# Vectors in Two Dimensions and Velocity Problems

> CIE IGCSE Additional Mathematics · 2025-2027
> Source: https://www.owlsprep.com/study/cie-0606-u13-vectors-in-two-dimensions-and/

This guide covers all 2D vector content for CIE IGCSE Add Maths 0606, including notation, operations, collinearity, velocity composition, and collision problems, aligned strictly to the 2025–2027 syllabus. No out-of-scope A-Level content is included.

**Prerequisites:** Basic algebra and coordinate geometry; Pythagoras' theorem and right triangle trigonometry

## Learning objectives

- Represent 2D vectors using all standard notation forms (arrow, bold, column, and i-j component form)
- Calculate magnitude, unit vectors, and perform addition, subtraction and scalar multiplication of 2D vectors
- Apply vector ratios to solve collinearity and position vector problems
- Resolve and compose velocities to find resultant velocity of moving objects
- Solve particle collision problems by equating position vectors at time t

## 2D Vector Notation and Core Operations

**2D Vector** — A quantity with both magnitude (size) and direction in the 2D plane, represented by horizontal (x) and vertical (y) components.

*Notation:* $\overrightarrow{AB}, \; \mathbf{v}, \; \begin{pmatrix}x \\ y\end{pmatrix}, \; a\mathbf{i} + b\mathbf{j}$

*Example:* The vector from origin O to point (3, 4) is written as $\overrightarrow{OP} = \mathbf{p}= \begin{pmatrix}3 \\ 4\end{pmatrix} = 3\mathbf{i} + 4\mathbf{j}$.

Syllabus 13.1 expects you to read and write vectors in component (i-j) form as well as column form, using correct vector notation. Here $\mathbf{i}$ is the unit vector in the positive $x$-direction and $\mathbf{j}$ is the unit vector in the positive $y$-direction, so $a\mathbf{i} + b\mathbf{j}$ means $a$ steps across and $b$ steps up. This is exactly the same object as the column vector $\begin{pmatrix}a \\ b\end{pmatrix}$, and you can switch freely between the two forms.

$$a\mathbf{i} + b\mathbf{j} = \begin{pmatrix}a \\ b\end{pmatrix}, \qquad 3\mathbf{i} - 2\mathbf{j} = \begin{pmatrix}3 \\ -2\end{pmatrix}$$

All vectors can be manipulated using three core operations: addition, subtraction, and scalar multiplication. The magnitude of a vector $\begin{pmatrix}x \\ y\end{pmatrix}$ is calculated using Pythagoras' theorem. A unit vector in the direction of $\mathbf{v}$ is found by dividing $\mathbf{v}$ by its magnitude.

$$|\mathbf{v}| = \sqrt{x^2 + y^2}, \quad \hat{\mathbf{v}} = \frac{1}{\sqrt{x^2 + y^2}}\begin{pmatrix}x \\ y\end{pmatrix}$$

**Worked example:** Given $\mathbf{a}= \begin{pmatrix}2 \\ 5\end{pmatrix}$ and $\mathbf{b}= \begin{pmatrix}-4 \\ 1\end{pmatrix}$, calculate: (a) $3\mathbf{a}- 2\mathbf{b}$, (b) the unit vector in the direction of $\mathbf{a}$.

1. Part (a): First perform scalar multiplication on each vector:

   $$3\mathbf{a}= 3\begin{pmatrix}2 \\ 5\end{pmatrix} = \begin{pmatrix}6 \\ 15\end{pmatrix}, \quad 2\mathbf{b}= 2\begin{pmatrix}-4 \\ 1\end{pmatrix} = \begin{pmatrix}-8 \\ 2\end{pmatrix}$$
2. Subtract the second vector from the first:

   $$3\mathbf{a}- 2\mathbf{b}= \begin{pmatrix}6 - (-8) \\ 15 - 2\end{pmatrix} = \begin{pmatrix}14 \\ 13\end{pmatrix}$$
3. Part (b): Calculate magnitude of $\mathbf{a}$ first:

   $$|\mathbf{a}| = \sqrt{2^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29}$$
4. Divide $\mathbf{a}$ by its magnitude to get the unit vector:

   $$\hat{\mathbf{a}} = \frac{1}{\sqrt{29}}\begin{pmatrix}2 \\ 5\end{pmatrix} = \begin{pmatrix}\frac{2}{\sqrt{29}} \\ \frac{5}{\sqrt{29}}\end{pmatrix}$$

