Topic 1 / 17basic

3D Space, Distance, Surfaces & Regions

Plot points in 3D using the right-hand rule, compute distance and midpoint, recognize spheres, and describe surfaces and regions with equations and inequalities.

Calculus 3 begins by extending coordinates from $(x, y)$ in the plane to $(x, y, z)$ in space. Three perpendicular axes meet at the origin. The right-hand rule fixes a consistent orientation, so distance, midpoint, sphere equations, and trace-based sketching all follow naturally.

Plotting in 3D and the right-hand rule

A point in 3D is written

(x, y, z)

. To plot, start at the origin

(0, 0, 0)

, move

x

along the

x

-axis, then

y

parallel to the

y

-axis, then

z

parallel to the

z

-axis. Negative coordinates move opposite. The **right-hand rule** orients the axes: point your right hand's fingers from the positive

x

-axis curling toward the positive

y

-axis; your thumb then points along the positive

z

-axis.

Key takeaways

$(x, y, z)$ uses three real numbers, one per axis.
Right-hand rule: $x \to y$ curl, thumb is $+ z$ .
Negative coordinates move the opposite direction along that axis.

Strategy tips

Check signs carefully — a missing minus sign on any coordinate puts the point in the wrong octant.
When sketching, draw axes lightly first, then plot $(x, y, 0)$ in the $x y$ -plane and rise/fall by $z$ .

Distance and midpoint formulas

For points

P (x_{1}, y_{1}, z_{1})

and

Q (x_{2}, y_{2}, z_{2})

, the distance is

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}

and the midpoint is

M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2}, \frac{z _{1} + z _{2}}{2})

. Distance comes from applying Pythagoras twice: once in the

x y

-plane, once vertically.

Key takeaways

Distance: subtract corresponding coordinates, square, add, take square root.
Midpoint: average each coordinate separately.
Both formulas reduce cleanly to the 2D versions when $z_{1} = z_{2} = 0$ .

Equation of a sphere

A sphere with center

(a, b, c)

and radius

r

satisfies

(x - a)^{2} + (y - b)^{2} + (z - c)^{2} = r^{2}

. The equation uses

r^{2}

, not

r

. To find the radius from a point on the sphere and a center, plug both into the distance formula. To go from a general second-degree equation to standard form, **complete the square** in each variable.

Key takeaways

Equation form: $(x - a)^{2} + (y - b)^{2} + (z - c)^{2} = r^{2}$ .
Right side is $r^{2}$ , not $r$ .
Use distance formula to find $r$ from a known point.

Describing surfaces and regions

Equations like

x = c

y = c

, or

z = c

describe planes parallel to a coordinate plane. A missing variable means the surface extends along that direction — for instance

x^{2} + y^{2} = r^{2}

in 3D is a circular **cylinder** along the

z

-axis. To picture a surface, fix one variable at a time (a **trace**) and sketch the resulting cross-section. Regions in 3D are described by inequalities such as

0 \leq z \leq 3

(slab) or

x^{2} + y^{2} \leq 4

(solid cylinder).

Key takeaways

$x = c$ is parallel to the $y z$ -plane; $y = c$ to $x z$ ; $z = c$ to $x y$ .
If a variable is missing, the surface extends along that axis.
Use traces $x = c$ , $y = c$ , $z = c$ to visualize surfaces.

Strategy tips

Start trace-sketching with $z = c$ traces, then check $x = c$ and $y = c$ if needed.
Sketch the region whenever possible — it makes setting up integrals later much easier.

Worked examples

Example 1

Find the distance between

P (1, - 2, 3)

and

Q (4, 0, 1)

, and the midpoint of

\overline{P Q}

1
Compute the differences of corresponding coordinates.
$x_{2} - x_{1} = 3, y_{2} - y_{1} = 2, z_{2} - z_{1} = - 2$
2
Square, add, take the square root.
$d = 3^{2} + 2^{2} + (- 2)^{2} = 9 + 4 + 4 = 17$
3
Average each coordinate for the midpoint.
$M = (\frac{1 + 4}{2}, \frac{- 2 + 0}{2}, \frac{3 + 1}{2}) = (\frac{5}{2}, - 1, 2)$

Answer

d = 17

M = (5/2, - 1, 2)

Example 2

Find the equation of the sphere passing through

(1, 2, - 1)

with center

(0, 1, 2)

1
Compute the radius using the distance formula.
$r = (1 - 0)^{2} + (2 - 1)^{2} + (- 1 - 2)^{2} = 1 + 1 + 9 = 11$
2
Plug center $(a, b, c) = (0, 1, 2)$ and $r^{2} = 11$ into the standard form.
$(x - 0)^{2} + (y - 1)^{2} + (z - 2)^{2} = 11$

Answer

x^{2} + (y - 1)^{2} + (z - 2)^{2} = 11

Example 3

Identify and describe the surface

x^{2} + y^{2} = 9

R^{3}

1
$z$ does not appear, so the surface extends along the $z$ -axis.
2
The trace at any $z = c$ is the circle $x^{2} + y^{2} = 9$ of radius $3$ centered at $(0, 0, c)$ .
3
Stacking these circles gives a circular cylinder of radius $3$ along the $z$ -axis.

Answer

A circular cylinder of radius

3

centered on the

z

-axis.

Interactive visualizations

Loading visualization…

Right-hand rule: rotate from +x toward +y; thumb gives +z. The orange dot is plotted at (2, 1, 1.5) by traveling along x, then y, then z.

Loading visualization…

center a0.00

center b0.00

center c0.00

radius r2.00

(x - 0.0)^{2} + (y - 0.0)^{2} + (z - 0.0)^{2} = 2. 0^{2}

Slide the sphere's center and radius. Notice the equation updates with

r^{2}

on the right.

Loading visualization…

Region defined by $x^2 + y^2 \le 4$ and $0 \le z \le 3$ — a solid cylinder of radius 2, height 3.

Formulas in this topic

Distance in 3D

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}

Distance between two points in 3-space.

Midpoint in 3D

M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2}, \frac{z _{1} + z _{2}}{2})

Midpoint of a segment in 3-space.

Sphere

(x - a)^{2} + (y - b)^{2} + (z - c)^{2} = r^{2}

Sphere with center (a,b,c) and radius r.

Circular cylinder along z-axis

x^{2} + y^{2} = r^{2}

When a variable is missing in 3D, the surface extrudes along that axis.

Plotting in 3D and the right-hand rule

Distance and midpoint formulas

Equation of a sphere

Describing surfaces and regions

Worked examples

Interactive visualizations

Formulas in this topic

Related topics