1.7: Tangent Planes and Normal Lines

Tangent Planes

Let $z = f(x,y)$ be a function of two variables. We can define a new function $F(x,y,z)$ of three variables by subtracting $z$. This has the condition

$$F(x,y,z) = 0.\nonumber$$

Now consider any curve defined parametrically by

$$x = x(t), \;\;\; y = y(t), \;\;\; z = z(t).\nonumber$$

We can write,

$$F(x(t), y(t), z(t)) = 0.\nonumber$$

Differentiating both sides with respect to $t$, and using the chain rule gives

$$F_x(x, y, z) x' + F_y(x, y, z) y' + F_z(x, y, z) z' = 0\nonumber$$

Notice that this is the dot product of the gradient function and the vector $\langle x',y',z'\rangle$,

$$\nabla F \cdot \langle x', y', z'\rangle = 0.\nonumber$$

In particular the gradient vector is orthogonal to the tangent line of any curve on the surface. This leads to:

Definition: Tangent Plane

Let $F(x,y,z)$ define a surface that is differentiable at a point $(x_0,y_0,z_0)$, then the tangent plane to $F ( x, y, z )$ at $( x_0, y_0, z_0)$ is the plane with normal vector

$$\nabla \, F(x_0,y_0,z_0) \nonumber$$

that passes through the point $(x_0,y_0,z_0)$. In particular, the equation of the tangent plane is

$$\nabla \, F(x_0,y_0,z_0) \cdot \langle x - x_0 , y - y_0 , z - z_0 \rangle = 0. \nonumber$$

Example $\PageIndex{1}$

Find the equation of the tangent plane to

$$z = 3x^2 - xy \nonumber$$

at the point $(1,2,1)$.

Solution

We let

$$F(x,y,z) = 3x^2 - xy - z\nonumber$$

then

$$\nabla F = \langle 6x - y, -x, -1\rangle . \nonumber$$

At the point $(1,2,1)$, the normal vector is

$$\nabla F(1,2,1) = \langle 4, -1, -1\rangle . \nonumber$$

Now use the point normal formula for a plan

$$\langle 4, -1, -1\rangle \cdot \langle x - 1, y - 2, z - 1\rangle = 0\nonumber$$

or

$$4(x - 1) - (y - 2) - (z - 1) = 0.\nonumber$$

Finally we get

$$4x - y - z = 1.\nonumber$$

Normal Lines

Given a vector and a point, there is a unique line parallel to that vector that passes through the point. In the context of surfaces, we have the gradient vector of the surface at a given point. This leads to the following definition.

Definition: Normal Line

Let $F(x,y,z)$ define a surface that is differentiable at a point $(x_0,y_0,z_0)$, then the normal line to $F(x,y,z)$ at $(x_0,y_0,z_0)$ is the line with normal vector

$$\nabla \, F(x_0,y_0,z_0) .\nonumber$$

that passes through the point $(x_0,y_0,z_0)$. In Particular the equation of the normal line is

$$x(t) = x_0 + F_x(x_0,y_0,z_0) t, \nonumber$$

$$y(t) = y_0 + F_y(x_0,y_0,z_0) t, \nonumber$$

$$z(t) = z_0 + F_z(x_0,y_0,z_0) t. \nonumber$$

Example $\PageIndex{2}$

Find the parametric equations for the normal line to

$$x^2yz - y + z - 7 = 0 \nonumber$$

at the point $(1,2,3)$.

Solution

We compute the gradient:

$$\nabla F = \langle 2xyz, x^2z - 1, x^2y + 1\rangle = \langle 12, 2, 3\rangle .\nonumber$$

Now use the formula to find

$$x(t) = 1 + 12t, \;\;\; y(t) = 2 + 2t, \;\;\; z(t) = 3 + 3t.\nonumber$$

The diagram below displays the surface and the normal line.

Angle of Inclination

Given a plane with normal vector n the angle of inclination, $q$ is defined by

$$\cos q = \dfrac{|\textbf{n} \cdot k|}{ ||\textbf{n} ||}. \nonumber$$

More generally, if $F(x,y,z) = 0$ is a surface, then the angle of inclination at the point $(x_0,y_0,z_0)$ is defined by the angle of inclination of the tangent plane at the point with

$$\cos\,q = \dfrac{ | \nabla F(x_0, y_0, z_0) \cdot \textbf{k}| }{|| \nabla F(x_0, y_0, z_0)||}. \nonumber$$

Example $\PageIndex{3}$

Find the angle of inclination of

$$\dfrac{x^2}{4} + \dfrac{y^2}{4} + \dfrac{z^2}{8} = 1\nonumber$$

at the point $(1,1,2)$.

Solution

First compute

$$\nabla F = \langle \dfrac{x}{2}, \dfrac{y}{2}, \dfrac{z}{4}\rangle .\nonumber$$

Now plug in to get

$$\nabla F(1,1,2) = \langle \dfrac{1}{2}, \dfrac{1}{2}, \dfrac{1}{2} \rangle .\nonumber$$

We have

$$|\langle \dfrac{1}{2} , \dfrac{1}{2} , \dfrac{1}{2} \rangle \cdot \hat{\textbf{k}} | = \dfrac{1}{2} .\nonumber$$

Also,

$$||\langle \dfrac{1}{2} , \dfrac{1}{2} , \dfrac{1}{2} \rangle || = \dfrac{\sqrt{3}}{2} .\nonumber$$

Hence

$$\cos q = \dfrac{\frac{1}{2}}{( \frac{\sqrt{3}}{2} )} = \dfrac{1}{\sqrt{3}} . \nonumber$$

So the angle of inclination is

$$q = \cos^{-1}(\dfrac{1}{\sqrt{3}}) = 0.955 \text{ radians} .\nonumber$$

The Tangent Line to a Curve

Example $\PageIndex{4}$

Find the tangent line to the curve of intersection of the sphere

$$x^2 + y^2 + z^2 = 30\nonumber$$

and the paraboloid

$$z = x^2 + y^2\nonumber$$

at the point $(1,2,5)$.

Solution

We find the gradient of the two surfaces at the point

$$\nabla(x^2 + y^2 + z^2) = \langle 2x, 2y, 2z\rangle = \langle 2, 4,10\rangle \nonumber$$

and

$$\nabla (x^2 + y^2 - z) = \langle 2x, 2y, -1\rangle = \langle 2, 4, -1\rangle .\nonumber$$

These two vectors will both be perpendicular to the tangent line to the curve at the point, hence their cross product will be parallel to this tangent line. We compute

$$\begin{vmatrix} \hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\ 2 & 4 & 10 \\ 2 & 4 & -1 \end{vmatrix} = -44 \hat{\textbf{i}} + 22 \hat{\textbf{j}}. \nonumber$$

Hence the equation of the tangent line is

$$x(t) = 1 - 44t y(t) = 2 + 22t z(t) = 5.\nonumber$$