Vectors

Q: What is the fundamental difference between a scalar and a vector?

A scalar has only magnitude (size), whereas a vector has both magnitude and direction.

Q: How is a 2D vector represented in column form and in unit vector form?

In column form as \(\begin{pmatrix} x \\ y \end{pmatrix}\) and in unit vector form as \(x\mathbf{i} + y\mathbf{j}\), where \(\mathbf{i} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\) and \(\mathbf{j} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\).

Q: How are vector addition, vector subtraction, and scalar multiplication carried out algebraically?

By adding, subtracting, or multiplying the corresponding horizontal and vertical components independently.

Q: What is the algebraic condition for two vectors to be parallel, and how do you prove three points are collinear?

Two vectors are parallel if \(\mathbf{a} = k\mathbf{b}\) for a scalar \(k \neq 0\). Three points \(A, B, C\) are collinear if \(\vec{AB} = k\vec{BC}\) and they share a common point.

Q: What is a position vector and how do you find the displacement vector \(\vec{AB}\) between two points?

A position vector \(\mathbf{a} = \vec{OA}\) gives the position of point \(A\) relative to origin \(O\); the displacement is \(\vec{AB} = \mathbf{b} - \mathbf{a}\).

Q: How do you calculate the magnitude of a vector \(\mathbf{a} = x\mathbf{i} + y\mathbf{j}\)?

\(|\mathbf{a}| = \sqrt{x^2 + y^2}\)

Q: How do you find a unit vector in the direction of a given non-zero vector \(\mathbf{a}\)?

Divide the vector by its magnitude: \(\mathbf{\hat{a}} = \frac{\mathbf{a}}{|\mathbf{a}|} = \frac{1}{\sqrt{x^2 + y^2}}\mathbf{a}\).

Q: How do you determine the direction angle or bearing of a 2D vector?

Use \(\tan\theta = \left|\frac{y}{x}\right|\) with a sketch to find the angle relative to an axis, or measure clockwise from North for a bearing.

Q: How do you find the position vector of a point dividing a line segment \(PQ\) in the ratio \(\lambda : \mu\)?

\(\vec{OR} = \vec{OP} + \frac{\lambda}{\lambda + \mu}\vec{PQ} = \frac{\mu\mathbf{p} + \lambda\mathbf{q}}{\lambda + \mu}\)

Mathematics 2210 · CCEA A Level · 10 min read · Updated: 31 Aug 2026

Welcome to 2D Vectors for AS 1 Pure Mathematics

Welcome to your study notes on Vectors! Vectors are one of the most powerful and intuitive tools in mathematics. Whether you are navigating a ship across the Irish Sea, designing computer graphics in video games, or calculating forces on a bridge, vectors provide the language to describe movement, direction, and position.

In CCEA AS 1: Pure Mathematics, we focus entirely on two-dimensional (2D) vectors. Don't worry if vectors seemed confusing at GCSE—we will break down every single concept step-by-step from the very beginning!

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1. What is a Vector? Representation & Notation

Scalars vs. Vectors

In mathematics and physics, quantities fall into two categories:

1. Scalar: A quantity that has magnitude (size) only. Examples include mass (\(5\text{ kg}\)), distance (\(10\text{ m}\)), temperature (\(18^\circ\text{C}\)), and time (\(12\text{ s}\)).
2. Vector: A quantity that has both magnitude (size) and direction. Examples include displacement (\(10\text{ m}\) North-East), velocity (\(20\text{ m s}^{-1}\) at \(045^\circ\)), and force (\(50\text{ N}\) downwards).

Vector Notation in CCEA Exams

Exam papers represent vectors in distinct ways, and understanding the notation is vital:

• In printed exam questions, vectors are shown in bold lowercase type, such as \(\mathbf{a}\), \(\mathbf{b}\), or \(\mathbf{r}\).
• In your handwritten exam working, you cannot easily write in bold, so you should underline your vector, such as \(\underline{a}\), or put an arrow on top, such as \(\vec{a}\).
• A directed line segment from a starting point \(A\) to an ending point \(B\) is written as \(\vec{AB}\) or \(\mathbf{\vec{AB}}\). The arrow points from start to finish.

