2D and 3D coordinate geometry

IB Mathematics AI HL· 45 min read

1. Distance, Midpoint and Section Formula★★☆☆☆⏱ 15 min

📘 Definition

Distance between two points

$d (P_{1}, P_{2})$

The shortest straight-line path between two points, derived from Pythagoras' theorem, for both 2D and 3D space.

Example:

Distance between $(1, 2)$ and $(4, 6)$ is $5$ units

All core distance and midpoint formulas extend naturally from 2D to 3D by adding a $z$ -coordinate term. The section formula finds the coordinates of a point that divides a line segment in a given ratio.

\begin{aligned} \text{2D distance: } d &= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \\ \text{3D distance: } d &= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} \\ \text{Section ratio } k:m \text{ (internal): } P &= \left(\frac{mx_1 + kx_2}{k+m}, \frac{my_1 + ky_2}{k+m}, \frac{mz_1 + kz_2}{k+m}\right) \end{aligned}

📐 Worked Example

Find the coordinates of the point that divides the segment joining $A (1, 2, - 3)$ and $B (3, 4, 5)$ in the ratio $2 : 1$ internally.

1
Identify input values: $A (x_{1}, y_{1}, z_{1}) = (1, 2, - 3)$ , $B (x_{2}, y_{2}, z_{2}) = (3, 4, 5)$ , ratio $k : m = 2 : 1$
2
Apply the section formula for each coordinate:
3
$x = \frac{(1 \times 1) + (2 \times 3)}{2 + 1} = \frac{7}{3}, \quad y = \frac{(1 \times 2) + (2 \times 4)}{3} = \frac{10}{3}, \quad z = \frac{(1 \times -3) + (2 \times 5)}{3} = \frac{7}{3}$
4
Final answer: the required point is $(\frac{7}{3}, \frac{10}{3}, \frac{7}{3})$

Exam tip:

Always confirm if the ratio requested is internal or external; external division flips the sign of the ratio term.

2. Equations of Straight Lines★★☆☆☆⏱ 20 min

📘 Definition

Straight line

$r = a + t d$

A set of points defined by a fixed initial point (with position vector $a$ ) and a constant direction vector $d$ that defines the line's orientation. The parameter $t$ scales $d$ to reach any point on the line.

In IB AI HL, you will mostly use parametric vector form for lines in 2D and 3D. To find the direction vector of a line passing through two points $A$ and $B$ , calculate $A B = B - A$ .

📐 Worked Example

Find the parametric equation of the line passing through $(2, - 1, 3)$ and parallel to the vector $2 i - j + 4 k$ .

1
Write the general form for parametric equations:
2
$x = x_0 + td_x, \quad y = y_0 + td_y, \quad z = z_0 + td_z$
3
Substitute the fixed point $(x_{0}, y_{0}, z_{0}) = (2, - 1, 3)$ and direction vector components $(d_{x}, d_{y}, d_{z}) = (2, - 1, 4)$
4
Final parametric equations:
5
$x = 2 + 2t, \quad y = -1 - t, \quad z = 3 + 4t, \quad t \in \mathbb{R}$

3. Intersection of Two Lines★★★☆☆⏱ 20 min

Two lines can be: (1) intersecting (meet at exactly one point), (2) parallel (never meet, direction vectors are scalar multiples), (3) coincident (the same line), or (4) skew (only in 3D: non-parallel and never intersect). To check for intersection, set parametric coordinates equal and solve for the parameters.

📐 Worked Example

Check if lines $L_{1} : x = 1 + s, y = 2 + 3 s, z = - s$ and $L_{2} : x = 2 t, y = 1 + t, z = 1 + 3 t$ intersect.

1
Set corresponding coordinates equal to get a system of equations:
2
$1 + s = 2t, \quad 2 + 3s = 1 + t, \quad -s = 1 + 3t$
3
Solve the first two equations: rearrange first to get $s = 2 t - 1$ , substitute into the second:
4
$2 + 3(2t - 1) = 1 + t \implies 5t = 2 \implies t = \frac{2}{5}, s = -\frac{1}{5}$
5
Check if the solution satisfies the third equation: Left side $- s = \frac{1}{5}$ , Right side $1 + 3 (\frac{2}{5}) = \frac{11}{5}$ . No match.
6
Conclusion: no consistent solution exists, so lines do not intersect.

✓ Quick check

If the lines above are not parallel, what are they?

