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In my problem set I have the following kind of exercises:

I am given a normal vector and a surface $F(x,y,z)=0$ and $N=(a,b,c)$ the normal vector.

I am asked to determine the points where the vector is normal to the surface.

First I tried to make $\bigtriangledown F(x,y,z)=N$ and to solve for $x,y,z$. However this was wrong.

After hours trying I tried to make the gradient linearly dependant (parallel) to the Normal vector, that is $\bigtriangledown F(x,y,z)=kN$ and solve for $x,y,z$ which all were a function of $k$ (4 variables and 3 equations sistem) and putting these values into the surface $F(x,y,z)=0$ (The fourth equation) and solving for $k$ and substituting it the points and this done the trick.

My question is: Why did this last procedure work? Because I think that making the gradient parallel to the normal vector given it is not what i was asked (Finding the point in the surface where that vector was normal to)

mmendina
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  • Well, what is your definition of “normal to a surface?” What determined whether or not a vector is normal to a surface? The gradient of $F$ is only one of an infinite number of possible normals at a point. – amd Apr 29 '20 at 20:21

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\begin{equation} F(x, y, z)=0 \end{equation} is a level curve of the function \begin{equation} F(x, y, z) \end{equation} and when you find the general statement for gradient of the level curve it is equivalent to the local maximum change direction of the function at level curve so, when you consider the relation between normal vector and gradient vector they are collinear because both are perpendicular to the linearly approximated neighborhood of point (consider it as a infinitesimally small plane). Apparently, normal vector and gradient vector doesn't necessarily need to be equal because gradient vector depends on the change of the function and spatial but normal vector is only spatial.

asd.123
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  • Sorry I didn't understand what you are saying, can you put it in simpler terms? Why should the gradient and the normal vector be collinear? – mmendina Apr 29 '20 at 12:58
  • @MartinMendina https://math.stackexchange.com/q/947604/613151 has the answer for your question – asd.123 Apr 29 '20 at 13:05
  • @MartinMendina you may also refer to https://mathoverflow.net/q/1977 for your last question. – asd.123 Apr 29 '20 at 13:26
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    Thank you for the comments, I realized that if they're linearly dependent so one must be a multiplous of the other, therefore they're aligned in the same straight line. – mmendina Apr 30 '20 at 00:20