Given a unit vector $\hat{u}$ and a vector $\vec{v}$ perpendicular to $\hat{u}$, we can rotate $\vec{v}$ by 90 degrees around $\hat{u}$ with the cross product $\hat{u} \times \vec{v}$. Since the cross product is linear in its first argument this can be rewritten as:
$$\hat{u} \times \vec{v} = (u_x\hat{x} + u_y\hat{y} + u_z\hat{z}) \times \vec{v} = u_x(\hat{x} \times \vec{v}) + u_y(\hat{y} \times \vec{v}) + u_z(\hat{z} \times \vec{v})$$
The geometric interpretation is: project $\vec{v}$ onto the 3 coordinate planes (YZ, ZX, and XY), then rotate each projection 90 degrees in its respective plane, then weight these rotated projections by the components of $\hat{u}$, then sum them all up to get the rotated vector $\vec{v}$.
What I'm looking for is some sort of intuition or geometric picture illustrating why this works. I know it works because I can show it with the algebra but I'd like a deeper understanding of this if possible.
Best Answer
Note that $ \mathbf{\hat{u}} $ and $ \mathbf{v} $ necessarily lie in a plane, so it's possible to decompose $ \mathbf{v} $ into components that lie strictly in that plane. Suppose $ \{ \mathbf{\hat{f}}_1, \mathbf{\hat{f}}_2 \} $ is a basis for that plane, and further simplify things by setting $ \mathbf{\hat{f}}_1 = \mathbf{\hat{u}} $. Then we have $$\mathbf{\hat{u}} = \mathbf{\hat{f}}_1$$ $$\mathbf{v} = v \mathbf{\hat{f}}_1 \cos\theta + v \mathbf{\hat{f}}_2 \sin\theta,$$ where $ v = \left\lVert {\mathbf{v}} \right\rVert $, and $ \mathbf{\hat{f}}_1 \cdot \mathbf{\hat{f}}_2 = 0 $.
With that change of basis made, you now have $$\mathbf{\hat{u}} \times \mathbf{v}= v \mathbf{\hat{f}}_1 \times \left( { \mathbf{\hat{f}}_1 \cos\theta + \mathbf{\hat{f}}_2 \sin\theta } \right)$$ $$= \left( { \mathbf{\hat{f}}_1 \times \mathbf{\hat{f}}_2 } \right) v \sin\theta.$$ This is a vector is in a direction perpendicular to the plane that $ \mathbf{\hat{u}} $ and $ \mathbf{v} $ lie in, and has the magnitude of the parallelogram formed by $ \mathbf{\hat{u}} $ and $ \mathbf{v} $.
You can now identify the coordinate expressions in your original question. First let's write $ \mathbf{\hat{w}} = \mathbf{\hat{f}}_1 \times \mathbf{\hat{f}}_2 $ for the direction of that vector perpendicular to the $ \mathbf{\hat{u}}, \mathbf{v} $ plane, so $$u_x (\mathbf{\hat{x}} \times \mathbf{\hat{v}}) +u_y (\mathbf{\hat{y}} \times \mathbf{\hat{v}}) +u_z (\mathbf{\hat{z}} \times \mathbf{\hat{v}})= \sin\theta \mathbf{\hat{w}}.$$ Since $ \sin\theta $ is the area of the paralellogram formed by $ \mathbf{\hat{u}}, \mathbf{\hat{v}} $, let's write this as $$\textrm{Area}(\mathbf{\hat{u}}, \mathbf{\hat{v}})=u_x (\mathbf{\hat{x}} \times \mathbf{\hat{v}}) \cdot \mathbf{\hat{w}} +u_y (\mathbf{\hat{y}} \times \mathbf{\hat{v}}) \cdot \mathbf{\hat{w}} +u_z (\mathbf{\hat{z}} \times \mathbf{\hat{v}}) \cdot \mathbf{\hat{w}},$$ or $$\textrm{Area}(\mathbf{\hat{u}}, \mathbf{\hat{v}})=u_x \begin{vmatrix} \mathbf{\hat{x}} & \mathbf{\hat{v}} & \mathbf{\hat{w}} \end{vmatrix} +u_y \begin{vmatrix} \mathbf{\hat{y}} & \mathbf{\hat{v}} & \mathbf{\hat{w}} \end{vmatrix} +u_z \begin{vmatrix} \mathbf{\hat{z}} & \mathbf{\hat{v}} & \mathbf{\hat{w}} \end{vmatrix}$$ Each of these triple product determinants represents the volume of the parallelopiped respectively formed by $ \mathbf{\hat{x}}, \mathbf{\hat{y}}, \mathbf{\hat{z}} $ on one edge, and $ \mathbf{\hat{v}}, \mathbf{\hat{w}} $ on the other edges.
We see that the cross product, which can be thought of as an parallelogram area computing machine, can also be recast as a parallopiped volume computing machine, if one of the vectors making extending the parallogram into a volume is a unit vector perpendicular to the plane containing the original two vectors. That is $$\textrm{Area}(\mathbf{\hat{u}}, \mathbf{\hat{v}}) = \textrm{Volume}(\mathbf{\hat{u}}, \mathbf{\hat{v}}, \mathbf{\hat{w}}) = \begin{vmatrix} \mathbf{\hat{u}} & \mathbf{\hat{v}} & \mathbf{\hat{w}} \end{vmatrix}.$$ or $$\textrm{Volume}(\mathbf{\hat{u}}, \mathbf{\hat{v}}, \mathbf{\hat{w}})= u_x \textrm{Volume}(\mathbf{\hat{x}}, \mathbf{\hat{v}}, \mathbf{\hat{w}})+ u_y \textrm{Volume}(\mathbf{\hat{y}}, \mathbf{\hat{v}}, \mathbf{\hat{w}})+ u_z \textrm{Volume}(\mathbf{\hat{z}}, \mathbf{\hat{v}}, \mathbf{\hat{w}}).$$ This volume function $ \textrm{Volume}(\mathbf{\hat{u}}, \mathbf{\hat{v}}, \mathbf{\hat{w}}) $ is multilinear. Of note here is that it is linear in the $ \mathbf{\hat{u}} $ parameter, so we see that the factors $ u_x, u_y, u_z $ represent the weightings of the total volume of this extended space with respect to the volume computation with respect to each of the $ \mathbf{\hat{x}}, \mathbf{\hat{y}}, \mathbf{\hat{z}} $ directions instead of $ \mathbf{\hat{u}} $.
Aside:
It doesn't actually matter for conceptualizing the geometry of the situtation, but if you want to compute $ \mathbf{\hat{f}}_2 $, or $ \theta $, you can do so by writing $$\mathbf{v} = (\mathbf{v} \cdot \mathbf{\hat{u}}) \mathbf{\hat{u}} + \mathbf{\hat{u}} \times \left( { \mathbf{v} \times \mathbf{\hat{u}} } \right),$$ so $$\cos\theta = \mathbf{\hat{v}} \cdot \mathbf{\hat{u}}$$ $$\mathbf{\hat{f}}_2 \sin\theta = \mathbf{\hat{u}} \times \left( { \mathbf{\hat{v}} \times \mathbf{\hat{u}} } \right).$$