Walkthrough

Distance to a plane follows its normal

A plane's normal vector points in the one direction perpendicular to the plane. That makes it the right direction for measuring the shortest distance from a point to the plane.

Choose any point $Q$ on the plane and let $P$ be the point outside it. The vector $\overrightarrow{QP}$ may be slanted, so keep only its scalar component along a nonzero normal vector $\mathbf n$:

For $P=(1,2,1)$ and the plane $x+y+z=1$, take $Q=(1,0,0)$ and $\mathbf n=\langle1,1,1\rangle$. Then $\overrightarrow{QP}=\langle0,2,1\rangle$, so the distance is $3/\sqrt3=\sqrt3$.

An equation can describe a whole shape

An equation in two variables usually leaves one degree of freedom. Its solutions form a curve in the plane. For example,

collects every point at distance $1$ from the origin, so it describes the unit circle.

An equation in three variables usually leaves two degrees of freedom. Its solutions form a surface in space. The equation

describes the unit sphere.

These are both level sets. A level set is the collection of inputs where a function has one fixed value. Thus $f(x,y)=k$ usually describes a level curve in $\mathbb R^2$, while $f(x,y,z)=k$ usually describes a level surface in $\mathbb R^3$.

Read a surface one slice at a time

A complicated three-dimensional equation becomes easier to picture when one coordinate is held constant. Each resulting cross-section is called a trace.

For a surface in $(x,y,z)$-space:

• an $x$-trace sets $x=a$ and lies in the plane $x=a$;
• a $y$-trace sets $y=b$ and lies in the plane $y=b$;
• a $z$-trace sets $z=c$ and lies in the horizontal plane $z=c$.

The fixed coordinate tells you both what to substitute and where the resulting curve lives. After finding several traces, fit the families together into one surface.

Two trace patterns worth recognizing

For the cone

horizontal traces are circles:

Their radius is $|c|$, so the circles widen at a constant rate above and below the origin. Vertical traces are pairs of lines at the center and hyperbolas away from it. Together, these slices force a double cone.

For the surface

fixing $x$ produces upward-opening parabolas, while fixing $y$ produces downward-opening parabolas. The surface therefore rises in one direction and falls in the perpendicular direction. This opposite curvature creates a hyperbolic paraboloid, usually called a saddle.

Distance from a point to a plane

The shortest route from a point to a plane is perpendicular to the plane. If $P$ has position vector $\mathbf p$, choose any point $Q$ on the plane with position vector $\mathbf q$. For a nonzero normal vector $\mathbf n$, the displacement from the plane to $P$ is

$$\overrightarrow{QP}=\mathbf p-\mathbf q.$$

Its signed component in the normal direction is

$$\operatorname{comp}_{\mathbf n}\overrightarrow{QP} =\frac{\overrightarrow{QP}\cdot\mathbf n}{\|\mathbf n\|}.$$

Distance cannot be negative, so take the absolute value:

$$\boxed{ d(P,\text{plane}) =\frac{|\overrightarrow{QP}\cdot\mathbf n|}{\|\mathbf n\|} }.$$

The choice of $Q$ does not matter. The vector between any two points in the plane is orthogonal to $\mathbf n$, so changing $Q$ adds nothing to the dot product.

For a plane in the form

$$ax+by+cz=d$$

and a point $P=(x_1,y_1,z_1)$, use the normal vector $\mathbf n=\langle a,b,c\rangle$. The projection formula becomes

$$\boxed{ d(P,\text{plane}) =\frac{|ax_1+by_1+cz_1-d|}{\sqrt{a^2+b^2+c^2}} }.$$

Example

Find the distance from

$$P=(1,2,1)$$

to the plane

$$x+y+z=1.$$

The plane has normal vector

$$\mathbf n=\langle1,1,1\rangle.$$

Choose $Q=(1,0,0)$ on the plane. Then

$$\overrightarrow{QP} =\langle1,2,1\rangle-\langle1,0,0\rangle =\langle0,2,1\rangle.$$

Now calculate the normal component:

$$\begin{aligned} \overrightarrow{QP}\cdot\mathbf n &=\langle0,2,1\rangle\cdot\langle1,1,1\rangle =3,\\ \|\mathbf n\|&=\sqrt{1^2+1^2+1^2}=\sqrt3. \end{aligned}$$

Therefore,

$$\boxed{ d(P,\text{plane}) =\frac{3}{\sqrt3}=\sqrt3 }.$$

The coordinate formula gives the same result directly:

$$\frac{|1+2+1-1|}{\sqrt{1^2+1^2+1^2}} =\sqrt3.$$

Level curves and level surfaces

Let $k$ be a constant. For a function of two variables, the equation

$$f(x,y)=k$$

selects all points in the plane where $f$ has the value $k$. This set is called a level curve when it is a curve.

An ordinary graph can be written this way too. If $y=g(x)$, define

$$f(x,y)=y-g(x).$$

Then

$$f(x,y)=0 \quad\Longleftrightarrow\quad y=g(x).$$

So a familiar graph is the zero level curve of a function of two variables.

For a function of three variables, the equation

$$f(x,y,z)=k$$

selects points in space. When the solution set has two dimensions, it is a level surface. For example,

$$x^2+y^2+z^2=1$$

is the sphere of radius $1$ centered at $(0,0,0)$.

The words “usually a curve” and “usually a surface” matter. A level set can collapse to a point, split into pieces, or even be empty. Its actual shape comes from the equation.

