I need help with this problem:
Let $F:\mathbb{R}^3$\ $\{\mathbf0\}\rightarrow\mathbb{R}$ be defined by $$F(\mathbf{r})=\frac{1}{r}, \quad \mathbf{r}\neq\mathbf0$$ where $\mathbf{r}=(x,y,z)$ and $r=\Vert\mathbf{r}\Vert=\sqrt{(x^2+y^2+x^2)}$. Prove that $$(grad\ F)(\mathbf{r})=-\frac{1}{r^3}\mathbf{r}$$ Show that $($grad $F)(a,b,c)$ is normal to the sphere $r^2=a^2+b^2+c^2$ at the point $(a,b,c)$ and pints in the direction of increasing values of F. Find the equation of the tangent plane to the sphere $(a,b,c)$.
I've already proved that $(grad\ F)(\mathbf{r})=-\frac{1}{r^3}\mathbf{r}$, but I don't know hot to show that the $(grad\ F)(a,b,c)$ is normal to the sphere. Should I use dot product? Also how do I find the equation of the tangent plane? Hope you can help me.
There are two facts about the gradient in this problem that you have to use. The first one, is that the gradient vector $\nabla F$ always points in the direction of maximum increase of $F$. The second one, is that the gradient vector is normal to the level surface of $F(x,y,z)$ at a point $(x,y,z)$. There are really good proofs here to the bottom of the page, under each blue note.
So, we have already solved half of the problem. Now to show that $\nabla F$ is perpendicular to the sphere $r^2 = a^2 + b^2 + c^2$ at a point $(a,b,c)$, we have to show that the sphere is the level surface of $F$ at the point $(a,b,c)$. The level surface of $F$ at a point $(a,b,c)$ is the surface given by the equation $F(x,y,z) = k$ where $k = F(a,b,c)$. So we have
$$ F(x,y,z) = F(a,b,c) \\ \frac{1}{\sqrt{x^2 + y^2 + z^2}} = \frac{1}{\sqrt{a^2 + b^2 + c^2}} \\ \sqrt{r^2} = \sqrt{a^2 + b^2 + c^2} \\ r^2 = a^2 + b^2 + c^2$$
So the sphere $r^2 = a^2 + b^2 + c^2$ (which is a sphere centered at the origin that has the point $(a,b,c)$ on it's surface) is indeed the level surface of $F$ at the point $(a,b,c)$ and so $\nabla F(a,b,c)$ is perpendicular to it.
To find the equation of the tangent plane to the sphere at the point $(a,b,c)$, you need a normal vector to the sphere in that point. We already have it: it's the gradient. The equation of a tangent plane to a surface at a point $P$ is given by:
$$ \vec n \cdot \left(\vec r - \vec P\right) = 0 $$ where $\vec P$ is the position vector of the point $P$ and $\vec n$ is the normal vector. So in our case, the equation for the plane is:
$$ \nabla F \cdot \left(\left\langle x, y, z \right\rangle - \left\langle a, b, c \right\rangle\right) = 0 $$ and then just carry the calculations on.