> **Exam tip:** Always rationalize denominators for unit vectors if required by the question, but leaving the denominator as a root is acceptable for most 0606 answers.

## Position Vectors, Section Ratios and Collinearity

**Position Vector** — A vector describing the position of a point P relative to a fixed origin O, with components equal to the coordinates of P.

*Notation:* $\overrightarrow{OP} = \mathbf{r}$

*Example:* Point P with coordinates (2, -3) has position vector $\mathbf{p}= \begin{pmatrix}2 \\ -3\end{pmatrix}$.

To find the position vector of a point dividing the line segment AB in the ratio $m:n$, you can use the ratio ("section") result $\mathbf{r}= \frac{n\mathbf{a}+ m\mathbf{b}}{m + n}$, where $\mathbf{a}$ and $\mathbf{b}$ are the position vectors of A and B respectively. Note this result is not printed in the 0606 List of formulas; in the exam you are usually expected to reach the same point by stepping along a base vector (see the base-vector method later in this guide), and the formula is best used as a quick check. Three points A, B, C are collinear if vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are scalar multiples of each other, meaning they lie on the same straight line.

**Worked example:** Points A and B have position vectors $\mathbf{a}= \begin{pmatrix}1 \\ 3\end{pmatrix}$ and $\mathbf{b}= \begin{pmatrix}7 \\ 15\end{pmatrix}$. (a) Find the position vector of point M that divides AB in the ratio 1:2. (b) Show that points A, B, and C with position vector $\mathbf{c}= \begin{pmatrix}9 \\ 19\end{pmatrix}$ are collinear.

1. Part (a): Apply the section formula with m=1, n=2:

   $$\mathbf{m}= \frac{2\mathbf{a}+ 1\mathbf{b}}{1 + 2} = \frac{2\begin{pmatrix}1 \\ 3\end{pmatrix} + \begin{pmatrix}7 \\ 15\end{pmatrix}}{3}$$
2. Simplify the numerator and divide by 3:

   $$\mathbf{m}= \frac{\begin{pmatrix}2 + 7 \\ 6 + 15\end{pmatrix}}{3} = \frac{\begin{pmatrix}9 \\ 21\end{pmatrix}}{3} = \begin{pmatrix}3 \\ 7\end{pmatrix}$$
3. Part (b): Calculate vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$:

   $$\overrightarrow{AB} = \mathbf{b}- \mathbf{a}= \begin{pmatrix}7 - 1 \\ 15 - 3\end{pmatrix} = \begin{pmatrix}6 \\ 12\end{pmatrix}, \quad \overrightarrow{AC} = \mathbf{c}- \mathbf{a}= \begin{pmatrix}9 - 1 \\ 19 - 3\end{pmatrix} = \begin{pmatrix}8 \\ 16\end{pmatrix}$$
4. Check if one is a scalar multiple of the other: $\overrightarrow{AC} = \frac{4}{3}\overrightarrow{AB}$. Since the vectors are parallel and share point A, A, B, C are collinear.

> **Exam tip:** When proving collinearity, always state that the vectors are parallel *and* share a common point; both conditions are required to get full marks.

## Composing and Resolving Velocities for Resultant Calculations

**Resultant Velocity** — The vector sum of two or more velocity vectors acting on an object, representing its actual motion relative to a fixed frame of reference.

*Notation:* $\mathbf{v}_{R}$

*Example:* A boat moving with velocity $\mathbf{v}_{b}$ in a river with current velocity $\mathbf{v}_{c}$ has resultant velocity $\mathbf{v}_{R} = \mathbf{v}_{b} + \mathbf{v}_{c}$.

