2D and 3D vectors, dot product and lines

IB Mathematics AA SL· 25 min read

1. Vectors in 2D and 3D Coordinates★★☆☆☆⏱ 10 min

📘 Definition

Vector

A quantity with both magnitude (size) and direction, represented as an ordered pair in 2D or triplet in 3D.

Example:

A displacement of 3 units right and 4 units up is written $(34)$ .

Vectors are translation-invariant: two vectors are equal if they have the same components, regardless of their starting point. A position vector is fixed relative to the origin, with components equal to the coordinates of its endpoint.

📐 Worked Example

Find the magnitude of the 3D position vector $a = 2 - 3 6$ .

1
The formula for the magnitude of a 3D vector is:
2
$|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$
3
Substitute the components of $a$ :
4
$|\vec{a}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49}$
5
Simplify to get the final magnitude:
6
$|\vec{a}| = 7$

2. The Dot Product of Two Vectors★★★☆☆⏱ 15 min

📘 Definition

Dot Product (Scalar Product)

$a \cdot b$

A scalar value calculated from two vectors, using either their components or their magnitudes and the angle between them.

Algebraic form (3D): $a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$
Geometric form: $a \cdot b = ∣ a ∣∣ b ∣ cos θ$ , where $θ$ is the angle between the vectors
Angle formula: $cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}$

📐 Worked Example

Find the angle between vectors $a = 12 - 1$ and $b = 3 - 1 2$ , correct to 1 decimal place.

1
First calculate the dot product:
2
$\vec{a} \cdot \vec{b} = (1)(3) + (2)(-1) + (-1)(2) = 3 - 2 - 2 = -1$
3
Calculate magnitudes of each vector:
4
$|\vec{a}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6} \approx 2.449, \quad |\vec{b}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{14} \approx 3.742$
5
Substitute into the angle formula:
6
$\cos\theta = \frac{-1}{\sqrt{6}\sqrt{14}} \approx -0.1091$
7
Take inverse cosine (angles are between 0° and 180°):
8
$\theta \approx 96.3^\circ$

3. Vector Equations of Lines★★★☆☆⏱ 15 min

Any line can be uniquely defined by one known point on the line and a direction vector that describes the line's orientation. This gives a flexible vector form that works for both 2D and 3D lines.

📘 Definition

Vector equation of a line

$r = a + t d$

$r$ = position vector of any point on the line, $a$ = position vector of a known point on the line, $d$ = direction vector, $t$ = scalar parameter.

📐 Worked Example

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

1
First calculate the direction vector from A to B:
2
$\vec{AB} = B - A = \begin{pmatrix} 4-1 \\ -1-2 \\ 2 - (-3) \end{pmatrix} = \begin{pmatrix} 3 \\ -3 \\ 5 \end{pmatrix}$
3
Use point A as the known fixed point, so $a = 12 - 3$
4
Write the full equation with parameter $t$ :
5
$\vec{r} = \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix} + t\begin{pmatrix} 3 \\ -3 \\ 5 \end{pmatrix}, \quad t \in \mathbb{R}$

4. Intersections and Angles Between Lines★★★★☆⏱ 20 min

To find if two lines intersect, set their position vectors equal and solve for the scalar parameters. In 3D space, there are three possible cases for two lines: intersecting, parallel, or skew.

📐 Worked Example

Classify lines $L_{1} : r = 121 + s - 1 11$ and $L_{2} : r = 214 + t 2 - 1 1$ .

1
Set components equal to get a system of equations:
2
$1 - s = 2 + 2t, \quad 2 + s = 1 - t, \quad 1 + s = 4 + t$
3
Solve the first two equations: add them to get $3 = 3 + t ⟹ t = 0$ , so $s = - 1$ .
4
Check if $s = - 1, t = 0$ satisfies the third equation: Left = $1 + (- 1) = 0$ , Right = $4 + 0 = 4$ . $0 \neq = 4$ , so no solution.
5
Check if direction vectors are parallel: $(- 1, 1, 1)$ is not a scalar multiple of $(2, - 1, 1)$ , so lines are skew.

To find the angle between two lines, use their direction vectors in the dot product formula, and take the acute angle (0° to 90°), since the angle between two lines is defined as the smallest angle between them.

5. Common Pitfalls

Wrong move:

Assuming all non-parallel 3D lines intersect

Why:

3D space allows non-parallel lines that never meet (skew lines)

Correct move:

Always check the third equation when solving for intersections of 3D lines to confirm the solution is consistent

Wrong move:

Not taking the absolute value of the dot product for the angle between two lines

Why:

The dot product can give a negative cosine, but the angle between two lines is always acute

Correct move:

Use $cos θ = \frac{∣ d _{1} \cdot d _{2} ∣}{∣ d _{1} ∣∣ d _{2} ∣}$ for the angle between lines

Wrong move:

Mixing up position vectors and direction vectors in the line equation

Why:

This leads to incorrect points on the line and wrong intersection results

Correct move:

Remember: $a$ is a fixed point on the line, $d$ is the direction vector multiplied by the parameter

Wrong move:

Calculating magnitude of a sum as $∣ a + b ∣ = ∣ a ∣ + ∣ b ∣$

Why:

This only holds for parallel vectors, it is not true for most vectors

Correct move:

First add the components of $a$ and $b$ , then calculate the magnitude of the resulting vector

Wrong move:

Writing the dot product as a vector instead of a scalar

Why:

The dot product always outputs a single number, not another vector

Correct move:

Confirm your result for a dot product is a single scalar value before proceeding to further calculations

6. Quick Reference Cheatsheet

Concept	Formula / Rule
Magnitude (2D)	$∣ a ∣ = a_{1}^{2} + a_{2}^{2}$
Magnitude (3D)	$∣ a ∣ = a_{1}^{2} + a_{2}^{2} + a_{3}^{2}$
Dot Product (algebraic)	$a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$
Dot Product (geometric)	$a \cdot b = ∣ a ∣∣ b ∣ cos θ$
Angle between vectors	$cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}$
Angle between lines	$cos θ = \frac{∣ d _{1} \cdot d _{2} ∣}{∣ d _{1} ∣∣ d _{2} ∣}$
Vector line equation	$r = a + t d$
Perpendicular vectors	$a \cdot b = 0$
Parallel vectors	$a = k b$ for scalar $k$
Skew lines (3D)	Non-parallel, no intersection

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
Find angle between two 3D vectors
2023 · 2
Intersection of two 3D lines
2021 · 1
Dot product for perpendicular vectors

Going deeper

official syllabusIB AA SL Geometry Topic GuideOfficial syllabus reference

What's Next

Mastering 2D and 3D vectors is a core foundation for all further geometry topics in IB AA SL. This subtopic regularly appears in both Paper 1 and Paper 2, often combined with trigonometry and coordinate geometry to create multi-part, multi-mark questions. The skills you learn here will also support any further study of mathematics, including vector calculus and 3D geometry for IB AA HL. Consistent practice of solving intersection and angle problems will build the speed and accuracy you need for exam day.