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.

Mathematics 2210 · CCEA AS Level · 6 min read · Updated: 31 Aug 2026

Welcome to 2D Vectors for AS 1 Pure Mathematics

Welcome to your study guide on Vectors for CCEA AS Level Pure Mathematics! Whether you find math straightforward or sometimes find it a bit daunting, this guide breaks down every core concept into clear, step-by-step pieces.

In everyday life, some quantities are completely described by just a single number (a magnitude), like your mass or the temperature outside. These are called scalars. But other quantities—like the displacement of an aircraft or a gust of wind—need both a magnitude (size) and a direction. That is exactly what a vector is!

Real-World Analogy: Imagine telling a friend, "Walk 5 miles." They won't know where they will end up because you didn't give a direction! If you say, "Walk 5 miles due North," you have just given them a vector.

1. Vector Representations and Notations

At AS Level, we focus on two-dimensional (2D) vectors that exist in the Cartesian \(xy\)-plane.

Unit Vector Form (\(\mathbf{i}\) and \(\mathbf{j}\))

We define two standard perpendicular unit vectors (vectors of length \(1\)):

• \(\mathbf{i}\) is a unit vector of length \(1\) in the direction of the positive \(x\)-axis.
• \(\mathbf{j}\) is a unit vector of length \(1\) in the direction of the positive \(y\)-axis.

Any vector \(\mathbf{r}\) moving \(x\) units horizontally and \(y\) units vertically is written as:

\(\mathbf{r} = x\mathbf{i} + y\mathbf{j}\)

Column Vector Form

You can also write vectors as a vertical column:

\(\mathbf{r} = \begin{pmatrix} x \\ y \end{pmatrix}\)

The top number is the horizontal change (\(x\)), and the bottom number is the vertical change (\(y\)).

Handwritten Notation Tip

In printed textbooks and exam papers, vectors are printed in bold (e.g., \(\mathbf{a}\), \(\mathbf{i}\), \(\mathbf{j}\)). In your handwritten exam scripts, you cannot write in bold, so you must underline your vector symbols to distinguish them from regular numbers:

\(\underline{a} = x\underline{i} + y\underline{j}\)

Key Takeaway: A vector \(\begin{pmatrix} 3 \\ -4 \end{pmatrix}\) is identical to \(3\mathbf{i} - 4\mathbf{j}\). Always remember to underline vectors in your written exam solutions!

2. Vector Operations

Addition and Subtraction

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

• Column form: \(\begin{pmatrix} x_1 \\ y_1 \end{pmatrix} \pm \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} x_1 \pm x_2 \\ y_1 \pm y_2 \end{pmatrix}\)
• Component form: \((x_1\mathbf{i} + y_1\mathbf{j}) \pm (x_2\mathbf{i} + y_2\mathbf{j}) = (x_1 \pm x_2)\mathbf{i} + (y_1 \pm y_2)\mathbf{j}\)

Scalar Multiplication

Multiplying a vector by a normal number (a scalar \(k\)) scales its length. If \(k\) is negative, it also reverses its direction:

\(k(x\mathbf{i} + y\mathbf{j}) = (kx)\mathbf{i} + (ky)\mathbf{j}\) or \(k\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} kx \\ ky \end{pmatrix}\)

Parallel Vectors and Collinearity

Two 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 scalar)

Collinear Points: Three distinct points \(A\), \(B\), and \(C\) lie on the same straight line (are collinear) if:

1. The vector \(\vec{AB}\) is parallel to \(\vec{BC}\) (meaning \(\vec{AB} = k\vec{BC}\)), and
2. Both vectors share the common point \(B\).

Key Takeaway: If two vectors have proportional components, they are parallel. If they also share a point, the points are collinear.

3. Magnitude, Unit Vectors, and Bearings

Magnitude (Length or Modulus)

The magnitude of a vector \(\mathbf{a} = x\mathbf{i} + y\mathbf{j} = \begin{pmatrix} x \\ y \end{pmatrix}\) is written as \(|\mathbf{a}|\). By Pythagoras' Theorem, it is the straight-line distance:

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

Note: Magnitude is always a positive scalar value.

Unit Vector in the Direction of \(\mathbf{a}\)

A unit vector has a magnitude of exactly \(1\). To find a unit vector \(\hat{\mathbf{a}}\) in the direction of \(\mathbf{a}\), divide the vector by its own magnitude:

\(\hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}\)

Direction and 3-Figure Bearings

The direction angle \(\theta\) made with the positive \(x\)-axis satisfies:

\(\tan\theta = \left|\frac{y}{x}\right|\)

In many CCEA exam questions, you may be asked to find the direction as a 3-figure compass bearing.

• Bearings are measured clockwise from North (the positive \(y\)-direction, \(\mathbf{j}\)).
• Step 1: Sketch the vector in the correct quadrant.
• Step 2: Use basic trigonometry to find the acute angle with a reference axis.
• Step 3: Calculate the full clockwise angle from North and write it as a 3-digit number (e.g., \(045^\circ\), \(120^\circ\), \(285^\circ\)).

Key Takeaway: Magnitude is found using Pythagoras (\(\sqrt{x^2+y^2}\)). A unit vector is simply the vector divided by its magnitude.

