Difficulty: Introductory | Prerequisites: Familiarity with 2D coordinate geometry, the Pythagorean theorem, and basic equation of a circle.
This topic extends everything you know about 2D coordinates into three dimensions. It is the foundation for all of Calculus III: once you can plot points, measure distances, and describe surfaces in 3D, you can move on to vectors, curves, and multivariable functions. If you are comfortable with the equation of a circle in 2D, you already have the intuition; the sphere equation is the same idea with an extra coordinate. Students who skip this section tend to struggle later with visualising vector operations and surfaces of integration.
The 3D coordinate system adds a z-axis to the familiar xy-plane, letting you locate any point in space with an ordered triple (x, y, z). The distance formula generalises to three dimensions by adding a z-term under the square root. A sphere is the set of all points at a fixed distance (the radius) from a centre point, and its equation looks just like a circle's equation with an extra squared term.
Three-dimensional coordinate system (3D Cartesian coordinates)
A system of three mutually perpendicular axes (x, y, z) intersecting at the origin O. Every point in space corresponds to a unique ordered triple (x, y, z). Think of it as adding a "height" axis to the flat xy-plane you already know.
Octant
One of the eight regions that the three coordinate planes divide space into. The first octant is where x > 0, y > 0, and z > 0. In simple terms, octants are the 3D equivalent of the four quadrants in 2D.
Distance formula (3D)
The distance between two points P₁(x₁, y₁, z₁) and P₂(x₂, y₂, z₂) is given by d = √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²]. Think of it as the Pythagorean theorem applied twice: once in the horizontal plane, then again to include the vertical separation.
Midpoint formula (3D)
The midpoint of the segment joining P₁(x₁, y₁, z₁) and P₂(x₂, y₂, z₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2, (z₁ + z₂)/2). In simple terms, average each coordinate separately.
Sphere
The set of all points in space that are at a fixed distance r (the radius) from a fixed point (h, k, l) called the centre. Think of it as a 3D circle: every point on the surface is exactly the same distance from the middle.
Standard equation of a sphere
The equation (x − h)² + (y − k)² + (z − l)² = r², where (h, k, l) is the centre and r is the radius. This is the 3D analogue of the circle equation (x − h)² + (y − k)² = r².
Coordinate planes
The three planes formed by pairs of axes: the xy-plane (z = 0), the yz-plane (x = 0), and the xz-plane (y = 0). These are the "walls" and "floor" of the coordinate system.
Three axes (x, y, z) meet at the origin at right angles
The right-hand rule determines axis orientation: curl fingers from positive x toward positive y, and your thumb points in the positive z direction
Together the axes create three coordinate planes:
xy-plane: all points where z = 0
yz-plane: all points where x = 0
xz-plane: all points where y = 0
These three planes carve space into eight octants; the first octant has all coordinates positive
A point P(a, b, c) is located by moving a units along the x-axis, b units parallel to the y-axis, and c units parallel to the z-axis
Example: the point (2, 3, 5) sits 2 units along x, 3 along y, and 5 units up
Derived by applying the Pythagorean theorem twice
First find the distance in the xy-plane, then use that as one leg and the z-difference as the other
Result: d = √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²]
A sphere with centre (h, k, l) and radius r satisfies (x − h)² + (y − k)² + (z − l)² = r²
To find the equation, you need the centre and the radius
If the sphere is described by "touching" a coordinate plane, the radius equals the distance from the centre to that plane:
Touches the xy-plane: r = |l| (the absolute value of the z-coordinate of the centre)
Touches the yz-plane: r = |h| (the absolute value of the x-coordinate)
Touches the xz-plane: r = |k| (the absolute value of the y-coordinate)
Example: a sphere centred at (−3, 2, 5) that touches the xy-plane has r = |5| = 5, so its equation is (x + 3)² + (y − 2)² + (z − 5)² = 25
When a sphere's equation is given in expanded form (x² + y² + z² + Dx + Ey + Fz + G = 0), complete the square in each variable to convert back to standard form
Group x-terms, y-terms, and z-terms separately, then add the needed constants to both sides
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\text{Midpoint} = \left(\frac{x_1 + x_2}{2},\; \frac{y_1 + y_2}{2},\; \frac{z_1 + z_2}{2}\right)GPS satellites locate a receiver by computing its distance from several known points in 3D space, which is exactly the distance formula at work. In computer graphics, every object on screen is defined by vertices in a 3D coordinate system, and bounding spheres (sphere equations) are used for fast collision detection in games and simulations.