To solve velocity problems, you can either add/subtract velocity vectors directly in column form, or resolve velocities into horizontal and vertical components if they are given as magnitude and direction. The magnitude of the resultant velocity gives the speed of the object, and its direction can be found using basic right triangle trigonometry (SOH-CAH-TOA).

**Worked example:** A plane flies due north with an airspeed of 200 km/h. A wind blows due east at 50 km/h. Calculate the magnitude of the resultant velocity of the plane relative to the ground, correct to 3 significant figures.

1. Represent the velocities as column vectors, taking east as positive x and north as positive y:

   $$\mathbf{v}_{p} = \begin{pmatrix}0 \\ 200\end{pmatrix}, \quad \mathbf{v}_{w} = \begin{pmatrix}50 \\ 0\end{pmatrix}$$
2. Add the vectors to find the resultant velocity:

   $$\mathbf{v}_{R} = \begin{pmatrix}0 + 50 \\ 200 + 0\end{pmatrix} = \begin{pmatrix}50 \\ 200\end{pmatrix}$$
3. Calculate the magnitude of the resultant vector using Pythagoras' theorem:

   $$|\mathbf{v}_{R}| = \sqrt{50^2 + 200^2} = \sqrt{2500 + 40000} = \sqrt{42500} \approx 206 \text{ km/h (3 s.f.)}$$

*Calculator:* allowed

## Solving Particle Collision Problems

For two moving particles, collision occurs if their position vectors are equal at the same time $t ≥ 0$. The position of a particle at time $t$ is given by $\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}$, where $\mathbf{r}_0$ is the initial position vector and $\mathbf{v}$ is the constant velocity vector of the particle.

**Worked example:** Particle P has initial position $\begin{pmatrix}2 \\ 1\end{pmatrix}$ and constant velocity $\begin{pmatrix}1 \\ 3\end{pmatrix}$ m/s. Particle Q has initial position $\begin{pmatrix}11 \\ 13\end{pmatrix}$ and constant velocity $\begin{pmatrix}-2 \\ -1\end{pmatrix}$ m/s. Show that the particles collide, and find the time of collision.

1. Write position vector functions for both particles at time t seconds:

   $$\mathbf{r}_P(t) = \begin{pmatrix}2 \\ 1\end{pmatrix} + t\begin{pmatrix}1 \\ 3\end{pmatrix} = \begin{pmatrix}2 + t \\ 1 + 3t\end{pmatrix}$$
2. Position vector for Q:

   $$\mathbf{r}_Q(t) = \begin{pmatrix}11 \\ 13\end{pmatrix} + t\begin{pmatrix}-2 \\ -1\end{pmatrix} = \begin{pmatrix}11 - 2t \\ 13 - t\end{pmatrix}$$
3. Equate the x-components to solve for t:

   $$2 + t = 11 - 2t \\ 3t = 9 \\ t = 3$$
4. Verify that the y-components are equal at t=3 to confirm collision:

   $$\text{y-component of P at t=3: } 1 + 3(3) = 10 \\ \text{y-component of Q at t=3: } 13 - 3 = 10$$
5. Both components are equal at t=3 ≥ 0, so the particles collide after 3 seconds.

> **Exam tip:** Always check that the time you calculate is non-negative; a negative time means the particles would have collided before the start of the motion, so no collision occurs in the context of the question.

## Expressing Vectors with Base Vectors and Finding Intersection Points

**Base Vectors** — Two non-parallel vectors from a common point, used to express every other vector in a diagram as a linear combination $p\mathbf{a} + q\mathbf{b}$.

*Notation:* $\overrightarrow{OA} = \mathbf{a}, \; \overrightarrow{OB} = \mathbf{b}$

*Example:* In triangle $OAB$, the side $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a}$ ("tip minus tail").

