Difficulty: Introductory | Prerequisites: Familiarity with 2D coordinate geometry and the Pythagorean theorem.
This section extends the flat (x, y) plane you already know into three dimensions by adding a z-axis. You will learn how to measure distances between points in 3D using a generalised Pythagorean formula, and how to write the equation of a sphere by treating it as "all points equidistant from a centre." If you can handle 2D distance, you can handle everything here.
This topic sits at the very start of Calculus III. Everything that follows in the course (vectors, curves, surfaces, integration in 3D) assumes you are comfortable navigating three-dimensional space and computing distances. If you are joining late, this is the place to begin.
R² (two-dimensional space)
The set of all ordered pairs (x, y) where x and y are real numbers. In simple terms, this is the ordinary flat plane you have been graphing on since algebra.
R³ (three-dimensional space)
The set of all ordered triples (x, y, z) where x, y and z are real numbers. Think of it as the flat plane with a vertical axis added so you can describe height or depth.
First octant
The region of R³ where x ≥ 0, y ≥ 0 and z ≥ 0. In simple terms, it is the 3D equivalent of the first quadrant: everything is positive.
Coordinate planes (xy-plane, xz-plane, yz-plane)
Flat surfaces in R³ defined by setting one coordinate to zero. The xy-plane is all points with z = 0, the xz-plane has y = 0, and the yz-plane has x = 0. Think of each as the "floor," "back wall" or "side wall" of your 3D coordinate system.
Distance formula (3D)
The formula √((x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²), which gives the straight-line distance between any two points in R³. It is the Pythagorean theorem applied twice.
Sphere (standard form)
The set of all points at a fixed distance r from a centre point (x₀, y₀, z₀). Its equation in standard form is (x − x₀)² + (y − y₀)² + (z − z₀)² = r².
Completing the square
An algebraic technique for rewriting a quadratic expression so you can read off the centre and radius of a sphere. You group each variable's terms, add the right constant to both sides, and factor into perfect squares.
The xy-plane is defined by z = 0. Every point on it looks like (x, y, 0).
The xz-plane is defined by y = 0.
The yz-plane is defined by x = 0.
The x-axis requires both y = 0 and z = 0.
The y-axis requires both x = 0 and z = 0.
The z-axis requires both x = 0 and y = 0.
The pattern: a plane is defined by setting the "missing" variable to zero. An axis is defined by setting the two variables that are not that axis to zero.
In R²: d = √((x₂ − x₁)² + (y₂ − y₁)²)
In R³: d = √((x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²)
Both are direct applications of the Pythagorean theorem. The 3D version applies the theorem twice: once in the base plane, once to reach the third dimension.
From a point (x₁, y₁, z₁):
Distance to the xy-plane = |z₁| (drop the point straight down to z = 0).
Distance to the xz-plane = |y₁|.
Distance to the yz-plane = |x₁|.
Distance to the x-axis = √(y₁² + z₁²) (the point's distance from (x₁, 0, 0)).
Distance to the y-axis = √(x₁² + z₁²).
Distance to the z-axis = √(x₁² + y₁²).
The trick: identify what the "closest point" on the plane or axis would be, then use the distance formula.
A sphere with centre (x₀, y₀, z₀) and radius r satisfies:
(x − x₀)² + (y − y₀)² + (z − z₀)² = r²
To find the equation, you need two pieces of information: the centre and the radius.
Given: 2x² + 12x + 2y² + 2z² + 6z = 4. Determine the centre and radius.
Divide everything by 2 so the squared terms have coefficient 1: x² + 6x + y² + z² + 3z = 2.
Complete the square for x: half of 6 is 3, and 3² = 9. Add 9 to both sides.
Complete the square for z: half of 3 is 3/2, and (3/2)² = 9/4. Add 9/4 to both sides.
The equation becomes (x + 3)² + y² + (z + 3/2)² = 2 + 9 + 9/4 = 53/4.
