# Equation of Normal Line

## Homework Statement

Find the equation of the normal line of function f(x,y) = x^2+y^2-1 through points (2,1,4)

## Homework Equations

(x-xo)/f_x = (y-yo)/f_y = (z-zo)/f_z

## The Attempt at a Solution

so...
f_x = 2x = 4
f_y = 2y = 2
f_z = ???

I don't know how to solve for f_z....
here is random stuff I was doing that *might* be related?
f(x,y) - z = 0
f(x,y) = z = 4

yeah.. any math guru wanna show this grasshopper the way?

If z = f(x,y) then z -1 = ##x^2 + y^2##. What kind of surface is this? Based on elementary geometry you should know what the normal line looks like. And if you get that far, you can solve this without any calculus at all. I recommend you do so.

That is probably not what your teacher had in mind. However, there is nothing like knowing the answer to help you solve the problem. (I'm not actually joking here -- if you have no idea what the answer should look like you are not ready to tackle a solution).

Given that I've told you how to express z, and provided a hint about what this line should look like, can you proceed now?

HallsofIvy
You don't "solve for $f_z$". Your "f" is NOT a function of z!
$z= f(x,y)= x^2+ y^2- 1$ is the same as $x^2+ y^2- z= 1$. We can think of that as a "level surface" for function $g(x, y, z)= x^2+ y^2- z$. Now, g is a function of x, y, and z with $g_x= 2x$, $g_y= 2y[/tex], and [itex]g_z= -1$. So $\nabla g= 2x\vec{i}+ 2y\vec{j}- \vec{k}$ is perpendicular to that surface at every (x, y, z). In particular, at (2, 1, 4) this is $2\vec{i}+ 4\vec{j}- \vec{k}$.
So the problem of "find the normal line" reduces to "find parametric equations of the line through (2, 1, 4) with direction given by $2\vec{i}+ 4\vec{j}- \vec{k}$". Can you do that?