Two Ways to Write a 2D Vector

Every 2D vector consists of a horizontal component (along the \(x\)-axis) and a vertical component (along the \(y\)-axis):

1. Column Vector Form:
\(\begin{pmatrix} x \\ y \end{pmatrix}\)
The top number \(x\) represents horizontal displacement (positive = right, negative = left).
The bottom number \(y\) represents vertical displacement (positive = up, negative = down).

2. Unit Vector Form (\(\mathbf{i}, \mathbf{j}\) notation):
\(x\mathbf{i} + y\mathbf{j}\)
Here, \(\mathbf{i}\) is a unit vector of length \(1\) in the positive \(x\)-direction: \(\mathbf{i} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\).
\(\mathbf{j}\) is a unit vector of length \(1\) in the positive \(y\)-direction: \(\mathbf{j} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\).

Example: A movement of \(4\) units right and \(3\) units down can be written as \(\begin{pmatrix} 4 \\ -3 \end{pmatrix}\) or \(4\mathbf{i} - 3\mathbf{j}\).

Key Takeaway: Vectors possess magnitude and direction. \(\begin{pmatrix} x \\ y \end{pmatrix}\) and \(x\mathbf{i} + y\mathbf{j}\) mean the exact same thing; you can freely switch between them depending on what feels clearest!

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2. Fundamental Operations on Vectors

Vector Addition and Subtraction

To add or subtract vectors algebraically, simply add or subtract their corresponding horizontal and vertical components:

\(\begin{pmatrix} x_1 \\ y_1 \end{pmatrix} + \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} x_1 + x_2 \\ y_1 + y_2 \end{pmatrix}\)
\(\begin{pmatrix} x_1 \\ y_1 \end{pmatrix} - \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} x_1 - x_2 \\ y_1 - y_2 \end{pmatrix}\)

In \(\mathbf{i}, \mathbf{j}\) form: \((x_1\mathbf{i} + y_1\mathbf{j}) + (x_2\mathbf{i} + y_2\mathbf{j}) = (x_1 + x_2)\mathbf{i} + (y_1 + y_2)\mathbf{j}\).

Geometric Meaning: The Triangle & Parallelogram Laws

Geometrically, adding two vectors means placing the tail of the second vector at the nose (head) of the first vector. The resultant vector goes directly from the very start to the very end:

\(\vec{AC} = \vec{AB} + \vec{BC}\)

Analogy: If you walk from point \(A\) to point \(B\), and then from point \(B\) to point \(C\), your overall journey (displacement) is directly from \(A\) to \(C\).

Scalar Multiplication

Multiplying a vector by a real number (a scalar) \(k\) scales each component:

\(k\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} kx \\ ky \end{pmatrix}\)

• If \(k > 0\), the magnitude changes by a factor of \(k\), but the direction remains identical.
• If \(k < 0\), the vector is scaled and its direction is completely reversed (\(180^\circ\) flip).

Parallel Vectors and Collinearity

• Parallel Vectors: Two non-zero vectors \(\mathbf{a}\) and \(\mathbf{b}\) are parallel if and only if one is a scalar multiple of the other:
\(\mathbf{a} = k\mathbf{b}\) where \(k\) is a non-zero constant (\(k \in \mathbb{R}\)).

• Collinear Points: Three distinct points \(A\), \(B\), and \(C\) lie on the same straight line (are collinear) if:
1. \(\vec{AB} = k\vec{BC}\) (or \(\vec{AB} = \lambda\vec{AC}\)), proving the vectors are parallel, and
2. They share a common point (such as point \(B\) or point \(A\)).

Key Takeaway: Parallel vectors are scalar multiples of each other (\(\mathbf{a} = k\mathbf{b}\)). To prove three points are on a single straight line, prove their vector segments are parallel and share a common point.

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3. Position Vectors & Displacement Between Points

What is a Position Vector?

