Question

Show that $\mathbf{\nabla }\mathbf{.}\left(U\times r\right)\mathbf{=}\mathbf{r}\mathbf{.}\left(\nabla \times U\right)$where U is a vector function of $\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}$and$\mathbf{r}\mathbf{=}\mathbf{xi}\mathbf{+}\mathbf{yj}\mathbf{+}\mathbf{zk}$ .

Short answer

It has been proved that $\nabla .(\mathrm{U}\times \mathrm{r})=\mathrm{r}.(\nabla \times \mathrm{U})$

Step-by-step solution

Given Information.

It has been given that U is a vector function of x,y and z and $\mathrm{r}=\mathrm{xi}+\mathrm{yj}+\mathrm{zk}$

Definition of vector.

A quantity that has magnitude as well as direction is called a vector. It is typically denoted by an arrow in which the head determines the direction of the vector and the length determines it magnitude.

Use the formula to evaluate the expression.

Use the identity mentioned below.

$\nabla .\left(\mathrm{U}\times \mathrm{V}\right)=\mathrm{V}.\left(\nabla \times \mathrm{U}\right)-\mathrm{U}.\left(\nabla \times \mathrm{V}\right)$

Now let$V\to r,$

$\nabla .\left(U\times r\right)=r.\left(\nabla \times U\right)-U.\left(\nabla \times r\right)$

But r is a radial vector field, so it doesn't curl at all, hence its curl should vanish.

$\nabla .\left(U\times r\right)=r.\left(\nabla \times U\right)$