# Vector analysis: understanding formulas for normal and tangent

I'm studying a chapter on vector analysis and I don't really understand why the following equations are switched for the two situations. (My course is not in English, so I apologize if I translated something wrong.)

Equations for the tangent plane and normal in a point on a surface

1. The equation for a tangent plane in $p(x_0, y_0, z_0)$ on a surface $\varphi(x,y,z) = 0$:

$$\left(\frac{\partial \varphi}{\partial x}\right)_p (x-x_0) + \left(\frac{\partial \varphi}{\partial y}\right)_p (y-y_0) + \left(\frac{\partial \varphi}{\partial z}\right)_p (z-z_0) = 0$$

1. The equation for the normal in $p(x_0, y_0, z_0)$ on a surface $\varphi(x,y,z) = 0$:

$$\frac{x-x_0}{\left(\frac{\partial \varphi}{\partial y}\right)_p} = \frac{y-y_0}{\left(\frac{\partial \varphi}{\partial y}\right)_p} = \frac{z-z_0}{\left(\frac{\partial \varphi}{\partial z}\right)_p}$$

Equations for the tangent and normal plane in a point on a curve

For $p(t=t_0) = \{f_1(t_0), f_2(t_0), f_3(t_0)\}$ with $t \in R$

1. The equation for the tangent in $p$:

$$\frac{x-x_0}{f'_1(t_0)} = \frac{y-y_0}{f'_2(t_0)} = \frac{z-z_0}{f'_3(t_0)}$$

1. The equation for the normal plane in $p$:

$$f'_1(t_0) (x-x_0) + f'_2(t_0)(y-y_0) + f'_3(t_0)(z-z_0) = 0$$

I tried to find a graphical explanation for this to understand how the equations were derived, but without much success.

The equation

$$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$$

describes an affine plane passing through the point $(x_0,y_0,z_0)$ whose normal vector is $(a,b,c)$ (so $(a,b,c)$ should be non-zero). The equations

$$\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$$

describe an affine line passing through $(x_0, y_0, z_0)$ whose direction vector is $(a,b,c)$ (this representation is possible only if $a,b$ and $c$ are non-zero).

In the first case, your surface $S$ is described as a level set of a single function $\varphi \colon \mathbb{R}^3 \rightarrow \mathbb{R}$ (so $S = \{ (x,y,z) \, | \, \varphi(x,y,z) = 0\}$). If $p = (x_0,y_0,z_0)$ is a point on the surface, then the gradient $\nabla \varphi|_p$ is perpendicular to the tangent plane of $S$ at $p$. Hence, the gradient is a normal vector for the tangent plane at $p$ and you get the resulting equation for the tangent plane. Similarly, the normal line is a line passing through $(x_0,y_0,z_0)$ in the direction of a normal to the tangent plane (which is $\nabla \varphi|_p$), hence the resulting equation for the normal line.

In the second case, the curve $\gamma$ is described as the image of a parametrization $\mathbf{f} \colon \mathbb{R} \rightarrow \mathbb{R}^3$ (so $\gamma = \mathbf{f}(\mathbb{R})$). The tangent line to $\gamma$ at $\mathbf{f} (t_0)$ is the line passing through $p = (f_1(t_0), f_2(t_0), f_3(t_0))$ in the direction of the velocity vector $\mathbf{f}'(t_0)$, hence the resulting equation for the tangent line. The normal plane to $\gamma$ at $p$ is the plane passing through $p$ whose normal vector is the velocity vector $\mathbf{f}'(t_0)$ (because the normal plane should be perpendicular to the velocity vector), hence the resulting equation for the plane.

• Thanks. Helped me a lot. An additional question though. Is it possible to find the equation for a normal line instead of a plan in the second situation and is it a similar equation? – Aaron Dec 28 '15 at 19:03
• Since the curve $\gamma$ is in $\mathbb{R}^3$, there isn't a notion of a "normal line" because there are infinitely many lines that are perpendicular to the tangent line to $\gamma$ but only one perpendicular plane. If the curve would be in $\mathbb{R}^2$, then you would have a single normal line. – levap Dec 28 '15 at 19:05

I would recommend reading/watching this Khan Academy article. In there, they discuss tangent planes, derive their equations, and have quite a bit of visual output.

This is also a good place to learn the Mathematics behind the equations. It is hard to just look at a graph and see where the equations are coming from. You want a combination of theory and graphs.