What is a vector in the context of IB MYP Mathematics?

In IB MYP Mathematics, a vector is a mathematical entity that has both magnitude and direction. It is often represented as an arrow in geometry, where the length of the arrow indicates the magnitude and the direction of the arrow shows the direction. Vectors are used to model physical quantities like force and velocity.

How do you add and subtract vectors in Geometry and Trigonometry?

To add vectors, you use the tip-to-tail method, where the tail of the second vector is placed at the tip of the first vector. The resultant vector is drawn from the tail of the first vector to the tip of the second vector. For subtraction, you add the negative of the vector you want to subtract. This involves reversing the direction of the vector being subtracted and then adding it using the tip-to-tail method.

What is the difference between scalar and vector quantities?

Scalar quantities have only magnitude, such as temperature or mass, while vector quantities have both magnitude and direction, such as displacement or velocity. In the IB MYP curriculum, understanding the distinction between these two types of quantities is crucial for solving problems in Geometry and Trigonometry.

How do you calculate the magnitude of a vector?

The magnitude of a vector can be calculated using the Pythagorean theorem. For a vector represented as (x, y) in a 2D plane, the magnitude is given by the formula √(x² + y²). This formula is derived from the distance formula in coordinate geometry and is essential for solving vector problems in the IB MYP Mathematics curriculum.

What role do vectors play in real-world applications?

Vectors are crucial in various real-world applications, including physics, engineering, and computer graphics. They are used to represent quantities like force, velocity, and acceleration. In the IB MYP curriculum, understanding vectors helps students model and solve real-world problems, enhancing their comprehension of Geometry and Trigonometry concepts.

Why is "Computing $\overrightarrow{AB}$ as $A - B$." a common mistake in Vectors?

Why it happens: Reading order. How to avoid it: AB goes FROM A TO B, so it's B - A (end minus start).

Why is "Forgetting the square root in magnitude." a common mistake in Vectors?

Why it happens: Quick computation. How to avoid it: Magnitude = √(x^2 + y^2). Without the root you have the squared magnitude.

Geometry and TrigonometryIB MYP Maths Extended Level

Vectors

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Vectors — IB MYP Mathematics (Extended): magnitude and direction

Vectors describe quantities with magnitude AND direction. This MYP Mathematics Extended note covers vector notation, addition, scalar multiplication, magnitude, and basic applications.

What you’ll learn

Mapped to the IB MYP Mathematics subject guide (2026 onwards).

MYP Mathematics A — Add, subtract, and scale vectors in column form.
MYP Mathematics A — Find the magnitude of a vector.
MYP Mathematics C — Express vectors clearly in standard notation.
MYP Mathematics D — Apply vectors to displacement, velocity, and force.

What is a vector?

Magnitude AND direction — represented by an arrow or column.

A vector has both:

Magnitude (size / length).
Direction.

Compare with a scalar — just a number, no direction (e.g. temperature, mass, time).

Notation. Three common ways to denote a vector:

Bold letter: $\mathbf{a}$ .
Underlined letter (handwritten): $\underline{a}$ .
Arrow notation: $\overrightarrow{AB}$ (from point $A$ to point $B$ ).

Column form (most common at MYP Extended):

$\mathbf{a} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$

means '3 units right, 2 units down'. Top number = x-displacement, bottom = y-displacement.

Magnitude (length) of a vector $\mathbf{a} = \begin{pmatrix} x \\ y \end{pmatrix}$ :

$|\mathbf{a}| = \sqrt{x^2 + y^2}.$

This is Pythagoras — same as the distance formula in coordinate geometry.

Worked example. $\mathbf{a} = \begin{pmatrix} 4 \\ -3 \end{pmatrix}$ . Find $|\mathbf{a}|$ .

$|\mathbf{a}| = \sqrt{16 + 9} = \sqrt{25} = 5$ .

The vector $\mathbf{a} = (4, -3)$ moves 4 right and 3 down. Its magnitude is $\sqrt{4^2 + 3^2} = 5$.