In most 0606 diagram questions you are given a triangle or parallelogram with two base vectors, for example $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. Every other vector in the figure is then written as a combination of $\mathbf{a}$ and $\mathbf{b}$. To reach a point on a line, travel from a known point along a known direction: the midpoint $M$ of $AB$ has $\overrightarrow{OM} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{1}{2}(\mathbf{a} + \mathbf{b})$.

More generally, if $M$ lies on $AB$ with $AM:MB = m:n$, step the fraction $\frac{m}{m+n}$ of the way along $\overrightarrow{AB}$. This is exactly the section formula written with base vectors:

$$\overrightarrow{OM} = \mathbf{a} + \frac{m}{m+n}(\mathbf{b} - \mathbf{a}) = \frac{n\mathbf{a} + m\mathbf{b}}{m+n}$$

To find where two lines cross, give the crossing point $P$ two names. If $P$ lies on a line through $O$ in direction $\mathbf{d}_1$, write $\overrightarrow{OP} = \lambda\mathbf{d}_1$. If $P$ also lies on a line through a point $C$ in direction $\mathbf{d}_2$, write $\overrightarrow{OP} = \overrightarrow{OC} + \mu\mathbf{d}_2$. Both describe the same point, so set them equal. Because $\mathbf{a}$ and $\mathbf{b}$ are not parallel, two combinations of $\mathbf{a}$ and $\mathbf{b}$ can be equal only when the amount of $\mathbf{a}$ agrees on both sides and the amount of $\mathbf{b}$ agrees on both sides. This gives two equations in the two unknowns $\lambda$ and $\mu$.

**Worked example:** In triangle $OAB$, $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. The point $D$ lies on $AB$ with $AD:DB = 1:2$, and $E$ is the midpoint of $OA$. The lines $OD$ and $BE$ intersect at $P$. Find $\overrightarrow{OP}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

1. Express the key points with the base vectors. $D$ divides $AB$ with $AD:DB = 1:2$, so step $\frac{1}{3}$ of the way from $A$ towards $B$:

   $$\overrightarrow{OD} = \mathbf{a} + \frac{1}{3}(\mathbf{b} - \mathbf{a}) = \frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b} = \frac{1}{3}(2\mathbf{a} + \mathbf{b})$$
2. $E$ is the midpoint of $OA$, so:

   $$\overrightarrow{OE} = \frac{1}{2}\mathbf{a}$$
3. $P$ lies on line $OD$, so write $\overrightarrow{OP}$ as a multiple $\lambda$ of $\overrightarrow{OD}$:

   $$\overrightarrow{OP} = \lambda\,\overrightarrow{OD} = \frac{2\lambda}{3}\mathbf{a} + \frac{\lambda}{3}\mathbf{b}$$
4. $P$ also lies on line $BE$. Start at $B$ and travel a fraction $\mu$ along $\overrightarrow{BE} = \overrightarrow{OE} - \overrightarrow{OB} = \frac{1}{2}\mathbf{a} - \mathbf{b}$:

   $$\overrightarrow{OP} = \mathbf{b} + \mu\left(\frac{1}{2}\mathbf{a} - \mathbf{b}\right) = \frac{\mu}{2}\mathbf{a} + (1 - \mu)\mathbf{b}$$
5. The two expressions are the same vector. Since $\mathbf{a}$ and $\mathbf{b}$ are not parallel, equate the $\mathbf{a}$-coefficients and the $\mathbf{b}$-coefficients separately:

   $$\frac{2\lambda}{3} = \frac{\mu}{2} \quad \text{(a-terms)} \\ \frac{\lambda}{3} = 1 - \mu \quad \text{(b-terms)}$$
6. From the a-equation, $\mu = \frac{4\lambda}{3}$. Substitute into the b-equation:

