Vectors in Two Dimensions – Cambridge IGCSE / A‑Level Additional Mathematics
Learning Objectives
After studying this section students will be able to:
- Write the position vector of any point in the plane.
- Convert between pair notation \((a,b)\) and component ( \(\mathbf{i},\mathbf{j}\) ) form.
- Express a vector in component form \(a\mathbf{i}+b\mathbf{j}\).
- Calculate the magnitude of a vector.
- Obtain a unit vector parallel to a given non‑zero vector and recognise the notation \(\hat{v}\).
- Perform vector addition, subtraction and scalar multiplication.
- Resolve a vector into its i‑ and j‑components from a given magnitude and direction.
- Apply vectors to simple kinematics (relative‑velocity) problems.
Syllabus Reference
Section 13 – “Vectors in the plane” (Cambridge IGCSE Additional Mathematics 0606). All content below is taken directly from the official syllabus and is required for the examination.
Notation & Definitions
| Symbol | Meaning |
|---|
| \(\mathbf{i}\), \(\mathbf{j}\) | Standard unit vectors. \(\mathbf{i}\) points along the positive x‑axis, \(\mathbf{j}\) points along the positive y‑axis. Both have magnitude 1. |
| \(\vec{OP}\) | Position vector of point \(P(x,y)\); the vector from the origin \(O(0,0)\) to \(P\). \(\displaystyle\vec{OP}=x\mathbf{i}+y\mathbf{j}\). |
| \(\overrightarrow{AB}\) | Vector from point \(A(x1,y1)\) to point \(B(x2,y2)\); \(\displaystyle\overrightarrow{AB}=(x2-x1)\mathbf{i}+(y2-y1)\mathbf{j}\). |
| \(|\vec{v}|\) | Magnitude (length) of vector \(\vec{v}=a\mathbf{i}+b\mathbf{j}\); \(\displaystyle|\vec{v}|=\sqrt{a^{2}+b^{2}}\) (always non‑negative). |
| \(\hat{v}\) | Unit vector having the same direction as \(\vec{v}\); \(\displaystyle\hat{v}= \frac{\vec{v}}{|\vec{v}|}= \frac{a}{\sqrt{a^{2}+b^{2}}}\mathbf{i}+ \frac{b}{\sqrt{a^{2}+b^{2}}}\mathbf{j}\). By definition \(|\hat{v}|=1\). |
Conversion Quick‑Reference
Both notations are accepted in the exam.
| Pair notation | Component notation |
|---|
| \((a,b)\) | \(a\mathbf{i}+b\mathbf{j}\) |
| \(\overrightarrow{AB}\) | \((xB-xA)\mathbf{i}+(yB-yA)\mathbf{j}\) |
Formula Box (quick reference)
| Concept | Formula |
|---|
| Position vector of \(P(x,y)\) | \(\displaystyle\vec{OP}=x\mathbf{i}+y\mathbf{j}\) |
| Vector between two points | \(\displaystyle\overrightarrow{AB}=(x2-x1)\mathbf{i}+(y2-y1)\mathbf{j}\) |
| Magnitude of \(\vec{v}=a\mathbf{i}+b\mathbf{j}\) | \(\displaystyle|\vec{v}|=\sqrt{a^{2}+b^{2}}\) |
| Unit vector in the direction of \(\vec{v}\) | \(\displaystyle\hat{v}= \frac{\vec{v}}{|\vec{v}|}= \frac{a}{\sqrt{a^{2}+b^{2}}}\mathbf{i}+ \frac{b}{\sqrt{a^{2}+b^{2}}}\mathbf{j}\) |
| Scalar multiplication | \(\displaystyle k\vec{v}=k a\mathbf{i}+k b\mathbf{j}\) |
| Vector addition / subtraction | \(\displaystyle\vec{u}+\vec{v}=(au+av)\mathbf{i}+(bu+bv)\mathbf{j}\)
\(\displaystyle\vec{u}-\vec{v}=(au-av)\mathbf{i}+(bu-bv)\mathbf{j}\) |
| Components from magnitude & direction \(\theta\) | \(\displaystyle\vec{v}=|\vec{v}|(\cos\theta\,\mathbf{i}+\sin\theta\,\mathbf{j})\) |
Step‑by‑Step Procedure: Obtaining a Parallel Unit Vector
- Write the given vector in component form \(\vec{v}=a\mathbf{i}+b\mathbf{j}\).
- Find its magnitude \(|\vec{v}|=\sqrt{a^{2}+b^{2}}\). Remember the magnitude is always positive.
- Divide each component by the magnitude:
\[
\hat{v}= \frac{a}{|\vec{v}|}\mathbf{i}+ \frac{b}{|\vec{v}|}\mathbf{j}.
\]
- Optional check: \(|\hat{v}|=1\). If the result does not simplify to 1, a sign or arithmetic error has occurred.
Worked Examples
Example 1 – Position vector and unit vector of \(\overrightarrow{AB}\)
Points: \(A(2,-1)\), \(B(5,3)\).