Sketching with traces

A trace is the intersection of a surface with a coordinate-parallel plane. To calculate one, hold a coordinate fixed:

$$x=a,\qquad y=b,\qquad\text{or}\qquad z=c.$$

This gives three families:

A reliable sketching process is:

1. Find the traces in the three coordinate planes by setting $x=0$, $y=0$, and $z=0$.
2. Try nearby values such as $\pm1$ and $\pm2$ to see how each family changes.
3. Identify each two-dimensional curve and record the plane it belongs to.
4. Use symmetry and the movement of the vertices or radii to assemble the surface.

Do not draw every trace in the same plane. For example, $x=1$ and $x=-1$ may give identical equations in $y$ and $z$, but they lie in two different parallel planes in space.

The double cone

Consider

$$\boxed{x^2+y^2-z^2=0}.$$

Equivalently,

$$x^2+y^2=z^2.$$

Horizontal traces

Set $z=c$. The trace equation is

$$\boxed{z=c\quad\Longrightarrow\quad x^2+y^2=c^2}.$$

This is a circle in the plane $z=c$, centered on the $z$-axis, with radius $|c|$.

• At $z=0$, the equation $x^2+y^2=0$ gives only the origin.
• At $z=\pm1$, the traces are circles of radius $1$.
• At $z=\pm2$, the traces are circles of radius $2$.

The same circle appears at heights $c$ and $-c$, so the surface is symmetric across the $xy$-plane. Because the radius grows in direct proportion to $|z|$, the sides are straight rather than curved outward.

Vertical traces

To examine the vertical cross-sections, set $y=c$. Then

$$x^2+c^2-z^2=0,$$

which can be rearranged as

$$\boxed{y=c\quad\Longrightarrow\quad z^2-x^2=c^2}.$$

At $c=0$, this factors:

$$z^2-x^2=(z-x)(z+x)=0,$$

so the central trace is the pair of lines

$$z=\pm x.$$

For $c\ne0$, the trace is a hyperbola opening in the positive and negative $z$-directions. For instance,

$$y=\pm1 \quad\Longrightarrow\quad z=\pm\sqrt{x^2+1},$$

and

$$y=\pm2 \quad\Longrightarrow\quad z=\pm\sqrt{x^2+4}.$$

The $x$-traces are identical after swapping $x$ and $y$:

$$x=c \quad\Longrightarrow\quad z^2-y^2=c^2.$$

These circular and hyperbolic slices fit together into a double circular cone around the $z$-axis.

Traces assemble the double cone

The hyperbolic paraboloid

Now consider

$$\boxed{z=y^2-x^2}.$$

The signs are the key. The $y^2$ term raises the surface as $|y|$ grows, while the $-x^2$ term lowers it as $|x|$ grows.

Horizontal traces

Set $z=c$. Then

$$\boxed{z=c\quad\Longrightarrow\quad y^2-x^2=c}.$$

The sign of $c$ decides how the hyperbola opens:

• If $c>0$, the hyperbola opens in the positive and negative $y$-directions.

• If $c=0$, the equation factors as

  $$y^2-x^2=(y-x)(y+x)=0,$$

  giving the two lines $y=\pm x$.

• If $c<0$, rewrite the equation as $x^2-y^2=-c$; the hyperbola opens in the positive and negative $x$-directions.

For example,

$$\begin{aligned} z=1&:\quad y^2-x^2=1,\\ z=2&:\quad y^2-x^2=2,\\ z=-1&:\quad x^2-y^2=1,\\ z=-2&:\quad x^2-y^2=2. \end{aligned}$$

The two crossing lines at $z=0$ are the boundary between the two opening directions.

Fixed-$x$ traces

Set $x=a$. The remaining variables satisfy

$$\boxed{x=a\quad\Longrightarrow\quad z=y^2-a^2}.$$

Each trace is an upward-opening parabola in the plane $x=a$. Its vertex is

$$(a,0,-a^2).$$

The first few traces are

$$\begin{aligned} x=0&:\quad z=y^2,\\ x=\pm1&:\quad z=y^2-1,\\ x=\pm2&:\quad z=y^2-4. \end{aligned}$$

As $|a|$ grows, the vertex moves downward.

Fixed-$y$ traces

Set $y=b$. This gives

$$\boxed{y=b\quad\Longrightarrow\quad z=b^2-x^2}.$$

Each trace is a downward-opening parabola in the plane $y=b$, with vertex

$$(0,b,b^2).$$

For example,

$$\begin{aligned} y=0&:\quad z=-x^2,\\ y=\pm1&:\quad z=1-x^2,\\ y=\pm2&:\quad z=4-x^2. \end{aligned}$$

As $|b|$ grows, the vertex moves upward.

Opposite parabolic traces create the saddle

The origin is the saddle point. Along the $y$-axis, where $x=0$, the surface has $z=y^2\ge0$. Along the $x$-axis, where $y=0$, it has $z=-x^2\le0$. Every neighborhood of the origin contains points above and below it, so the origin is neither a local maximum nor a local minimum.

Quadric surfaces

A quadric surface is a level surface described by a polynomial of degree at most $2$ in $x$, $y$, and $z$:

$$P(x,y,z)=k, \qquad \deg P\le2.$$

Both surfaces in this note are quadrics:

$$x^2+y^2-z^2=0$$

is a double cone, while

$$y^2-x^2-z=0$$

is the hyperbolic paraboloid $z=y^2-x^2$.

Recognizing a standard name is useful, but traces explain why the shape has that name. They also keep working when a surface has been shifted, stretched, or oriented in a less familiar way.