How would one find the equation for the normal line to a 3-dimensional equation at a given point?

Question

I have searched long and hard for this, but all that is discussed are tangent planes to a 3-D equation. I am aware of how to do derivates. If anyone could provide me with an equation, or a pointer in the right direction it would be much appreciated.

Answer 1

A tangent line is characteristic of a curve - which isn't going to be defined by a single equation. Most commonly, a 3-D curve will be defined by parametric equations $(x(t),y(t),z(t))$, and the tangent is then a line in the direction of $(x'(t),y'(t),z'(t))$ - which, again, wouldn't be described as a single equation.

I recommend looking for a calculus textbook's treatment of vector calculus. Before the material with functions of multiple variables and multiple integrals, there should be a chapter dealing with curves and the like. (In Stewart, on a shelf near me, that's chapter 11)

[Added in edit]
For the normal line to a surface, we take the perpendicular to the tangent plane. The normal to a plane $a(x-x_0)+b(y-y_0)+c(z-z_0)$ at the point $(x_0,y_0,z_0)$ has vector equation $(x_0,y_0,z_0)+t(a,b,c)$ - the vector difference of two points on the line is a multiple of the vector of coefficients of the plane's equation. Coordinate-wise, that's $(x_0+ta,y_0+tb,z_0+tc)$. Of course, $t$ is a dummy variable for the parameter here; any convenient name that doesn't conflict with existing variable names can be chosen.