PlanetPhysics/Position Vector

In the space $$\mathbb{R}^3$$, the vector $$\vec{r} \;:=\; (x,\,y,\,z) \;=\; x\vec{i}+y\vec{j}+z\vec{k}$$ directed from the origin to a point \,$$(x,\,y,\,z)$$\, is the position vector of this point.\, When the point is variable, $$\vec{r}$$ represents a vector field and its length $$r \;:=\; \sqrt{x^2+y^2+z^2}$$ a scalar field.

The simple formulae

are valid, where $$\vec{r}^0$$ is the unit vector having the direction of $$\vec{r}$$.
 * $$\nabla\!\cdot\vec{r} = 3$$
 * $$\nabla\!\times\!\vec{r} = \vec{0}$$
 * $$\nabla r = \frac{\vec{r}}{r} = \vec{r}^0$$
 * $$\nabla\frac{1}{r} = -\frac{\vec{r}}{r^3} = -\frac{\vec{r}^0}{r^2}$$
 * $$\nabla^2\frac{1}{r} = 0$$

If\, $$\vec$$\, is a constant vector,\, $$\vec{U}\!\!:\mathbb{R}^3\to\mathbb{R}^3$$\, a vector function and\, $$f\!\!:\mathbb{R}\to\mathbb{R}$$\, is a twice differentiable function, then the formulae

hold.
 * $$\nabla(\vec\cdot\!\vec{r}) = \vec$$
 * $$\nabla\cdot(\vec\times\vec{r}) = 0$$
 * $$(\vec{U}\!\cdot\!\nabla)\vec{r} = \vec{U}$$
 * $$(\vec{U}\!\times\!\nabla)\!\cdot\!\vec{r} = 0$$
 * $$(\vec{U}\!\times\!\nabla)\!\times\!\vec{r} = -2\vec{U}$$
 * $$\nabla f(r) = f'(r)\,\vec{r}^0$$
 * $$\nabla^2f(r) = f''(r)\!+\frac{2}{r}f'(r)$$