# Why gradient vector is perpendicular to the plane

I know what gradient vector or $\nabla F$ is and I know how to prove that it is orthogonal to the surface (using calculation - not intuitive).
In a particular case, in which we have a three variable function, I want to know why the gradient vector is perpendicular as mentioned. I mean, not in theoretical terms, but intuitive.

• math.harvard.edu/archive/21a_spring_09/PDF/… – dato datuashvili Dec 9 '13 at 8:10
• math.stackexchange.com/questions/122222/… – dato datuashvili Dec 9 '13 at 8:11
• I'm not sure what you count as intuitive. Do you find it intuitive that direction of greatest rate of change of $F$ should be perpendicular to the level sets of $F$ (i.e. the direction in which there is no change)? If so, then $\nabla F$ points in the direction of the greatest rate of change and the surface is defined by $F=C$ a level set. – Matt Dec 9 '13 at 8:12
• @Matt thanks, by intuitive, I mean why $∇F$ is perpendicular to the plane which is tangent to the surface F. two other comments above, give a theoretical proof on what I want to know, which I knew in advance. I am looking for a visual (or intuitive) method to make the concept near to my mind. thanks. – Sida Dec 9 '13 at 8:17
• See this question and may answer: math.stackexchange.com/questions/401845/… – MathOverview Mar 27 '15 at 14:06

Let $F:\mathbb{R}^3 \to \mathbb{R}$ be a function of three variables. Say we are looking at the surface defined by $F(x,y,z) = 0$. By definition, the total derivative of $F$ at $p \in \mathbb{R}^3$ is the best linear approximation to $F$ near $p = (a,b,c)$, in other words

$$dF\big|_p (\Delta x,\Delta y,\Delta z)= \dfrac{\partial F}{\partial x} \Delta x + \dfrac{\partial F}{\partial y} \Delta y +\dfrac{\partial F}{\partial z} \Delta z$$

and

$$F(a+\Delta x,b +\Delta y,c + \Delta z) \approx F(p)+dF\big|_p (\Delta x,\Delta y,\Delta z)$$

This is really what the multivariable derivative is all about. Now we can notice that

$$dF\big|_p (\Delta x,\Delta y,\Delta z)= \langle\dfrac{\partial F}{\partial x}, \dfrac{\partial F}{\partial y},\dfrac{\partial F}{\partial z}\rangle \cdot \langle \Delta x,\Delta y,\Delta z\rangle$$

Now defining $\nabla F = \langle\dfrac{\partial F}{\partial x}, \dfrac{\partial F}{\partial y},\dfrac{\partial F}{\partial z}\rangle$, we see that

$$dF\big|_p (\textbf{v})= \nabla F\cdot \textbf{v}$$

This one formula packages a lot of mathematics.

Hopefully this gives you some more intuition about what the gradient does. In words, to see how much $F(\textbf{p})$ changes when you move away a little bit to $\textbf{p}+\textbf{v}$, just dot product $\nabla F\big|_p$ with $\textbf{v}$.

Armed with this intuitive understanding of the gradient, we can see why it must be perpendicular to the level curves of $F$ quite intuitively.

If $p$ is a point of the surface $F(x,y,z) = 0$, then the tangent vectors $\textbf{v}$ to the surface must satisfy $dF\big|_p(v) = 0$, because moving in the direction of the surface should not change the value of $F$ much since the value of $F$ is constant on the surface. Translating this into a gradient statement we see that $\nabla F\big|_p \cdot \textbf{v} = 0$ for each tangent vector to the surface.

This just says that $\nabla F\big|_p$ is perpendicular to the tangent plane to the surface at $p$!

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Brak}{\left\langle #1\right\rangle}$Arguably this is not a question of calculus, but of linear algebra. Let $x = (x_{1}, \dots, x_{n})$ denote Cartesian coordinates on $\Reals^{n}$. A function $\phi:\Reals^{n} \to \Reals$ is linear if and only if there exists a vector $a = (a_{1}, \dots, a_{n})$ such that $$\phi(x) = a_{1} x_{1} + \dots + a_{n} x_{n} = \sum_{i=1}^{n} a_{i} x_{i} = \Brak{a, x}.$$ If $a \neq 0$, then $\phi(x) = 0$ if and only if $\Brak{a, x} = 0$, if and only if $x$ is orthogonal to $a$. That is, $\phi$ decomposes $\Reals^{n}$ into the orthogonal direct sum of the line spanned by $a$ and the kernel of $\phi$ (a.k.a. the hyperplane orthogonal to $a$).

This orthogonality is "why" the gradient is orthogonal to the level set.

Steven Gubkin's excellent answer contains a proper explanation. In the spirit of geometric intuition: Let $F:\Reals^{n} \to \Reals$ be continuously-differentiable, $p$ a point with $\nabla F(p) \neq 0$, and $\Sigma$ the level set of $F$ through $p$.

"Zooming in" on $\Reals^{n}$ at the point $p$ causes $F$ to "look more and more like" its first-order approximation $$L_{p}(x) = F(p) + \Brak{\nabla F(p), x - p},$$ and causes $\Sigma$ to "look more and more like" its tangent space at $p$, a.k.a., the level set of $L_{p}$ through $p$, a.k.a., the kernel of $\phi(x) = \Brak{\nabla F(p), x}$ translated by $p$. Since $\nabla F(p)$ is orthogonal to $\ker \phi$, "the gradient is orthogonal to the level set".