   $$\frac{\lambda}{3} = 1 - \frac{4\lambda}{3} \;\Rightarrow\; \frac{\lambda}{3} + \frac{4\lambda}{3} = 1 \;\Rightarrow\; \frac{5\lambda}{3} = 1 \;\Rightarrow\; \lambda = \frac{3}{5}$$
7. Then $\mu = \frac{4\lambda}{3} = \frac{4}{5}$. Substitute $\lambda = \frac{3}{5}$ back into $\overrightarrow{OP} = \frac{2\lambda}{3}\mathbf{a} + \frac{\lambda}{3}\mathbf{b}$:

   $$\overrightarrow{OP} = \frac{2}{3}\cdot\frac{3}{5}\mathbf{a} + \frac{1}{3}\cdot\frac{3}{5}\mathbf{b} = \frac{2}{5}\mathbf{a} + \frac{1}{5}\mathbf{b} = \frac{1}{5}(2\mathbf{a} + \mathbf{b})$$
8. Check with the other line: $\overrightarrow{OP} = \frac{\mu}{2}\mathbf{a} + (1 - \mu)\mathbf{b} = \frac{1}{2}\cdot\frac{4}{5}\mathbf{a} + \left(1 - \frac{4}{5}\right)\mathbf{b} = \frac{2}{5}\mathbf{a} + \frac{1}{5}\mathbf{b}$. Both routes agree, confirming $\overrightarrow{OP} = \frac{1}{5}(2\mathbf{a} + \mathbf{b})$.

> **Exam tip:** When two lines meet, write $\overrightarrow{OP}$ along each line with its own scalar ($\lambda$ and $\mu$). Because $\mathbf{a}$ and $\mathbf{b}$ are not parallel, you may equate the $\mathbf{a}$-terms and the $\mathbf{b}$-terms separately to obtain two equations — this is the move examiners reward.

## Finding a Vector of Given Magnitude in a Given Direction

A common exam task gives you a direction and a required length, and asks for the matching vector. The idea is simple: first shrink the direction vector to length 1 (the unit vector), then stretch it to the length you want. For a direction vector $\mathbf{d}$ and a required magnitude $m$, the vector of magnitude $m$ in the same direction is $m\,\frac{\mathbf{d}}{|\mathbf{d}|}$. For the same length in the opposite direction, attach a minus sign.

$$\mathbf{u} = \pm\, m\,\frac{\mathbf{d}}{|\mathbf{d}|}, \quad \text{with } +\ \text{for the same direction and } -\ \text{for the opposite direction}$$

**Worked example:** Find the vector of magnitude 8.5 in the same direction as $\begin{pmatrix}3 \\ 4\end{pmatrix}$.

1. Find the magnitude of the direction vector:

   $$\left|\begin{pmatrix}3 \\ 4\end{pmatrix}\right| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$$
2. Divide by the magnitude to get the unit vector:

   $$\hat{\mathbf{d}} = \frac{1}{5}\begin{pmatrix}3 \\ 4\end{pmatrix} = \begin{pmatrix}0.6 \\ 0.8\end{pmatrix}$$
3. Multiply the unit vector by the required magnitude 8.5:

   $$8.5\begin{pmatrix}0.6 \\ 0.8\end{pmatrix} = \begin{pmatrix}5.1 \\ 6.8\end{pmatrix}$$
4. Check: $\sqrt{5.1^2 + 6.8^2} = \sqrt{26.01 + 46.24} = \sqrt{72.25} = 8.5$, as required.

**Worked example:** Find the vector of magnitude 26 in the direction opposite to $\begin{pmatrix}5 \\ -12\end{pmatrix}$.

1. Find the magnitude of the direction vector:

   $$\left|\begin{pmatrix}5 \\ -12\end{pmatrix}\right| = \sqrt{5^2 + (-12)^2} = \sqrt{169} = 13$$
2. The unit vector is $\frac{1}{13}\begin{pmatrix}5 \\ -12\end{pmatrix}$. For the opposite direction, multiply by $-26$:

   $$-26\cdot\frac{1}{13}\begin{pmatrix}5 \\ -12\end{pmatrix} = -2\begin{pmatrix}5 \\ -12\end{pmatrix} = \begin{pmatrix}-10 \\ 24\end{pmatrix}$$
3. Check: $\sqrt{(-10)^2 + 24^2} = \sqrt{100 + 576} = \sqrt{676} = 26$, and $\begin{pmatrix}-10 \\ 24\end{pmatrix} = -2\begin{pmatrix}5 \\ -12\end{pmatrix}$ is a negative multiple, confirming the opposite direction.