4. Position Vectors and Dividing a Line Segment

Position Vectors

The position vector of a point \(P(x, y)\) is the vector displacement from the origin \(O(0,0)\) to \(P\):

\(\vec{OP} = \mathbf{p} = x\mathbf{i} + y\mathbf{j} = \begin{pmatrix} x \\ y \end{pmatrix}\)

Displacement Between Two Points

To travel from point \(A\) (position vector \(\mathbf{a}\)) to point \(B\) (position vector \(\mathbf{b}\)), we go via the origin:

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

Memory Trick: To find \(\vec{AB}\), think "End minus Start" (\(\mathbf{b} - \mathbf{a}\)).

The distance between points \(A\) and \(B\) is the magnitude of this displacement:

\(|\vec{AB}| = |\mathbf{b} - \mathbf{a}| = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}\)

Ratio Division of a Line Segment

Suppose point \(R\) divides the line segment \(PQ\) internally in the ratio \(m : n\).

This means the distance from \(P\) to \(R\) represents \(\frac{m}{m+n}\) of the total vector \(\vec{PQ}\).

The position vector of \(R\) is given by:

\(\vec{OR} = \vec{OP} + \frac{m}{m+n}\vec{PQ}\)

\(\vec{OR} = \mathbf{p} + \frac{m}{m+n}(\mathbf{q} - \mathbf{p}) = \frac{n\mathbf{p} + m\mathbf{q}}{m+n}\)

Step-by-Step Example:
Find the position vector of point \(R\) which divides the line joining \(P(1, 2)\) and \(Q(7, 11)\) in the ratio \(1 : 2\).
1. Here, \(\mathbf{p} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}\), \(\mathbf{q} = \begin{pmatrix} 7 \\ 11 \end{pmatrix}\), \(m = 1\), and \(n = 2\).
2. Total parts = \(m + n = 1 + 2 = 3\).
3. \(\vec{PQ} = \mathbf{q} - \mathbf{p} = \begin{pmatrix} 7 - 1 \\ 11 - 2 \end{pmatrix} = \begin{pmatrix} 6 \\ 9 \end{pmatrix}\).
4. \(\vec{OR} = \mathbf{p} + \frac{1}{3}\vec{PQ} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + \frac{1}{3}\begin{pmatrix} 6 \\ 9 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 5 \end{pmatrix}\).

Key Takeaway: \(\vec{AB} = \mathbf{b} - \mathbf{a}\). For ratio division \(m:n\), remember the fraction of the total journey is \(\frac{m}{m+n}\).

5. Common Pitfalls to Avoid in the Exam

• Reversing Displacement: A common error is writing \(\vec{AB} = \mathbf{a} - \mathbf{b}\). Always do Destination minus Origin (\(\mathbf{b} - \mathbf{a}\)).
• Ratio Fractions: When a line is split in the ratio \(2 : 3\), the fraction along the line is \(\frac{2}{2+3} = \frac{2}{5}\), not \(\frac{2}{3}\).
• Bearing Confusions: Do not simply write down the angle from the horizontal \(x\)-axis as your bearing. Bearings must always be measured clockwise from the positive \(y\)-axis (North).
• Forgetting to Underline: Remember to underline vectors (\(\underline{a}\), \(\underline{i}\), \(\underline{j}\)) in your handwritten work.
• Premature Rounding: Keep exact values or surds in your intermediate calculation steps. Unless specified otherwise, give non-exact numerical answers rounded to 3 significant figures.

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 are 2D vectors represented in unit vector form and column vector form?

In unit vector form, {r} = x\mathbf{i} + y\mathbf{j}\), where \(\mathbf{i}\) and \(\mathbf{j}\) are standard unit vectors along the positive \(x\)- and \(y\)-axes; in column vector form, \(\mathbf{r} = \begin{pmatrix} x \\ y \end{pmatrix}\).

How are vector addition, subtraction, and scalar multiplication performed algebraically?

Vectors are added or subtracted by combining corresponding components: \(\begin{pmatrix} x_1 \\ y_1 \end{pmatrix} \pm \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} x_1 \pm x_2 \\ y_1 \pm y_2 \end{pmatrix}\). Scalar multiplication scales each component: \(k\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} kx \\ ky \end{pmatrix}\).

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

Two vectors \(\mathbf{a}\) and \(\mathbf{b}\) are parallel if \(\mathbf{a} = k\mathbf{b}\) for a non-zero scalar \(k\). Points \(A\), \(B\), and \(C\) are collinear if \(\vec{AB} = k\vec{BC}\) and they share the common point \(B\).

How is the magnitude of a 2D vector \(\mathbf{a} = x\mathbf{i} + y\mathbf{j}\) calculated?

The magnitude is calculated using Pythagoras' Theorem: \(|\mathbf{a}| = \sqrt{x^2 + y^2}\).

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

Divide the vector by its magnitude: \(\hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}\).

How is the direction of a vector determined as a 3-figure compass bearing?

Sketch the vector, use trigonometry to find the angle with a reference axis, and determine the clockwise angle measured from North (the positive \(y\)-axis) expressed as a 3-digit number.

How is the displacement vector \(\vec{AB}\) calculated from the position vectors of \(A\) and \(B\)?

\(\vec{AB} = \mathbf{b} - \mathbf{a}\) (position vector of the destination minus position vector of the start).

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

\(\vec{OR} = \mathbf{p} + \frac{m}{m+n}(\mathbf{q} - \mathbf{p}) = \frac{n\mathbf{p} + m\mathbf{q}}{m+n}\).

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