Vector = magnitude + direction.
Column form: $\begin{pmatrix} x \\ y \end{pmatrix}$ .
Magnitude: $\sqrt{x^2 + y^2}$ .

Vector arithmetic

Add components; scale by scalars.

Vector addition — add component-wise:

$\begin{pmatrix} a \\ b \end{pmatrix} + \begin{pmatrix} c \\ d \end{pmatrix} = \begin{pmatrix} a + c \\ b + d \end{pmatrix}.$

Geometrically: place vectors head-to-tail. The resultant goes from start of first to end of second.

Subtraction is addition of the NEGATIVE:

$\mathbf{a} - \mathbf{b} = \mathbf{a} + (-\mathbf{b}) = \begin{pmatrix} a - c \\ b - d \end{pmatrix}.$

Scalar multiplication — multiply each component:

$k \mathbf{a} = \begin{pmatrix} ka \\ kb \end{pmatrix}.$

$k > 0$ : same direction, magnitude scaled by $k$ .
$k < 0$ : opposite direction, magnitude scaled by $|k|$ .
$k = 0$ : zero vector.

Worked example. $\mathbf{a} = \begin{pmatrix} 3 \\ -1 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -2 \\ 4 \end{pmatrix}$ .

$\mathbf{a} + \mathbf{b} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$ .
$\mathbf{a} - \mathbf{b} = \begin{pmatrix} 5 \\ -5 \end{pmatrix}$ .
$3\mathbf{a} = \begin{pmatrix} 9 \\ -3 \end{pmatrix}$ .
$\mathbf{a} + 2\mathbf{b} = \begin{pmatrix} 3 \\ -1 \end{pmatrix} + \begin{pmatrix} -4 \\ 8 \end{pmatrix} = \begin{pmatrix} -1 \\ 7 \end{pmatrix}$ .

Add/subtract: component-wise.
$k \mathbf{a}$ : multiply each component by $k$ .
$|\mathbf{a}|$ : Pythagoras on the components.

Vector applications

Displacement, velocity, force.

Vectors appear whenever 'direction matters':

Displacement. $\overrightarrow{AB}$ is the vector from point $A$ to point $B$ . Computed as $\begin{pmatrix} x_B - x_A \\ y_B - y_A \end{pmatrix}$ .

Velocity. A car moving at $60\,\text{km/h}$ EAST has velocity vector $\begin{pmatrix} 60 \\ 0 \end{pmatrix}$ . North-east at $60$ : $\begin{pmatrix} 60 \cos 45° \\ 60 \sin 45° \end{pmatrix} = \begin{pmatrix} 42.4 \\ 42.4 \end{pmatrix}$ .

Force. Two ropes pull a sled. The combined force is the VECTOR SUM of the two individual forces — this is why ropes pulling at an angle don't simply add up to the sum of their magnitudes.

Worked example. A boat sails 8 km east then 6 km north. Find the displacement vector from start, its magnitude, and its direction.

Displacement: $\begin{pmatrix} 8 \\ 6 \end{pmatrix}$ .
Magnitude: $\sqrt{64 + 36} = \sqrt{100} = 10$ km.
Direction (bearing): $\tan \theta = 8/6 \Rightarrow \theta \approx 53°$ from north (measuring east of north).

This is a standard MYP criterion-D problem — translate a real situation into a vector and interpret the answer.

Displacement: $\overrightarrow{AB} = B - A$ .
Velocity, force: also vectors.
Real contexts: bearings, navigation, mechanics.

How it’s examined

Criterion A: vector operations and magnitude. Criterion C: well-formatted column vectors. Criterion D: real bearing / displacement problems.

Sources: IB MYP Mathematics subject guide (IBO, official). Last reviewed 2026-05-25.

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Step-by-step worked examples — Vectors

Step-by-step solutions to past-paper-style questions on vectors, written exactly the way a tutor would explain them at the board.

