2D and 3D coordinate geometry

IB Mathematics: Analysis and Approaches SL· 6 min read

1. Core Formulas for 2D and 3D Coordinates★★☆☆☆⏱ 15 min

All core coordinate geometry concepts extend naturally from 2D to 3D by adding a third perpendicular axis, the $z$ -axis. Every point in 3D space is represented by three coordinates $(x, y, z)$ , compared to two coordinates $(x, y)$ in 2D.

📘 Definition

Distance between two points

$P (x_{1}, y_{1}, z_{1}), Q (x_{2}, y_{2}, z_{2})$

The straight-line distance between two points is an extension of Pythagoras' theorem, adding an extra term for the $z$ -coordinate difference in 3D.

Example:

Distance from $(0, 0, 0)$ to $(1, 2, 2)$ is $e x t D i s t an ce = e x t 3$

📐 Worked Example

Find the distance between $P (2, - 1, 4)$ and $Q (- 3, 2, 6)$ , and the coordinates of their midpoint.

1
Recall the 3D distance formula:
2
$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$
3
Substitute the coordinates of the two points:
4
$d = \sqrt{(-3-2)^2 + (2-(-1))^2 + (6-4)^2} = \sqrt{(-5)^2 + 3^2 + 2^2}$
5
Simplify to get the final distance:
6
$d = \sqrt{25 + 9 + 4} = \sqrt{38} \approx 6.16$
7
The midpoint is the average of each coordinate, so:
8
$M = \left(\frac{2 + (-3)}{2}, \frac{-1 + 2}{2}, \frac{4 + 6}{2}\right) = (-0.5, 0.5, 5)$

2. Equations of Lines in 2D and 3D★★★☆☆⏱ 20 min

Lines can be expressed in multiple forms. In 2D, the common form is $y = m x + c$ , but this does not work for 3D lines. All lines in both 2D and 3D can be written in vector form.

📘 Definition

Vector equation of a line

A line passing through a point with position vector $a$ in the direction of direction vector $d$ is given by $r = a + t d$ , where $t \in R$ is a scalar parameter.

📐 Worked Example

Find the vector equation of the line passing through $(1, 3, - 2)$ and $(4, - 1, 1)$ .

1
Calculate the direction vector by subtracting the position vectors of the two points:
2
$\mathbf{d} = (4 - 1, -1 - 3, 1 - (-2)) = (3, -4, 3)$
3
Choose the first point as the position vector $a$ :
4
$\mathbf{a} = (1, 3, -2)$
5
Substitute into the general vector line equation:
6
$\mathbf{r} = \begin{pmatrix}1 \\ 3 \\ -2\end{pmatrix} + t\begin{pmatrix}3 \\ -4 \\ 3\end{pmatrix}, \quad t \in \mathbb{R}$

3. Equations of Planes in 3D★★★★☆⏱ 20 min

A plane is a flat 2D surface extending infinitely through 3D space. It is most commonly written in scalar product form or Cartesian form for exam problems.

📘 Definition

Scalar product form of a plane

A plane with normal vector $n$ (perpendicular to the plane) passing through a point with position vector $a$ is given by $r \cdot n = d$ , where $d = a \cdot n$ is a constant.

📐 Worked Example

Find the Cartesian equation of the plane passing through $(2, 0, 1)$ with normal vector $n = (3, - 1, 4)$ .

1
Start with the scalar product form $r \cdot n = d$ :
2
Calculate $d$ by substituting the given point into the equation:
3
$d = (2)(3) + (0)(-1) + (1)(4) = 6 + 0 + 4 = 10$
4
Let $r = (x, y, z)$ to get the Cartesian form:
5
$3x - y + 4z = 10$

4. Intersections and Angles★★★★☆⏱ 25 min

Common exam problems ask for the intersection of a line and a plane, and the angle between two lines or between a line and a plane. Both problems rely on substitution and the scalar product formula.

📐 Worked Example

Find the intersection point of the line $r = 13 - 2 + t 3 - 4 3$ and the plane $3 x - y + 4 z = 10$ .

1
Write the parametric coordinates of the line: $x = 1 + 3 t$ , $y = 3 - 4 t$ , $z = - 2 + 3 t$
2
Substitute these into the plane equation:
3
$3(1 + 3t) - (3 - 4t) + 4(-2 + 3t) = 10$
4
Expand and solve for $t$ :
5
$3 + 9t - 3 + 4t - 8 + 12t = 10 \implies 25t = 18 \implies t = \frac{18}{25}$
6
Substitute $t$ back to get the intersection point:
7
$\left(\frac{79}{25}, \frac{3}{25}, \frac{4}{25}\right) = (3.16, 0.12, 0.16)$

5. Common Pitfalls

Wrong move:

Omitting the $z$ -coordinate term when calculating distance in 3D

Why:

Students often forget to extend the 2D distance formula to 3D, leading to incorrect results

Correct move:

Always add the $(z_{2} - z_{1})^{2}$ term when calculating distance between two 3D points

Wrong move:

Using a point on the line as the direction vector, instead of subtracting points

Why:

Students often confuse the coordinates of the second point with the direction vector for a line through two points

Correct move:

Always calculate direction vector as $d = Point 2 - Point 1$ for a line through two points

Wrong move:

Using the same parameter $t$ for two lines when checking for intersection

Why:

Parameters are independent for each line, so forcing $t$ to be the same leads to incorrectly concluding lines are skew

Correct move:

Use different parameters (e.g. $t$ for line 1, $s$ for line 2) when solving for intersection of two lines

Wrong move:

Using $cos θ = \frac{∣ d \cdot n ∣}{∣ d ∣∣ n ∣}$ for the angle between a line and plane

Why:

The angle between the direction vector and normal is the complement of the required angle

Correct move:

Use $sin θ = \frac{∣ d \cdot n ∣}{∣ d ∣∣ n ∣}$ for the angle $θ$ between a line and plane

6. Quick Reference Cheatsheet

Concept	2D Formula	3D Formula
Distance between two points	$(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$	$(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$
Midpoint of two points	$(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$	$(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2}, \frac{z _{1} + z _{2}}{2})$
Line equation	$y = m x + c$ or $r = a + t d$	$r = a + t d$
Plane equation	Not applicable	$a x + b y + cz = d$ , $r \cdot n = d$
Angle between two lines	$cos θ = \frac{∣ d _{1} \cdot d _{2} ∣}{∣ d _{1} ∣∣ d _{2} ∣}$	$cos θ = \frac{∣ d _{1} \cdot d _{2} ∣}{∣ d _{1} ∣∣ d _{2} ∣}$
Angle between line and plane	Not applicable	$sin θ = \frac{∣ d \cdot n ∣}{∣ d ∣∣ n ∣}$

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.

2022 · 1
3D line intersection problem
2023 · 2
Distance from point to plane calculation

What's Next

2D and 3D coordinate geometry is the foundation for all geometric problem solving in IB AA SL. This topic is heavily linked to vector operations, which you use constantly to write equations and solve intersection problems. Mastery of these core formulas and methods is critical, as coordinate geometry appears in both Paper 1 and Paper 2, often combined with trigonometry or calculus to create multi-part exam questions. After mastering this sub-topic, you will build on this knowledge to solve more advanced problems including distances from points to lines and planes, and applications of coordinate geometry to real-world modelling problems.