Students often think the radius of a sphere touching the xy-plane is the full distance from the centre to the origin. It is not. The radius equals only the perpendicular distance from the centre to the plane, which is the absolute value of the relevant single coordinate.
Students sometimes confuse which coordinate gives the distance to which plane. Remember: the xy-plane is the "floor" at z = 0, so the distance to it is |z|. The yz-plane is the "wall" at x = 0, so the distance is |x|. The xz-plane is at y = 0, distance = |y|.
When completing the square, a common error is forgetting to add the completing constant to both sides of the equation, which produces an incorrect radius.
Some students mix up the signs in the standard form. The centre of (x + 3)² + (y − 2)² + (z − 5)² = 25 is (−3, 2, 5), not (3, −2, −5). The sign flips.
⚠️ Expect questions that give you a centre and ask for the sphere equation under a geometric condition ("touches the xy-plane", "passes through the origin", "tangent to a coordinate plane"). Know instantly which coordinate gives the radius.
⚠️ Completing the square to recover centre and radius from an expanded equation is a classic exam question. Practise until it is automatic.
⚠️ The distance formula is used constantly in later sections (for vector magnitudes, distances between points and planes, etc.). Errors here cascade.
True or False: The distance from the point (4, −1, 7) to the xz-plane is 7.
Fill in the blank: A sphere centred at (0, 0, 3) that touches the xy-plane has radius ___.
True or False: The first octant contains the point (−1, 2, 3).
Fill in the blank: The midpoint of (2, 4, 6) and (8, 0, −2) is ___.
True or False: The equation (x − 1)² + (y + 2)² + (z − 3)² = 16 describes a sphere of radius 16.
Answers: 1. False (it is 1, the absolute value of the y-coordinate). 2. r = 3. 3. False (x is negative). 4. (5, 2, 2). 5. False (the radius is 4, since r² = 16).
Q: Find the equation of the sphere with centre (−3, 2, 5) that touches the xy-plane.
A: The distance from the centre to the xy-plane is |5| = 5, so r = 5. The equation is (x + 3)² + (y − 2)² + (z − 5)² = 25.
Q: Find the equation of the sphere with centre (−3, 2, 5) that touches the yz-plane.
A: The distance from the centre to the yz-plane is |−3| = 3, so r = 3. The equation is (x + 3)² + (y − 2)² + (z − 5)² = 9.
Q: Find the equation of the sphere with centre (−3, 2, 5) that touches the xz-plane.
A: The distance from the centre to the xz-plane is |2| = 2, so r = 2. The equation is (x + 3)² + (y − 2)² + (z − 5)² = 4.
Q: A sphere has the equation x² + y² + z² − 6x + 4y − 2z = 11. Find its centre and radius.
A: Complete the square: (x² − 6x + 9) + (y² + 4y + 4) + (z² − 2z + 1) = 11 + 9 + 4 + 1 = 25. So (x − 3)² + (y + 2)² + (z − 1)² = 25. Centre is (3, −2, 1) and radius is 5.
Q: What is the distance between the points (1, −2, 4) and (3, 0, −1)?
A: d = √[(3 − 1)² + (0 − (−2))² + (−1 − 4)²] = √[4 + 4 + 25] = √33.
The distance formula reappears immediately in §12.2 when you compute the magnitude (length) of a vector. The sphere equation connects to level surfaces in §12.6: a sphere is a level surface of f(x, y, z) = (x − h)² + (y − k)² + (z − l)². Understanding coordinate planes is essential for setting up limits of integration in triple integrals later in the course.
3D coordinates, three-dimensional space, Cartesian coordinates, R3, ordered triple, xyz-axes, octants, first octant, distance formula 3D, Pythagorean theorem 3D, midpoint formula 3D, equation of a sphere, sphere standard form, completing the square sphere, tangent to coordinate plane, sphere touching plane, radius from centre to plane, level surface, Math 215, Calculus III, Stewart §12.1