# Thread: Vector Analysis

1. ## Vector Analysis

Let $f(x,y,z) = x^2 + 2y^2 + 3z^2$ and let S be the isotimic surface: $f = 1$. Find all the points $(x,y,z)$ on S that have tangent planes with normals $(1,1,1)$.

I'm not even sure where to begin...

2. We may use the fact that the normal to any point on the level curve of $f$ is parallel to $\nabla f$ at that point. We know this because

$\frac{d}{dt}f(\mathbf{r}(t))=\nabla f(\mathbf{r}(t))\cdot \mathbf{r'}(t)=0$

for a differentiable curve $\mathbf{r}$ on the surface $f(x,y,z)=c$, which means that $\nabla f$ is perpendicular to the curve and thus to the plane itself.

In our case, we are to find all those points of $f(x,y,z)=1$ such that

$\nabla f=\left$

for some constant $c$.