The same idea builds a velocity from a speed and a direction: velocity = speed $\times$ unit direction vector. Divide the direction vector by its magnitude, then multiply by the speed.

**Worked example:** A particle moves with speed 20 m/s in the direction of $\begin{pmatrix}-1 \\ 2\end{pmatrix}$. Find its velocity as a column vector.

1. Find the magnitude of the direction vector:

   $$\left|\begin{pmatrix}-1 \\ 2\end{pmatrix}\right| = \sqrt{(-1)^2 + 2^2} = \sqrt{5}$$
2. Multiply the unit vector by the speed 20:

   $$20\cdot\frac{1}{\sqrt{5}}\begin{pmatrix}-1 \\ 2\end{pmatrix} = \frac{20}{\sqrt{5}}\begin{pmatrix}-1 \\ 2\end{pmatrix}$$
3. Rationalise $\frac{20}{\sqrt{5}} = \frac{20\sqrt{5}}{5} = 4\sqrt{5}$, so the velocity is:

   $$4\sqrt{5}\begin{pmatrix}-1 \\ 2\end{pmatrix} = \begin{pmatrix}-4\sqrt{5} \\ 8\sqrt{5}\end{pmatrix}$$
4. Check the speed: $\sqrt{(-4\sqrt{5})^2 + (8\sqrt{5})^2} = \sqrt{80 + 320} = \sqrt{400} = 20$ m/s, as required.

> **Exam tip:** Always find the unit vector first (divide the direction vector by its own magnitude), then scale it by the required length. A vector 'in the opposite direction' just needs a minus sign; the magnitude you are given always stays positive.

## Finding the Direction (Angle or Bearing) of a Resultant Velocity

Adding velocity vectors gives the resultant velocity; its magnitude is the speed, and its direction is often asked for as an angle or a compass bearing. Once you have the resultant $\begin{pmatrix}v_x \\ v_y\end{pmatrix}$ (with $v_x$ the east component and $v_y$ the north component), find the base angle it makes with a chosen axis using right-angled trigonometry, then use the signs of $v_x$ and $v_y$ to place it in the correct quadrant.

$$\tan\theta = \left|\frac{v_y}{v_x}\right| \quad \Rightarrow \quad \theta = \tan^{-1}\left|\frac{v_y}{v_x}\right|$$

**Worked example:** A model boat is driven with velocity $\begin{pmatrix}3 \\ 5\end{pmatrix}$ m/s while the current adds velocity $\begin{pmatrix}4 \\ -1\end{pmatrix}$ m/s. Taking east as positive $x$ and north as positive $y$, find the resulting speed and the bearing on which the boat travels, giving the bearing to the nearest degree.

1. Add the velocities to find the resultant:

   $$\mathbf{v} = \begin{pmatrix}3 \\ 5\end{pmatrix} + \begin{pmatrix}4 \\ -1\end{pmatrix} = \begin{pmatrix}7 \\ 4\end{pmatrix}$$
2. The speed is the magnitude of the resultant:

   $$|\mathbf{v}| = \sqrt{7^2 + 4^2} = \sqrt{49 + 16} = \sqrt{65} ≈ 8.06 \text{ m/s}$$
3. Identify the quadrant: the east component $7 > 0$ and the north component $4 > 0$, so the resultant points into the north-east quadrant. Find the base angle above the east direction:

   $$\tan\theta = \left|\frac{4}{7}\right| \;\Rightarrow\; \theta = \tan^{-1}\frac{4}{7} ≈ 29.7^{\circ}\text{ (north of east)}$$
4. Convert to a bearing (measured clockwise from north). Due east is $090^{\circ}$, and the resultant is $29.7^{\circ}$ north of east:

   $$\text{bearing} = 090^{\circ} - 29.7^{\circ} = 060.3^{\circ} ≈ 060^{\circ}$$
5. So the boat travels at about 8.06 m/s on a bearing of 060°. (Equivalently, the angle east of north is $\tan^{-1}\frac{7}{4} ≈ 60.3^{\circ}$, giving the same bearing.)

