# From 3D -normal vector to $N$ -dimensional normal -vector?

I am studying Griffiths "Introduction to electrodynamism", page 7 and chat here. I want to understand the normal vector in more general situations and also here in 3D. What is wrong below? Have I misunderstood the $\nabla$ or the plane -equation or something else?

More about n -dimensional normal -vector in Wikipedia here.

I want to fast calculate and verify the normal vector with the 3D -object and then understand the case with n -dimensional object. I think the general case is with $\nabla$ so I created the implicit form for the surface i.e. $\bar F= \hat i + 2 \hat j + 3\hat k$ so $\nabla \bar F=(1,2,3)$ but manually I got $\hat n = \frac{1}{7}(6,3,2)$. Something wrong, what? Not pointing in the same direction.

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...with slightly different configuration, I got $\hat n=\frac{1}{7}(6,3,-2)$ but wanting to do this kind of things faster, still nothing like the $\nabla \bar F$. Is the implicit form of the plane right? Yes it should be, not seeing it yet. – hhh Sep 6 '12 at 14:36
What is this manual method you are using? – Karolis Juodelė Sep 6 '12 at 14:44

If a hypersurface in $\mathbb{R}^n$ is implicitly defined by the equation $F(x_1, \dots, x_n) = d$ for some $d \in \mathbb{R}$, then you can consider it as a level curve of the function $F$. As you mention, $\nabla F$ is normal to the surface, so, provided you can determine $F$, you can find the normal vector at a point on the surface.

In your situation, the surface is defined by an equation $F(x, y, z) = c$. Furthermore, it is a plane, so $F(x, y, z) = ax + by + cz$, where $a, b, c \in \mathbb{R}$ are yet to be determined. From the figure, we know that $(1, 0, 0)$, $(0, 2, 0)$ and $(0, 0, 3)$ are on the surface, and therefore must satisfy $F(x, y, z) = d$. Substituting in each point, one by one, we get:

\begin{align*} F(1, 0, 0) &= d \Rightarrow a(1) + b(0) + c(0) = d \Rightarrow a = d,\\ F(0, 2, 0) &= d \Rightarrow a(0) + b(2) + c(0) = d \Rightarrow 2b = d \Rightarrow b = \dfrac{1}{2}d,\\ F(0, 0, 3) &= d \Rightarrow a(0) + b(0) + c(3) = d \Rightarrow 3c = d \Rightarrow c = \dfrac{1}{3}d. \end{align*}

Now replacing $a, b,$ and $c$ in the equation for $F$, we obtain:

$$dx + \frac{1}{2}dy + \frac{1}{3}dz = d.$$

If $d = 0$, then every element $(x, y, z) \in \mathbb{R}^3$ would satisfy the equation and hence be on the surface. As this is not the case, $d \neq 0$ so we can divide both sides of the above equation by $d$, leaving us with:

$$x + \frac{1}{2}y + \frac{1}{3}z = 1.$$

If you like, you can multiply both sides by $6$ so that all the coefficients are integers, in which case the equation is $6x + 3y + 2z = 6$. Therefore $F(x, y, z) = 6x + 3y + 2z$, so $\nabla F = (\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}) = (6, 3, 2)$ is a vector normal to the surface. As $\|\nabla F\| = \sqrt{6^2 + 3^2 + 2^2} = \sqrt{49} = 7$, $\frac{1}{\|\nabla F\|}\nabla F = \frac{1}{7}(6, 3, 2)$ is a unit normal to the surface.

Note, as this surface is a plane in $\mathbb{R}^3$, you can obtain the normal in a quicker way.

If we can find two linearly independent vectors in the plane, then their cross product will be normal to the plane. We can use the points given to find two such vectors. As $(1, 0, 0)$ and $(0, 2, 0)$ are in the plane, the vector $(0, 2, 0) - (1, 0, 0) = (-1, 2, 0)$ from $(1, 0, 0)$ to $(0, 2, 0)$ is in the plane. Likewise $(0, 3, 0) - (1, 0, 0) = (-1, 0, 3)$ is also in the plane. Clearly these two vectors are linearly independent. Therefore:

$$(-1, 2, 0) \times (-1, 0, 3) = \left| \begin{array}{ccc} i & j & k \\ -1 & 2 & 0 \\ -1 & 0 & 3 \end{array} \right| = 6i + 3j + 2k = (6, 3, 2),$$

as before.

As for your concerns in higher dimensions, the approach I outlined is the standard approach, except that you usually know $F(x_1, \dots, x_n)$ and $d$, so all that you need to do is calculate $\nabla F = (\frac{\partial F}{\partial x_1}, \dots, \frac{\partial F}{\partial x_n})$. Most of the time $\nabla F$ will not be a constant vector as its components can be functions - in fact, the only time $\nabla F$ is constant is when your surface is a hyperplane in $\mathbb{R}^n$. Either way, $\nabla F$ is a vector which (can) vary from point to point; this is what we call a vector field. More formally, a vector field on a hypersurface $\Sigma$ is a function $\Sigma \to \mathbb{R}^n$; $\nabla F$ is such a function.

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For a level surface, $F(x,y,z)=0$, the gradient $\nabla F$ gives a formula for a normal to the surface at each point.

The surface you have graphed above looks to be a plane which passes through the points: $(1,0,0)$, $(0,2,0)$, and $(0,0,3)$. An equation for this plane is $6x+3y+2z-6=0$. So if we let $F(x,y,z)=6x+3y+2z-6$ then $\nabla F=6\hat{i}+3\hat{j}+2\hat{k}$ (the components are just the coefficients of $x,y,z$). Since $\nabla F$ is constant, this plane has the same normal everywhere (in fact this happens because it is a plane).

Of course, any non-zero multiple is also a normal so any vector $\hat{n} = 6t\hat{i}+3t\hat{j}+2t\hat{k}$ where $t \not=0$ is also normal to the plane. If we normalize these vectors we are left with two choices: $\pm\frac{1}{7}\left(6\hat{i}+3\hat{j}+2\hat{k}\right)$. These are the upward and downward unit normals for this plane sometimes called "orientations" for the plane.

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