A position vector specifies the location of a point relative to a fixed origin \(O(0, 0)\).
If point \(A\) has coordinates \((x_1, y_1)\), its position vector relative to the origin is:

\(\vec{OA} = \mathbf{a} = x_1\mathbf{i} + y_1\mathbf{j} = \begin{pmatrix} x_1 \\ y_1 \end{pmatrix}\)

Displacement Between Two Points (\(\vec{AB}\))

To travel from point \(A\) to point \(B\), you can go via the origin \(O\): first from \(A\) to \(O\), then from \(O\) to \(B\):

\(\vec{AB} = \vec{AO} + \vec{OB} = -\vec{OA} + \vec{OB} = \vec{OB} - \vec{OA}\)

\(\vec{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix}\)

Memory Aid for Displacement:

"Destination minus Origin" or "End minus Start": To find \(\vec{AB}\), take the position vector of the end point \(\mathbf{b}\) and subtract the position vector of the start point \(\mathbf{a}\).

Worked Example: Finding Displacement

Points \(P\) and \(Q\) have position vectors \(\mathbf{p} = 3\mathbf{i} - 2\mathbf{j}\) and \(\mathbf{q} = -5\mathbf{i} + 4\mathbf{j}\) relative to origin \(O\). Find the vector \(\vec{PQ}\).

Solution:
\(\vec{PQ} = \mathbf{q} - \mathbf{p}\)
\(\vec{PQ} = (-5\mathbf{i} + 4\mathbf{j}) - (3\mathbf{i} - 2\mathbf{j})\)
\(\vec{PQ} = (-5 - 3)\mathbf{i} + (4 - (-2))\mathbf{j} = -8\mathbf{i} + 6\mathbf{j} = \begin{pmatrix} -8 \\ 6 \end{pmatrix}\)

Key Takeaway: Never confuse coordinates with vectors. The displacement vector from \(A\) to \(B\) is always \(\vec{AB} = \mathbf{b} - \mathbf{a}\).

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4. Magnitude, Distance, Unit Vectors, and Direction

Magnitude (Length) of a Vector

Using Pythagoras' theorem, the magnitude (or length) of vector \(\mathbf{a} = x\mathbf{i} + y\mathbf{j} = \begin{pmatrix} x \\ y \end{pmatrix}\) is denoted by \(|\mathbf{a}|\):

\(|\mathbf{a}| = \sqrt{x^2 + y^2}\)

Distance Between Two Points

The straight-line distance between two points \(A(x_1, y_1)\) and \(B(x_2, y_2)\) is the magnitude of the displacement vector \(\vec{AB}\):

\(d(A, B) = |\vec{AB}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Unit Vectors

A unit vector is defined as any vector with a magnitude of exactly \(1\).
To create a unit vector in the same direction as any non-zero vector \(\mathbf{a}\), we divide the vector by its own magnitude. This is denoted \(\mathbf{\hat{a}}\) ("a-hat"):

\(\mathbf{\hat{a}} = \frac{\mathbf{a}}{|\mathbf{a}|} = \frac{1}{\sqrt{x^2 + y^2}}\begin{pmatrix} x \\ y \end{pmatrix}\)

Worked Example: Finding a Unit Vector

Find a unit vector in the direction of \(\mathbf{v} = 6\mathbf{i} - 8\mathbf{j}\).

Step 1: Calculate the magnitude of \(\mathbf{v}\):
\(|\mathbf{v}| = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10\)

Step 2: Divide \(\mathbf{v}\) by its magnitude:
\(\mathbf{\hat{v}} = \frac{1}{10}(6\mathbf{i} - 8\mathbf{j}) = \frac{6}{10}\mathbf{i} - \frac{8}{10}\mathbf{j} = \frac{3}{5}\mathbf{i} - \frac{4}{5}\mathbf{j}\) (or \(\begin{pmatrix} 0.6 \\ -0.8 \end{pmatrix}\))

Direction and Angles

The direction of a vector is described by the angle it makes with standard axes:

• Angle with the positive \(x\)-axis (\(\mathbf{i}\) direction):
For \(\mathbf{a} = x\mathbf{i} + y\mathbf{j}\), the basic angle satisfies \(\tan\theta = \left|\frac{y}{x}\right|\).
Always draw a quick sketch to ensure you place \(\theta\) in the correct quadrant!

• Bearings:
A bearing is measured clockwise from North (the positive \(y\)-axis / \(\mathbf{j}\) direction) and given as a 3-figure angle (e.g., \(060^\circ\), \(225^\circ\)).