(On a tangent, as it were, squinting at this picture "explains" the implicit function theorem for real-valued functions: If $\nabla F(p) \neq 0$, the level set of the first-order approximation of $F$ at $p$ is an affine hyperplane, which "must be tangent" to the level set of $F$ through $p$. That is, the level set of $F$ at $p$ must be a manifold of dimension $(n - 1)$ in some neighborhood of $p$. The implicit function theorem for mappings $F:\Reals^{n} \to \Reals^{m}$ (with $m \leq n$) has a similar interpretation: If $Df(p)$ has rank $m$, then the level set $\Sigma$ of $F$ through $p$ "looks like" the level set of the first-order approximation $L_{p}(x) = F(p) + Df(p)(x - p)$, an affine space of dimension $(n - m)$, which "must be tangent" to $\Sigma$ at $p$; that is, $\Sigma$ "must be an $(n - m)$-manifold near $p$".)

I've been thinking on this for awhile now (well, I first learned of this fact about 15 years ago, so I'd say awhile), and I now had a bit of a thought. So, first thing's first, we're discussing a tangent plane, so we're talking about a surface, let's just call it $z = f(x,y)$. In this case, our Function will be $F(x,y,z) = f(x,y) - z = 0$, and our gradient will look something like $\nabla F = (\partial f/ \partial x, \partial f / \partial y, -1)$. So what does the tangent plane look like? Say we are at a point $p_0 = (x_0, y_0, f(x_0,y_0))$. Given the definition of a derivative, any vector that lies on our tangent plane will start from $p_0$, then go out in some direction. Say we increase $x$ by $t$ and leave $y$ constant, how much does $z$ change?... $t\cdot\partial f/\partial x$ of course (if this is not clear, go meditate on the definition of the [partial] derivative for awhile). So this vector will be in the direction $(t, 0, t\cdot \partial f/\partial x)$. Similarly, increasing y, we go off in a direction $(0, t, t\cdot \partial f/\partial y)$. It seems that both of these guys are orthogonal to $\nabla F$, so far so good. These two vectors certainly span a two dimensional space, so they take care of our whole tangent plane, and our gradient really is orthogonal to it. If you don't like the form $z = f(x,y)$, then consult the implicit function theorem and just rotate the space a little bit for those pesky places where $\nabla F$ vanishes. I don't know if this is the answer you need, but this is the explanation I'm satisfied with. Happy to answer any follow up questions, or receive any criticism.

• very good explanation. for someone new to this, i want to point out that ∇F is orthogonal to "both the guys" since the dot product is clearly zero. – dayum Sep 7 '18 at 4:35

An intuitive explanation, could be as follows.

Take a function of two variables $z=f(x,y)$.
Plot two level curves $z=z_0$ and $z=z_0+\Delta z$.
Take a point on the first line $(x_0,y_0,z_0)$, and consider the tangent to the line in that point.
Suppose that $f(x,y)$ is "smooth" enough and that $\Delta z$ is enough "small" so that the tangent will not vary "abruptly" from one line to the other: that means that it will remain constant (apart from a $O(\Delta z)$) along the normal to the two curves.
That given, the path along the normal will be shortest one to pass from $z_0$ to $z_0+\Delta z$. Therefore $\Delta z / \Delta s$ will be the highest.

If then you have instead three variables, and thus level surfaces, the "visual representation" above does not change in substance.

• Apart from a what? Not sure what $O/\Delta z$ is. – Joseph Garvin Feb 3 '18 at 1:25
• @JosephGarvin: sorry, it was a typo : corrected. Thanks for signalling it. – G Cab Feb 3 '18 at 15:35
• Ah, thanks for correcting, but I don't know what $O(\Delta z)$ is either ;) Is this big-O notation, from e.g. algorithms? – Joseph Garvin Feb 3 '18 at 22:33
• @JosephGarvin, yes, it's big-O, to tell that the difference between the two tangents is $t_o+t_1\Delta z+\cdots$ – G Cab Feb 3 '18 at 22:45

I am using 3 dimensions in my proof, but it easily extends into higher dimensions.

Proof

First, we need to define the tangent plane to a specific point $$(x, y, z)$$. Suppose our surface satisfies the equation $$f(x) + f(y) + f(z) = k$$ for a constant c. Note that $$f(x+\delta_1) + f(y+\delta_2)+f(z+\delta_3)$$ approaches $$c$$ for small enough deltas. But, $$f(x+\delta_1) = f(x) + \delta_1 f'(x)$$ Therefore, the tangent plane is defined as $$[f(x) + \delta_1 f'(x)] + [g(y) + \delta_2 g'(y)] + [h(z) + \delta_3 h'(z)] = k$$ Or, $$f'(x)a + g'(y)b + h'(z)c = 0$$ I have replaced the $$\delta$$s with $$(a, b, c)$$ to show that the plane extends to infinity (as before, the $$\delta$$s tended towards $$0$$). It is well known that the normal vector to this plane is $$\langle f'(x),\ g'(y),\ h'(z)\rangle$$

Proof Suppose $$(a, b, c)$$ satisfies $$k_1a + k_2b + k_3c = 0.$$ Note that $$\langle k_1, k_2, k_3\rangle \cdot \langle a, b, c\rangle = k_1a + k_2b + k_3c = 0$$ Hence they are orthogonal. $$\blacksquare$$

Simply put: The gradient at a point p(a,b) is the greatest rate of change of z(a,b). The level curve has constant value z. Therefor for these two "lines" to satisfy there definition, they must be perpendicular to one another.

• But why do they have to be perpendicular? I don't see any immediate reason why this should be so. Why can't there be a scenario where we actually get greater increase by going at, say, $60$ degree angle to the constant value surface? – Wojowu Jun 10 '16 at 7:42