Centre = (−3, 0, −3/2), radius = √(53/4).
d_{\text{2D}} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}d_{\text{3D}} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}(x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2 = r^2 \quad \text{(sphere, standard form)}Completing the square pattern: for x² + bx, add (b/2)² to both sides, then factor as (x + b/2)².
The 3D distance formula is the basis of GPS positioning: a receiver computes its distance from several satellites and triangulates its location. The equation of a sphere describes everything from weather radar coverage (all points within range of a station) to collision detection in game physics and CAD software.
Students often think the distance from a point to the xy-plane involves plugging into the full distance formula. It does not. The distance is simply the absolute value of the z-coordinate.
When completing the square, students frequently forget to divide the entire equation first so the squared-term coefficients are 1. If you skip that step, the factoring will not work.
Some students confuse the centre signs: in (x + 3)² the centre coordinate is −3, not +3. The standard form uses subtraction, so a plus sign inside the parenthesis means the coordinate is negative.
Students sometimes forget that the equation of a sphere uses r², not r, on the right-hand side. If the equation gives you r² = 53/4, the radius is √(53/4), not 53/4 itself.
⚠️ Distance-to-axis and distance-to-plane problems appear frequently. Know the shortcuts (|z₁| for xy-plane, √(y₁² + z₁²) for the x-axis, etc.) so you do not waste time on the full formula.
⚠️ Completing the square to convert a sphere equation to standard form is a classic exam question. Practise until the algebra is fast and clean.
⚠️ You may be asked to determine whether a given equation represents a sphere (all squared terms have the same coefficient, no cross terms) or something else.
True or false: The equation of the xz-plane is x = 0. (False, it is y = 0.)
Fill in the blank: The distance from (2, −5, 7) to the xy-plane is ___. (7)
True or false: The first octant includes points where z can be negative. (False, all three coordinates must be ≥ 0.)
Fill in the blank: The distance from (1, 2, 3) to the z-axis is ___. (√(1² + 2²) = √5)
True or false: If (x − 1)² + (y + 2)² + (z − 3)² = 16, the radius is 16. (False, the radius is √16 = 4.)
Q: Find the distance between the points (3, −1, 4) and (−2, 5, 1).
A: d = √((−2 − 3)² + (5 − (−1))² + (1 − 4)²) = √(25 + 36 + 9) = √70.
Q: What is the distance from the point (4, −3, 8) to the y-axis?
A: The closest point on the y-axis is (0, −3, 0). Distance = √(4² + 0² + 8²) = √(16 + 64) = √80 = 4√5.
Q: Write the equation of a sphere with centre (2, −1, 5) and radius 3.
A: (x − 2)² + (y + 1)² + (z − 5)² = 9.
Q: Rewrite x² − 4x + y² + 6y + z² + 2z = 11 in standard form. What are the centre and radius?
A: Complete the square for each variable. x² − 4x + 4 + y² + 6y + 9 + z² + 2z + 1 = 11 + 4 + 9 + 1 = 25. So (x − 2)² + (y + 3)² + (z + 1)² = 25. Centre = (2, −3, −1), radius = 5.
Q: Describe the set of all points in R³ that satisfy x² + y² + z² ≤ 4. Is this a sphere?
A: It is a solid ball (not just the surface) of radius 2 centred at the origin. The sphere itself is the boundary where x² + y² + z² = 4.
The distance formula reappears in section 12.2 as the magnitude (length) of a vector. The sphere equation is used in later chapters when setting up triple integrals in spherical coordinates. Completing the square also comes back when you classify quadric surfaces in section 12.6.
3D coordinate system, three-dimensional space, R², R³, Cartesian coordinates, distance formula, Pythagorean theorem in 3D, first octant, coordinate planes, xy-plane, xz-plane, yz-plane, equation of a sphere, standard form sphere, completing the square, centre and radius, quadric surfaces, Math 126, Calculus III, UW Calculus