What is the classification of the lines in the example?
- Coincident
- Skew
- Perpendicular
Reveal answer
Skew —
Skew lines only exist in 3D space, and are defined as non-parallel lines that never intersect.

4. Angle Between Two Lines★★★☆☆⏱ 15 min

The angle between two lines is defined as the smallest (acute) angle between their direction vectors. We use the dot product formula, with an absolute value to ensure we get an acute angle between $0^{\circ}$ and $9 0^{\circ}$ .

\cos\theta = \frac{|\mathbf{d_1} \cdot \mathbf{d_2}|}{|\mathbf{d_1}| |\mathbf{d_2}|}

📐 Worked Example

Find the angle between two lines with direction vectors $d_{1} = 21 - 1$ and $d_{2} = 1 - 1 1$ .

1
Calculate the dot product of the two direction vectors:
2
$\mathbf{d_1} \cdot \mathbf{d_2} = (2)(1) + (1)(-1) + (-1)(1) = 2 - 1 - 1 = 0$
3
Substitute into the angle formula: $cos θ = \frac{∣0∣}{∣ d _{1} ∣∣ d _{2} ∣} = 0$
4
Therefore $θ = 9 0^{\circ}$ , so the lines are perpendicular.

Exam tip:

Exam questions always ask for the acute angle between two lines, so never leave an obtuse angle as your final answer.

5. Common Pitfalls

Wrong move:

Reversing the weights in the section formula for a ratio $A P : P B$

Why:

You mixed up which point gets which weight, leading to incorrect coordinates

Correct move:

If $P$ divides $A B$ in ratio $k : m$ , the coordinates are $P = \frac{m A + k B}{k + m}$ , so the weight for $A$ matches the ratio term for $B$

Wrong move:

Skipping the check for the third coordinate when testing 3D line intersections

Why:

The first two equations will always have a solution, even for skew lines that do not intersect

Correct move:

Always substitute your solved parameters into all three equations to confirm consistency before concluding lines intersect

Wrong move:

Forgetting the absolute value in the angle between two lines formula

Why:

You will calculate the obtuse angle instead of the required acute angle between the lines

Correct move:

Always take the absolute value of the dot product, and give the acute angle as your final answer

Wrong move:

Assuming all non-intersecting 3D lines are parallel

Why:

Skew lines are a common case of non-intersecting, non-parallel lines unique to 3D space

Correct move:

Always check if direction vectors are scalar multiples before classifying non-intersecting lines as parallel

Wrong move:

Subtracting mismatched coordinates when calculating distance

Why:

You swapped $x$ and $y$ terms, leading to an incorrect distance calculation

Correct move:

Label points clearly, and always subtract $x_{1}$ from $x_{2}$ and $y_{1}$ from $y_{2}$ before squaring

6. Quick Reference Cheatsheet

Concept	2D Formula	3D Formula
Distance between $P_{1}, P_{2}$	$(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$	$(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$
Midpoint of $P_{1} P_{2}$	$(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$	$(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2}, \frac{z _{1} + z _{2}}{2})$
Point dividing ratio $k : m$	$(\frac{m x _{1} + k x _{2}}{k + m}, \frac{m y _{1} + k y _{2}}{k + m})$	$(\frac{m x _{1} + k x _{2}}{k + m}, \frac{m y _{1} + k y _{2}}{k + m}, \frac{m z _{1} + k z _{2}}{k + m})$
Vector line equation	$r = a + t d$	$r = a + t d$
Angle between two lines	$cos θ = \frac{∣ d _{1} \cdot d _{2} ∣}{∣ d _{1} ∣∣ d _{2} ∣}$	$cos θ = \frac{∣ d _{1} \cdot d _{2} ∣}{∣ d _{1} ∣∣ d _{2} ∣}$

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.

2023 · 1
Intersection of 3D lines
2022 · 2
Distance and angle between lines
2021 · 1
Section formula problem

What's Next

Mastering 2D and 3D coordinate geometry is the foundation for all remaining topics in the IB AI HL Geometry and Trigonometry unit. The skills you learned here, especially working with lines, direction vectors and solving systems for intersections, are directly extended to work with planes in 3D space, finding intersections between lines and planes, and calculating volumes and surface areas of complex 3D shapes. These concepts also transfer directly to vector kinematics, where you will use line equations to model motion of particles in 2D and 3D space.