Key Takeaway: Magnitude is found via Pythagoras: \(|\mathbf{a}| = \sqrt{x^2+y^2}\). A unit vector in the direction of \(\mathbf{a}\) is simply \(\frac{\mathbf{a}}{|\mathbf{a}|}\).

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5. Geometric Problems & Ratio Division

Dividing a Line Segment in a Given Ratio

Suppose point \(R\) lies on the line segment connecting point \(P\) to point \(Q\), dividing the line in the ratio \(\lambda : \mu\) (or \(m : n\)).

To find the position vector \(\vec{OR}\) (or \(\mathbf{r}\)):

1. The fraction of the way along the line segment from \(P\) to \(Q\) is \(\frac{\lambda}{\lambda + \mu}\).
2. Start at \(P\) and travel this fractional displacement along \(\vec{PQ}\):

\(\vec{OR} = \vec{OP} + \frac{\lambda}{\lambda + \mu}\vec{PQ}\)

Substituting \(\vec{PQ} = \mathbf{q} - \mathbf{p}\):

\(\vec{OR} = \mathbf{p} + \frac{\lambda}{\lambda + \mu}(\mathbf{q} - \mathbf{p}) = \frac{\mu\mathbf{p} + \lambda\mathbf{q}}{\lambda + \mu}\)

Special Case (Midpoint): When \(M\) is the midpoint, the ratio is \(1:1\):
\(\vec{OM} = \frac{\mathbf{p} + \mathbf{q}}{2}\)

Worked Example: Ratio Division

Point \(A\) has position vector \(\mathbf{a} = 2\mathbf{i} + 5\mathbf{j}\) and point \(B\) has position vector \(\mathbf{b} = 11\mathbf{i} - 7\mathbf{j}\). Point \(P\) divides the line segment \(AB\) in the ratio \(1 : 2\). Find the position vector of \(P\).

Solution:
The total number of parts is \(1 + 2 = 3\). So, point \(P\) is \(\frac{1}{3}\) of the way from \(A\) to \(B\).

Step 1: Find displacement \(\vec{AB}\):
\(\vec{AB} = \mathbf{b} - \mathbf{a} = (11\mathbf{i} - 7\mathbf{j}) - (2\mathbf{i} + 5\mathbf{j}) = 9\mathbf{i} - 12\mathbf{j}\)

Step 2: Find \(\vec{OP}\):
\(\vec{OP} = \vec{OA} + \frac{1}{3}\vec{AB}\)
\(\vec{OP} = (2\mathbf{i} + 5\mathbf{j}) + \frac{1}{3}(9\mathbf{i} - 12\mathbf{j})\)
\(\vec{OP} = (2\mathbf{i} + 5\mathbf{j}) + (3\mathbf{i} - 4\mathbf{j}) = 5\mathbf{i} + \mathbf{j}\)

Geometric Proofs Using Vectors

Vectors provide elegant ways to prove geometric properties:

• Parallelogram: To prove quadrilateral \(ABCD\) is a parallelogram, show that opposite sides are equal vectors: \(\vec{AB} = \vec{DC}\) (which proves they are equal in length and parallel).
• Trapezium: To prove quadrilateral \(ABCD\) is a trapezium, show that one pair of opposite sides are parallel scalar multiples: \(\vec{AB} = k\vec{DC}\) where \(k \neq 1\).

Key Takeaway: When dividing a line in ratio \(\lambda : \mu\), use the fraction \(\frac{\lambda}{\lambda + \mu}\). Never just divide by \(\lambda\) or \(\mu\) alone!

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6. Common Pitfalls to Avoid in CCEA AS 1 Exams

Examiner reports highlight recurring errors that cost students easy marks. Review these carefully:

❌ Pitfall 1: Incorrect Displacement Direction (\(\vec{AB} = \mathbf{a} - \mathbf{b}\))
Correction: \(\vec{AB}\) is strictly \(\mathbf{b} - \mathbf{a}\) ("destination minus origin"). Writing \(\mathbf{a} - \mathbf{b}\) gives the vector in the opposite direction (\(\vec{BA}\)).