> **Exam tip:** Bearings are measured clockwise from north and written with three figures, e.g. 060°. Always note the signs of the east and north components first so the angle sits in the correct quadrant — a bare $\tan^{-1}$ only ever returns an acute angle.

*Calculator:* allowed

## Solving for Scalars by Equating Components (Like Vectors)

Syllabus 13.3 lists "equating like vectors" as a required skill. A common exam task gives two fixed vectors and a target vector, then asks for the scalars that combine the two into the target. Because the $\mathbf{i}$ and $\mathbf{j}$ directions are independent, two vectors are equal only when their $\mathbf{i}$-components are equal and, separately, their $\mathbf{j}$-components are equal. Writing one side as $(\ldots)\mathbf{i} + (\ldots)\mathbf{j}$ and matching it to the target therefore turns a single vector equation into two ordinary simultaneous equations in the unknown scalars. This is a pure component (coordinate) technique — the numbers are actual $x$- and $y$-components, not the geometric base vectors $\mathbf{a}$ and $\mathbf{b}$ used earlier.

$$x_1\mathbf{i} + y_1\mathbf{j} = x_2\mathbf{i} + y_2\mathbf{j} \iff x_1 = x_2 \ \text{and} \ y_1 = y_2$$

**Worked example:** The vectors $\mathbf{p} = 2\mathbf{i} + \mathbf{j}$ and $\mathbf{q} = 3\mathbf{i} - \mathbf{j}$ are given. Find the scalars $a$ and $b$ such that $a\mathbf{p} + b\mathbf{q} = 12\mathbf{i} + \mathbf{j}$.

1. Substitute the vectors and collect the $\mathbf{i}$ terms and the $\mathbf{j}$ terms separately:

   $$a(2\mathbf{i} + \mathbf{j}) + b(3\mathbf{i} - \mathbf{j}) = (2a + 3b)\mathbf{i} + (a - b)\mathbf{j}$$
2. This must equal $12\mathbf{i} + \mathbf{j}$. Equate like vectors — the $\mathbf{i}$ coefficient on each side, then the $\mathbf{j}$ coefficient on each side:

   $$2a + 3b = 12 \quad (\mathbf{i}\text{-terms}) \\ a - b = 1 \quad (\mathbf{j}\text{-terms})$$
3. Solve the simultaneous equations. From the $\mathbf{j}$-equation, $a = b + 1$; substitute into the $\mathbf{i}$-equation:

   $$2(b + 1) + 3b = 12 \;\Rightarrow\; 5b + 2 = 12 \;\Rightarrow\; b = 2$$
4. Back-substitute to find $a$:

   $$a = b + 1 = 3$$
5. Check by substituting $a = 3$, $b = 2$ into the original: $3(2\mathbf{i} + \mathbf{j}) + 2(3\mathbf{i} - \mathbf{j}) = (6 + 6)\mathbf{i} + (3 - 2)\mathbf{j} = 12\mathbf{i} + \mathbf{j}$, as required. So $a = 3$ and $b = 2$.

> **Exam tip:** Set out the two component equations clearly, labelling which is the $\mathbf{i}$-equation and which is the $\mathbf{j}$-equation, then solve them as ordinary simultaneous equations. Always substitute your scalars back into the original vector equation to check.

