finding the minimum or maximum value subject to two constraints

Question

This is from my textbook

I have two questions:

1. it says that "since C lies on both constraint surfaces, $∇g(x_0, y_0,z_0)$ and $∇h(x_0, y_0,z_0)$ are both orthogonal to $C$ at $(x_0, y_0,z_0)$". From my understanding, $∇g(x_0, y_0,z_0)$ and $∇h(x_0, y_0,z_0)$ are both orthogonal to its level surface, because they are orthogonal to its level surface, so they are also orthogonal to the curve that is in the level surface at $(x_0, y_0,z_0)$, is my understanding correct?

2. According to the picture,$∇g$ and $∇h$ are not even orthogonal to their level surfaces? And I don't understand why a curve can have two different orthogonal vector at one point?

Answer 1

1. Your understanding seems correct.
2. The picture is misleading. You are correct in that the gradient of a level surface should be orthogonal to the tangent plane of the surface at the point where we are calculating the gradient. As for your second concern, Suppose your curve $C$ has direction vector $(3,4,5)$ at a point $p$ on it, any vector $(x,y,z)$ satisfying $3x+4y+5z=0$ is an orthogonal vector to $C$ at $p$. So there is a while plane's worth of orthogonal vectors. A good rule of thumb is that if you are an $k$ dimensional object in $n$ dimensional space, you will generically have an $n-k$ dimensional subspace of orthogonal vectors. In our case, a curve has dimension $1$ and it lives in say $\mathbb R^3$, so there is a $2$ dimensional space (aka plane) worth of orthogonal vectors at most points on the curve.