3D coordinate geometry

IB Mathematics: Analysis and Approaches HL· Topic 3.7 3D coordinate geometry· 15 min read

1. Core Concepts & Basic Formulas★★☆☆☆⏱ 3 min

In 3D coordinate geometry, any point is represented by an ordered triple $(x, y, z)$ along three mutually perpendicular x, y, z axes that intersect at the origin $(0, 0, 0)$ . Most 2D coordinate geometry formulas extend naturally to 3D by adding a third coordinate term.

📘 Definition

Distance between two points

For $P_{1} (x_{1}, y_{1}, z_{1})$ and $P_{2} (x_{2}, y_{2}, z_{2})$

The straight-line distance between two points, derived from Pythagoras' theorem extended to 3D.

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

📘 Definition

Midpoint of a line segment

The midpoint is the average of each coordinate of the two endpoints.

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

📐 Worked Example

Find the distance between $P (1, 2, - 3)$ and $Q (4, - 2, 1)$ , and the midpoint of segment $P Q$ .

1
Identify coordinates and calculate differences:
2
$\Delta x = 4-1=3, \Delta y = -2-2=-4, \Delta z = 1 - (-3)=4$
3
Substitute into the distance formula:
4
$d = \sqrt{3^2 + (-4)^2 + 4^2} = \sqrt{9+16+16} = \sqrt{41} \approx 6.40$
5
Calculate midpoint by averaging coordinates:
6
$M\left(\frac{1+4}{2}, \frac{2+(-2)}{2}, \frac{-3+1}{2}\right) = (2.5, 0, -1)$

Exam tip:

Always double-check the sign of coordinates when calculating differences, especially for negative values.

2. Equations of Lines and Planes★★★☆☆⏱ 5 min

We can represent lines and planes in 3D space using both vector and Cartesian forms, both of which are commonly used in IB exams.

📘 Definition

Vector equation of a line

A line through point $A$ (position vector $a$ ) with direction vector $d$ is written as:

\vec{r} = \vec{a} + t\vec{d}, \quad t \in \mathbb{R}

📘 Definition

Cartesian equation of a plane

A plane with normal vector $n = n_{x} n_{y} n_{z}$ and constant $d$ is written as:

n_x x + n_y y + n_z z = d

📐 Worked Example

Find the Cartesian equation of the plane that contains the point $(2, 0, - 1)$ and has normal vector $3 - 1 2$ .

1
We use the form $n_{x} x + n_{y} y + n_{z} z = d$ , with $n_{x} = 3, n_{y} = - 1, n_{z} = 2$ .
2
Substitute the point into the equation to solve for $d$ :
3
$d = (3)(2) + (-1)(0) + (2)(-1) = 6 - 2 = 4$
4
Write the final equation:
5
$3x - y + 2z = 4$

Exam tip:

If given three points on a plane, calculate two direction vectors in the plane, then take their cross product to get the normal vector before finding the equation.

3. Intersections and Angles★★★★☆⏱ 4 min

A common exam question asks for the intersection of a line and a plane, or the angle between two lines, two planes, or a line and a plane. To find an intersection, substitute the parametric equation of the line into the plane equation and solve for the parameter $t$ .

📐 Worked Example

Find the intersection point of the line $r = 123 + t 1 - 1 1$ and the plane $x + 2 y - z = 4$ .

1
Write parametric coordinates from the line equation: $x = 1 + t, y = 2 - t, z = 3 + t$ .
2
Substitute into the plane equation:
3
$(1+t) + 2(2-t) - (3+t) = 4$
4
Simplify and solve for $t$ :
5
$2 - 2t = 4 \implies t = -1$
6
Substitute $t = - 1$ back to get coordinates:
7
$x=0, y=3, z=2 \implies \text{Intersection: } (0, 3, 2)$

For angles, the angle between two geometric objects equals the angle between their relevant vectors (normals for planes, direction vectors for lines), calculated using the dot product formula: $cos θ = \frac{∣ u \cdot v ∣}{∣ u ∣∣ v ∣}$ .

4. Shortest Distance Calculations★★★★☆⏱ 3 min

The most frequently tested distance calculation is the shortest distance from a point to a plane. The formula for distance from point $(x_{0}, y_{0}, z_{0})$ to plane $a x + b y + cz + d = 0$ is:

D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}

📐 Worked Example

Find the shortest distance from $(1, 1, 2)$ to the plane $2 x - y + 3 z - 8 = 0$ .

1
Identify values: $a = 2, b = - 1, c = 3, d = - 8, x_{0} = 1, y_{0} = 1, z_{0} = 2$ .
2
Calculate the absolute value of the numerator:
3
$|2(1) - 1(1) + 3(2) - 8| = |-1| = 1$
4
Calculate the denominator:
5
$\sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{14}$
6
Final distance:
7
$D = \frac{1}{\sqrt{14}} = \frac{\sqrt{14}}{14} \approx 0.267$

5. Common Pitfalls

Wrong move:

Dividing by zero when writing Cartesian line equations if a direction vector component is zero

Why:

If a direction vector component is zero, the corresponding coordinate is constant, so division by zero is undefined

Correct move:

Write the coordinate as a constant, e.g. if $d_{x} = 0$ , add $x = a_{x}$ to the equation instead of dividing

Wrong move:

Using direction vectors instead of normal vectors to calculate the angle between two planes

Why:

The angle between two planes is defined as the angle between their normals, not direction vectors in the planes

Correct move:

Always use the normal vectors of both planes when calculating the angle between two planes

Wrong move:

Dropping the absolute value in the distance from point to plane formula

Why:

The numerator can be negative, leading to a negative distance which is incorrect and loses marks

Correct move:

Always keep the absolute value around the numerator to ensure a positive distance

Wrong move:

Assuming all non-parallel lines in 3D must intersect

Why:

Unlike 2D, 3D space allows skew lines: non-parallel lines that never intersect

Correct move:

Always check for a consistent solution when finding intersections; no solution means lines are skew

6. Quick Reference Cheatsheet

Concept	Formula
Distance between 2 points	$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$
Midpoint	$(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2}, \frac{z _{1} + z _{2}}{2})$
Vector line equation	$r = a + t d$
Plane (Cartesian form)	$n_{x} x + n_{y} y + n_{z} z = d$
Distance from point to plane	$\frac{∣ a x _{0} + b y _{0} + c z _{0} + d ∣}{a ^{2} + b ^{2} + c ^{2}}$
Angle between vectors	$cos θ = \frac{∣ u \cdot v ∣}{∣ u ∣∣ v ∣}$

7. Frequently Asked

Do I need to memorize the distance from a point to a plane formula?

Yes, this formula is not included in the IB formula booklet, so you must memorize it for exams.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · 1
Intersection of line and plane
2024 · 2
Distance from point to plane
2023 · 1
Angle between two planes

Going deeper

referenceIB AA HL Formula BookletBasic vector formulas are provided here, distance from point to plane is not

What's Next

Mastering 3D coordinate geometry is essential for solving multi-part vector questions that frequently appear as extended response questions in both IB AA HL Paper 1 and Paper 2. This topic is almost always combined with vector algebra concepts like cross product to create complex problems that test multiple skills. A solid foundation here also prepares you for further study of 3D transformations, volumes of revolution, and vector calculus at university level. Use the links below to build on the concepts you learned here.