1Find the magnitude
Getting started• magnitude
Question
Find $|\mathbf{a}|$ for $\mathbf{a} = \begin{pmatrix} -5 \\ 12 \end{pmatrix}$ .
Step-by-step solution
1. Step 1
  $|\mathbf{a}| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169}$ .
2. Step 2
  $= 13$ .
Answer
$13$
2Add vectors
Building confidence• addition
Question
$\mathbf{a} = \begin{pmatrix} 4 \\ -1 \end{pmatrix}$ , $\mathbf{b} = \begin{pmatrix} 2 \\ 6 \end{pmatrix}$ . Find $\mathbf{a} + \mathbf{b}$ .
Step-by-step solution
1. Step 1
  Add component-wise: $4 + 2 = 6$ ; $-1 + 6 = 5$ .
2. Step 2
  $\mathbf{a} + \mathbf{b} = \begin{pmatrix} 6 \\ 5 \end{pmatrix}$ .
Answer
$\begin{pmatrix} 6 \\ 5 \end{pmatrix}$
3Scalar multiplication
Building confidence• scalar
Question
Compute $3 \mathbf{a} - 2\mathbf{b}$ where $\mathbf{a} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}$ .
Step-by-step solution
1. Step 1
  $3\mathbf{a} = \begin{pmatrix} 3 \\ 12 \end{pmatrix}$ and $2\mathbf{b} = \begin{pmatrix} -4 \\ 6 \end{pmatrix}$ .
2. Step 2
  Subtract: $\begin{pmatrix} 3 - (-4) \\ 12 - 6 \end{pmatrix} = \begin{pmatrix} 7 \\ 6 \end{pmatrix}$ .
Answer
$\begin{pmatrix} 7 \\ 6 \end{pmatrix}$
4Find displacement vector
Stretch• displacement
Question
Points $A(2, 1)$ and $B(7, 13)$ . Find $\overrightarrow{AB}$ and its magnitude.
Step-by-step solution
1. Step 1
  $\overrightarrow{AB} = \begin{pmatrix} 7 - 2 \\ 13 - 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 12 \end{pmatrix}$ .
2. Step 2
  $|\overrightarrow{AB}| = \sqrt{25 + 144} = 13$ .
Answer
$\overrightarrow{AB} = \begin{pmatrix} 5 \\ 12 \end{pmatrix}$ , magnitude $13$ .

Key Definitions and Keywords — Vectors

Definitions to memorise and the exact keywords mark schemes credit for vectors answers — sharpened from recent examiner reports for the 2026 IB MYP Mathematics (Extended) sitting.

Vector
Examiner keyword
A quantity with both magnitude and direction.
Magnitude of a vector
Examiner keyword
$|\mathbf{a}| = \sqrt{x^2 + y^2}$ — the length of the vector.
Scalar
Examiner keyword
A number (no direction), e.g. mass or time.
Displacement vector
$\overrightarrow{AB}$ — the vector from point $A$ to point $B$ . Computed as $B - A$ .

Common Mistakes and Misconceptions — Vectors

The traps other students keep falling into on vectors questions — taken from recent IB MYP Mathematics (Extended) examiner reports and mark schemes — and how to avoid them.

✕Computing $\overrightarrow{AB}$ as $A - B$ .
Why it happens
Reading order.
How to avoid it
$\overrightarrow{AB}$ goes FROM A TO B, so it's $B - A$ (end minus start).
✕Forgetting the square root in magnitude.
Why it happens
Quick computation.
How to avoid it
Magnitude = $\sqrt{x^2 + y^2}$ . Without the root you have the squared magnitude.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

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Vectors — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Vectors

Short Study Notes in the form of Slides

Short Study Notes — Vectors

Detailed Notes

What you’ll learn

How it’s examined

1Find the magnitude

2Add vectors

3Scalar multiplication

4Find displacement vector

Vector

Magnitude of a vector

Scalar

Displacement vector

✕Computing AB→\overrightarrow{AB}AB as A−BA - BA−B.

✕Forgetting the square root in magnitude.

Practice questions

Past paper style quiz

Assess your understanding

Vectors — frequently asked questions

✕Computing $\overrightarrow{AB}$ as $A - B$ .