## Common pitfalls

- **Wrong:** Confusing vector subtraction order when calculating $\overrightarrow{AB}$
  - Why it fails: Many students calculate $\overrightarrow{AB}$ as $\mathbf{a}- \mathbf{b}$ instead of $\mathbf{b}- \mathbf{a}$, leading to incorrect direction and component values.
  - Correct: Remember $\overrightarrow{AB} = \text{position of B} - \text{position of A} = \mathbf{b}- \mathbf{a}$.
- **Wrong:** Forgetting to divide by the sum of the ratio parts in the section formula
  - Why it fails: Students often write $\mathbf{r}= n\mathbf{a}+ m\mathbf{b}$ for the ratio m:n, omitting division by m+n, leading to position vectors with incorrect magnitude.
  - Correct: Always divide the weighted sum of position vectors by the total of the ratio parts: $\mathbf{r}= \frac{n\mathbf{a}+ m\mathbf{b}}{m + n}$.
- **Wrong:** Only showing vectors are parallel when proving collinearity
  - Why it fails: Parallel vectors could be on separate lines; you need to confirm they share a common point to prove collinearity.
  - Correct: State both that one vector is a scalar multiple of the other *and* that they share a common point (e.g. point A) to get full marks.
- **Wrong:** Equating velocity vectors to solve collision problems
  - Why it fails: Collision requires equal position at the same time, not equal velocity; equal velocity only means the particles are moving at the same speed and direction.
  - Correct: Write position vector functions for both particles in terms of t, then equate components and solve for t, verifying both components match at the same t.
- **Wrong:** Confusing scalar multiplication with the dot product
  - Why it fails: Some students reach for A-Level dot product rules when asked for scalar multiplication, which is not in the 0606 syllabus.
  - Correct: Scalar multiplication only multiplies each component of the vector by the scalar: $k\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}kx \\ ky\end{pmatrix}$; no dot product is needed for any 0606 vector question.

## Cheatsheet

| Concept | Formula/Rule | Example |
| --- | --- | --- |
| Vector Magnitude | $\|\begin{pmatrix}x \\ y\end{pmatrix}\| = \sqrt{x^2 + y^2}$ | $\|\begin{pmatrix}3 \\ 4\end{pmatrix}\| = 5$ |
| Unit Vector | $\hat{\mathbf{v}} = \frac{\mathbf{v}}{\|\mathbf{v}\|}$ | Unit vector of $\begin{pmatrix}3 \\ 4\end{pmatrix}$ is $\begin{pmatrix}3/5 \\ 4/5\end{pmatrix}$ |
| Section Ratio | Point dividing AB in ratio m:n: $\frac{n\mathbf{a}+ m\mathbf{b}}{m + n}$ | Ratio 1:2 gives $\frac{2\mathbf{a}+ \mathbf{b}}{3}$ |
| Collinearity Check | A,B,C collinear if $\overrightarrow{AB} = k\overrightarrow{AC}$ for some scalar k | $\overrightarrow{AB} = \begin{pmatrix}2 \\ 4\end{pmatrix}, \overrightarrow{AC} = \begin{pmatrix}4 \\ 8\end{pmatrix} = 2\overrightarrow{AB}$ → collinear |
| Position at time t | $\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}$ | Initial position $\begin{pmatrix}1 \\ 2\end{pmatrix}$, velocity $\begin{pmatrix}3 \\ 1\end{pmatrix}$ → $\mathbf{r}(t) = \begin{pmatrix}1 + 3t \\ 2 + t\end{pmatrix}$ |
| Resultant Velocity | $\mathbf{v}_{R} = \mathbf{v}_{1} + \mathbf{v}_{2}$ | Boat velocity + river current = total velocity relative to ground |

## What's next

Now that you have mastered 2D vectors and velocity problems for CIE IGCSE Additional Mathematics 0606, you can move on to other advanced topics that build on vector reasoning, such as kinematics with variable acceleration and further calculus applications. Make sure to practice past paper vector questions to familiarize yourself with the exam's phrasing and mark scheme requirements, as vector problems often carry 4-6 marks per question. Pay special attention to showing all steps for collinearity and collision proofs, as partial marks are awarded for correct intermediate calculations even if your final answer is wrong.

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