❌ Pitfall 2: Forgetting to Normalise a Unit Vector
Correction: If asked to find a unit vector parallel to \(\mathbf{a}\), remember to divide by its magnitude: \(\frac{\mathbf{a}}{|\mathbf{a}|}\). Leaving it as just \(\mathbf{a}\) gives a vector of magnitude \(|\mathbf{a}|\), not \(1\).

❌ Pitfall 3: Blindly using \(\arctan\left(\frac{y}{x}\right)\) for Directions & Bearings
Correction: A calculator will return negative angles or acute angles that may belong in another quadrant. Always draw a small coordinate diagram showing the vector components to determine the exact angle from the \(x\)-axis or the 3-figure bearing clockwise from North.

❌ Pitfall 4: Misinterpreting Ratios
Correction: If a point divides a line in the ratio \(2 : 3\), the fractions of the line are \(\frac{2}{5}\) and \(\frac{3}{5}\), not \(\frac{2}{3}\).

❌ Pitfall 5: Missing Vector Notation
Correction: In intermediate algebra, keep vectors clearly distinguished from scalars by using underlines (\(\underline{a}\)) or arrow notations (\(\vec{a}\)). Do not drop notation and treat vectors as plain scalars.

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Quick Summary & Formula Checklist

• Components: \(\mathbf{a} = \begin{pmatrix} x \\ y \end{pmatrix} = x\mathbf{i} + y\mathbf{j}\)
• Displacement: \(\vec{AB} = \vec{OB} - \vec{OA} = \mathbf{b} - \mathbf{a}\)
• Magnitude: \(|\mathbf{a}| = \sqrt{x^2 + y^2}\)
• Unit Vector: \(\mathbf{\hat{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}\)
• Parallel Condition: \(\mathbf{a} = k\mathbf{b}\) for non-zero scalar \(k\)
• Collinear Condition: \(\vec{AB} = k\vec{BC}\) with shared point \(B\)
• Ratio Division: \(\vec{OR} = \vec{OP} + \frac{\lambda}{\lambda + \mu}\vec{PQ}\)

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the fundamental difference between a scalar and a vector?

A scalar has only magnitude (size), whereas a vector has both magnitude and direction.

How is a 2D vector represented in column form and in unit vector form?

In column form as \(\begin{pmatrix} x \\ y \end{pmatrix}\) and in unit vector form as \(x\mathbf{i} + y\mathbf{j}\), where \(\mathbf{i} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\) and \(\mathbf{j} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\).

How are vector addition, vector subtraction, and scalar multiplication carried out algebraically?

By adding, subtracting, or multiplying the corresponding horizontal and vertical components independently.

What is the algebraic condition for two vectors to be parallel, and how do you prove three points are collinear?

Two vectors are parallel if \(\mathbf{a} = k\mathbf{b}\) for a scalar \(k \neq 0\). Three points \(A, B, C\) are collinear if \(\vec{AB} = k\vec{BC}\) and they share a common point.

What is a position vector and how do you find the displacement vector \(\vec{AB}\) between two points?

A position vector \(\mathbf{a} = \vec{OA}\) gives the position of point \(A\) relative to origin \(O\); the displacement is \(\vec{AB} = \mathbf{b} - \mathbf{a}\).

How do you calculate the magnitude of a vector \(\mathbf{a} = x\mathbf{i} + y\mathbf{j}\)?

\(|\mathbf{a}| = \sqrt{x^2 + y^2}\)

How do you find a unit vector in the direction of a given non-zero vector \(\mathbf{a}\)?

Divide the vector by its magnitude: \(\mathbf{\hat{a}} = \frac{\mathbf{a}}{|\mathbf{a}|} = \frac{1}{\sqrt{x^2 + y^2}}\mathbf{a}\).

How do you determine the direction angle or bearing of a 2D vector?

Use \(\tan\theta = \left|\frac{y}{x}\right|\) with a sketch to find the angle relative to an axis, or measure clockwise from North for a bearing.

How do you find the position vector of a point dividing a line segment \(PQ\) in the ratio \(\lambda : \mu\)?

\(\vec{OR} = \vec{OP} + \frac{\lambda}{\lambda + \mu}\vec{PQ} = \frac{\mu\mathbf{p} + \lambda\mathbf{q}}{\lambda + \mu}\)

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