Question: A glaciologist models ice flow with vectors $\mathbf{u}$ and $\mathbf{w}$, where $\|\mathbf{u}\| = 5$, $\|\mathbf{w}\| = 7$, and the angle between them is $60^\circ$. Compute the magnitude of $\mathbf{u} \times \mathbf{w}$.

["Title: How Glaciologists Compute Ice Flow with Vectors: A Glaciologist Models Flow Using Cross Products", "In glaciology, understanding the movement of ice is crucial for predicting glacier behavior, ice sheet dynamics, and global sea-level rise. One powerful mathematical tool used by glaciologists is vector analysis—particularly the cross product—to model how ice deforms and flows under stress. When modeling ice flow with vectors (\mathbf{u}) and (\mathbf{w}), one key quantity is the magnitude of their cross product: (|\mathbf{u} \ imes \mathbf{w}|). This measure captures the area of the parallelogram spanned by the two vectors and represents the component of ice motion perpendicular to their directions.", "Given:\n- Magnitude of vector (\mathbf{u}): (|\mathbf{u}| = 5)\n- Magnitude of vector (\mathbf{w}): (|\mathbf{w}| = 7)\n- Angle between (\mathbf{u}) and (\mathbf{w}): (60^\circ)", "The magnitude of the cross product is computed using the formula:\n[\n|\mathbf{u} \ imes \mathbf{w}| = |\mathbf{u}| |\mathbf{w}| \sin\ heta\n]\nSubstituting the given values:\n[\n|\mathbf{u} \ imes \mathbf{w}| = 5 \cdot 7 \cdot \sin(60^\circ)\n]\nSince (\sin(60^\circ) = \frac{\sqrt{3}}{2}),\n[\n|\mathbf{u} \ imes \mathbf{w}| = 35 \cdot \frac{\sqrt{3}}{2} = \frac{35\sqrt{3}}{2}\n]", "Thus, the magnitude of the cross product is (\frac{35\sqrt{3}}{2}), a key physical quantity encoding the transforming potential of ice flow in glaciological models.", "This calculation exemplifies how vector mathematics enables precise analysis of ice dynamics. For glaciologists, such computations are integral to understanding shear convergence, ice deformation, and stress distributions—vital for improving predictions of glacial response to climate change."]