The cross product $a \times b$ is defined as a vector $c$ that is perpendicular (orthogonal) to both $a$ and $b,$ with a direction given by the right-hand rule and a magnitude equal to the area of the parallelogram that the vectors span.
I have trouble understanding why $\|\vec{r}_u \,\Delta u \times \vec{r}_v \, \Delta v\| = \|\vec{r}_u \times \vec{r}_v \| \, \Delta u \, \Delta v $ where the $\text{“}\|\text{''}$ denotes vector norm and $\times$ is the cross product. I'm not understanding how vector norms preserve closure under scalar multiplication.
How do we prove that $\|\Delta u \, \vec{r}_u \times \Delta v \,\vec{r}_v \| = \|\vec{r}_u \times \vec{r}_v\| \, \Delta u \, \Delta v$?
I think $ r_u, r_v$ are vectors $ \in \mathbb{R^3 }$ and $\Delta u, \Delta v \in \mathbb{R}$?
Is $\Delta u \cdot \Delta v \in \mathbb{R}$ and the product of two real numbers?
For context, I saw the equality $\|\vec{r}_u \, \Delta u \times \vec{r}_v \, \Delta v\| = \|\vec{r}_u \times \vec{r}_v \| \, \Delta u \, \Delta v$ below for calculating surface integrals and the derivation of the volume element:
Yes $\Delta u$ and $\Delta v$ are scalar quantities.
Indeed by definition:
$$\vec a \times \vec b = (a_yb_z-a_zb_y)\hat e_1 + (a_zb_x-a_xb_z)\hat e_2 + (a_xb_y-a_yb_x)\hat e_3$$
therefore:
$$s\vec a \times t\vec b = st(a_yb_z-a_zb_y)\hat e_1 + st(a_zb_x-a_xb_z)\hat e_2 + st(a_xb_y-a_yb_x)\hat e_3=st(\vec a \times \vec b)$$