ imes \mathbf{u} = \langle -1, -2, 1 \rangle ). Compute its magnitude:", "[\n| \mathbf{w} \ imes \mathbf{u} | = \sqrt{(-1)^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}\n]", "---", "### Conclusion\nThe maximum value of ( \left| \mathbf{v} \cdot (\mathbf{w} \ imes \mathbf{u}) \right| ) for ( |\mathbf{v}| = 1 ) is ( \sqrt{6} ). This occurs when ( \mathbf{v} ) is a unit vector parallel to ( \mathbf{w} \ imes \mathbf{u} = \langle -1, -2, 1 \rangle ).", "---", "### Practical Insight\nThis result is widely applicable in physics, computer graphics, and geometry. For example, in rigid body motion and orientation, maximizing this expression helps determine extremal volumes or detection of maximal perpendicular projections.", "Understanding how to maximize scalar triple products empowers deeper insight into spatial relationships defined by vectors.", "---", "Key Takeaway:\n[\n\boxed{\max \left| \mathbf{v} \cdot (\mathbf{w} \ imes \mathbf{u}) \right| = \sqrt{6} \quad \ ext{when} \quad |\mathbf{v}| = 1, \quad \mathbf{v} \parallel \langle -1, -2, 1 \rangle}\n]"]