I'm studying multivariable calculus at the moment. I recently worked with a problem where one had to find the vector that describes the tangential direction $$\vec{v}$$ for a specific point of intersection $$\vec{a}$$ between two scalar valued functions, let's call them $$f$$ and $$g$$ respectively.

What I used in order to solve the problem was to evaluate the following: $$\vec{v}=(\nabla f \times \nabla g)(\vec{a})$$.

I'd pretty new to the concept of gradients as well. But I do understand that the gradient is a normal vector to the level surface for a specific point in space. However, I don't really understand why the tangential vector has to be normal to both functions at the point of intersection. I assume this might be a question involving linear algebra more than multivariable calculus itself. Still, I don't just want to take this idea for granted, and I really want to understand as to why this is the case.

I'd be glad for any ideas that can help me deepen my understanding of this.

• I think you should deepen your understanding of cross-product in $\mathbb{R}^3$. Commented Jan 21, 2022 at 14:52
• @Tanamas Be careful with your terminology. It doesn't make sense to talk about "the intersection of two functions"; I suspect what you're asking about is the intersection of the level surfaces of two functions. Similarly, it doesn't make sense to say that the tangential vector is "normal to both functions"; I'm actually not sure what you have in mind here. Commented Jan 21, 2022 at 15:03
• @BenGrossmann Thank you for your comment. Why is there a distinction between intersection of two functions and level surfaces, if I may ask? In my eyes, when I think of intersection of two functions, I see that as an intersection in the R^2 plane, but maybe that's not the way I should be viewing it. Commented Jan 21, 2022 at 15:17
• A function is not a graph. For example, the function $f(x) = x^2$ is not the same thing as the graph $y = x^2$, the function $f$ is a rule that takes an input $x$ to its output $x^2$. Note that the function $g(x,y) = x^2 - y$ is also not the same thing as the level curve $g(x,y) = 0$. Notice that $y = f(x)$ and $g(x,y) = 0$ are equations using these functions that describe same curve, but neither of these functions is the curve. Commented Jan 21, 2022 at 15:22

As you said, $$\nabla f(\vec a)$$ and $$\nabla g(\vec a)$$ are the normal vectors to the level sets of the functions $$f$$ and $$g$$ that pass through $$\vec a$$.
In order for a vector to be parallel to the level set of $$f$$, it needs to be orthogonal to $$\nabla f$$. Similarly, a vector is parallel to the level set of $$g$$ when it is orthogonal to $$\nabla g$$. A vector that is parallel to the intersection of these level sets is parallel to both of these surfaces, and is therefore orthogonal to $$\nabla f$$ and orthogonal to $$\nabla g$$, which is why it is parallel to the cross product $$\nabla f \times \nabla g$$.