Math Core

Single-variable calculus lives in the plane, but most of the world is three-dimensional: a drone's position, the temperature at a point in a room, the surface of a lens. Everything in this course is built on one simple device, a coordinate system that attaches three numbers to every point in space. This lesson sets up that system and the two basic measurements, distance and midpoint, that you'll use constantly.

Three axes and the right-hand rule

Start with a point $O$ , the origin, and three mutually perpendicular number lines through it: the $x$ -axis, the $y$ -axis and the $z$ -axis. There are two mirror-image ways to arrange them, and mathematicians and physicists agree to use the right-handed one: point the fingers of your right hand along the positive $x$ -axis and curl them toward the positive $y$ -axis; your thumb then points along the positive $z$ -axis.

A point $P$ in space gets the ordered triple $(a, b, c)$ if you can reach it from the origin by moving $a$ units parallel to the $x$ -axis, then $b$ units parallel to the $y$ -axis, then $c$ units parallel to the $z$ -axis. The set of all such triples is written $\mathbb{R}^3$ .

On paper we draw space with an oblique projection: the $y$ - and $z$ -axes are drawn at a right angle, and the $x$ -axis points diagonally down and to the left, "out of the page" toward you. The dashed path below shows how to locate the point $(2, 3, 4)$ .

Locating P(2, 3, 4): move 2 units along x, 3 units parallel to y, then 4 units up. The point (2, 3, 0) directly below P is its projection onto the xy-plane.

Coordinate planes and octants

Each pair of axes spans a coordinate plane:

plane	contains	equation
$xy$ -plane	$x$ - and $y$ -axes	$z = 0$
$yz$ -plane	$y$ - and $z$ -axes	$x = 0$
$xz$ -plane	$x$ - and $z$ -axes	$y = 0$

The three coordinate planes divide space into eight octants. The one where all three coordinates are positive is the first octant; the others are usually described by their sign patterns, such as " $x < 0$ , $y > 0$ , $z < 0$ ."

Dropping a perpendicular from $P(a, b, c)$ to a coordinate plane gives its projection onto that plane: $(a, b, 0)$ on the $xy$ -plane, $(0, b, c)$ on the $yz$ -plane, and $(a, 0, c)$ on the $xz$ -plane.

Common mistake

An equation means different things in different dimensions. In $\mathbb{R}^2$ , $y = 3$ is a line. In $\mathbb{R}^3$ , $y = 3$ is the set of all points $(x, 3, z)$ with $x$ and $z$ free: a plane parallel to the $xz$ -plane. Likewise $x^2 + y^2 = 1$ is a circle in the plane but an infinite circular cylinder in space, because $z$ is unrestricted. Always ask which variables are missing.

Distance and midpoint

To find the distance between $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$ , build a box with $P_1$ and $P_2$ at opposite corners. Its edges have lengths $|x_2 - x_1|$ , $|y_2 - y_1|$ and $|z_2 - z_1|$ . The Pythagorean theorem in the base gives the diagonal of the bottom face, $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ , and applying it once more with the vertical edge gives the space diagonal.

Distance and midpoint in space

The distance between $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$ is

|P_1P_2| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}.

The midpoint of segment $P_1P_2$ is

M = \left( \frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2},\ \frac{z_1 + z_2}{2} \right).

Worked example: Distance and midpoint

Find the distance between $P(1, -2, 3)$ and $Q(4, 2, -9)$ , and the midpoint of $\overline{PQ}$ .

Solution. The coordinate differences are $3$ , $4$ and $-12$ , so

|PQ| = \sqrt{3^2 + 4^2 + (-12)^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13.

The midpoint is $\left( \dfrac{1 + 4}{2}, \dfrac{-2 + 2}{2}, \dfrac{3 + (-9)}{2} \right) = \left( \dfrac52, 0, -3 \right)$ .

Spheres

A sphere is the set of points at a fixed distance $r$ (the radius) from a fixed point $C(h, k, l)$ (the center). Writing $|PC| = r$ with the distance formula and squaring gives its equation.

Definition

Sphere

The sphere with center $(h, k, l)$ and radius $r > 0$ has equation

(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2.

Expanding shows that every sphere has an equation of the form $x^2 + y^2 + z^2 + Gx + Hy + Iz + J = 0$ . Going backward, you complete the square in each variable separately. The result $(x-h)^2 + (y-k)^2 + (z-l)^2 = N$ is a sphere if $N > 0$ , a single point if $N = 0$ , and empty if $N < 0$ .

Worked example: Recognizing a sphere

Show that $x^2 + y^2 + z^2 - 4x + 6y - 2z - 2 = 0$ is a sphere, and find its center and radius.

Solution. Group the terms by variable and complete each square:

\begin{aligned} (x^2 - 4x + 4) + (y^2 + 6y + 9) + (z^2 - 2z + 1) &= 2 + 4 + 9 + 1 \\ (x - 2)^2 + (y + 3)^2 + (z - 1)^2 &= 16. \end{aligned}

Since $16 > 0$ , this is a sphere with center $(2, -3, 1)$ and radius $4$ .

Worked example: Describing a region

Describe the set of points satisfying $1 \le x^2 + y^2 + z^2 \le 4$ and $z \ge 0$ .

Solution. The quantity $\sqrt{x^2 + y^2 + z^2}$ is the distance from $(x, y, z)$ to the origin, so the first condition says that distance is between $1$ and $2$ . That is the solid shell between the spheres of radius $1$ and $2$ centered at the origin. The condition $z \ge 0$ keeps only the part on or above the $xy$ -plane. The region is the upper half of a thick spherical shell, like half of a hollow ball with walls $1$ unit thick.

Tip

The distance from $P(a, b, c)$ to a coordinate plane is just the absolute value of the missing coordinate: the distance to the $xy$ -plane is $|c|$ . The distance to a coordinate axis uses the other two coordinates: the distance to the $z$ -axis is $\sqrt{a^2 + b^2}$ , since the nearest point on that axis is $(0, 0, c)$ .

Practice

Practice 1

Find the distance between $(3, -1, 2)$ and $(-1, 1, 6)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Find the projection of the point $(4, -5, 7)$ onto the $xz$ -plane.

Enter a point like (2, -3)

Practice 3

Find the midpoint of the segment joining $(-3, 4, 1)$ and $(5, -2, 8)$ .

Enter a point like (2, -3)

Practice 4

Find the distance from the point $(2, -6, 5)$ to the $z$ -axis.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Find the center of the sphere $x^2 + y^2 + z^2 + 8x - 2y + 4z + 5 = 0$ .

Enter a point like (2, -3)

Practice 6

A sphere has a diameter with endpoints $A(2, 1, -3)$ and $B(6, -5, 1)$ . Its equation is $(x - 4)^2 + (y + 2)^2 + (z + 1)^2 = r^2$ . Find $r^2$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Which set does the equation $x^2 + z^2 = 9$ describe in $\mathbb{R}^3$ ?

Practice 8

The points $A(3, -2, 1)$ , $B(5, 0, 2)$ and $C(4, -4, 3)$ form a triangle. Which describes it?