- Vector \(\overrightarrow{AB}= (5-2)\mathbf{i}+(3-(-1))\mathbf{j}=3\mathbf{i}+4\mathbf{j}\).
- Magnitude \(|\overrightarrow{AB}|=\sqrt{3^{2}+4^{2}}=5\).
- Unit vector \(\displaystyle\hat{u}= \frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}\).
Example 2 – Vector subtraction
Given \(\vec{p}=4\mathbf{i}+2\mathbf{j}\) and \(\vec{q}=1\mathbf{i}+5\mathbf{j}\), find \(\vec{p}-\vec{q}\) and its unit vector.
- \(\vec{p}-\vec{q}= (4-1)\mathbf{i}+(2-5)\mathbf{j}=3\mathbf{i}-3\mathbf{j}\).
- Magnitude \(|\vec{p}-\vec{q}|=\sqrt{3^{2}+(-3)^{2}}=\sqrt{18}=3\sqrt{2}\).
- Unit vector \(\displaystyle\widehat{(\vec{p}-\vec{q})}= \frac{3}{3\sqrt{2}}\mathbf{i}+ \frac{-3}{3\sqrt{2}}\mathbf{j}= \frac{1}{\sqrt{2}}\mathbf{i}-\frac{1}{\sqrt{2}}\mathbf{j}\).
Example 3 – Resolving a vector from magnitude & direction
A wind blows at \(10\;\text{km h}^{-1}\) making an angle of \(30^{\circ}\) south of east.
- Take east as the positive x‑direction, south as the negative y‑direction.
- Components: \(\displaystyle\vec{W}=10\bigl(\cos30^{\circ}\,\mathbf{i}+\sin(-30^{\circ})\,\mathbf{j}\bigr)
=10\left(\frac{\sqrt3}{2}\mathbf{i}-\frac12\mathbf{j}\right)=5\sqrt3\,\mathbf{i}-5\,\mathbf{j}\).
- Unit vector in the wind’s direction:
\(\displaystyle\hat{W}= \cos30^{\circ}\,\mathbf{i}+\sin(-30^{\circ})\,\mathbf{j}
=\frac{\sqrt3}{2}\mathbf{i}-\frac12\mathbf{j}\).
Example 4 – Relative‑velocity (composition of vectors)
Boat B sails north at \(5\;\text{km h}^{-1}\). River R flows east at \(3\;\text{km h}^{-1}\). Find the boat’s speed and direction relative to the bank.
- \(\vec{v}{\text{boat}}=5\mathbf{j},\qquad \vec{v}{\text{river}}=3\mathbf{i}\).
- Resultant \(\vec{v}_{\text{result}}=3\mathbf{i}+5\mathbf{j}\).
- Speed \(|\vec{v}_{\text{result}}|=\sqrt{3^{2}+5^{2}}=\sqrt{34}\approx5.83\;\text{km h}^{-1}\).
- Direction \(\theta=\tan^{-1}\!\left(\dfrac{5}{3}\right)\approx59^{\circ}\) north of east.
Example 5 – Unit vector from a given vector (negative components)
Given \(\vec{v}= -6\mathbf{i}+8\mathbf{j}\), find the parallel unit vector.
- Magnitude \(|\vec{v}|=\sqrt{(-6)^{2}+8^{2}}=10\).
- \(\displaystyle\hat{v}= -\frac{6}{10}\mathbf{i}+\frac{8}{10}\mathbf{j}= -\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}\).
Practice Questions
- Find the position vector of the point \(C(-4,\,7)\).
- Given \(\vec{v}= -6\mathbf{i}+8\mathbf{j}\), determine a unit vector parallel to \(\vec{v}\).
- The points \(P(1,2)\) and \(Q(4,k)\) are such that the unit vector from \(P\) to \(Q\) is \(\displaystyle\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}\). Find the possible value(s) of \(k\).
- Show that \(\vec{a}=2\mathbf{i}+2\mathbf{j}\) and \(\vec{b}=5\mathbf{i}+5\mathbf{j}\) are parallel and write the common unit vector.
- Two forces act on a point: \(\vec{F}{1}=4\mathbf{i}+3\mathbf{j}\) and \(\vec{F}{2}= -2\mathbf{i}+6\mathbf{j}\). Find the resultant force and its unit vector.
- A particle moves with velocity \(\vec{v}=7\mathbf{i}+24\mathbf{j}\) m s\(^{-1}\). Determine the speed and the direction (angle measured anticlockwise from the positive x‑axis) to the nearest degree.
- In a relative‑motion problem, a plane flies north‑west (i.e. \(45^{\circ}\) west of north) at \(200\;\text{km h}^{-1}\). A wind blows due east at \(50\;\text{km h}^{-1}\). Find the ground speed and the bearing of the plane (bearing measured clockwise from north).
- A wind of \(12\;\text{km h}^{-1}\) blows \(20^{\circ}\) north of west. Write the wind vector in component form and give the corresponding unit vector.
Answer Key
| Q. | Answer |
|---|
| 1 | \(\displaystyle\vec{OC}= -4\mathbf{i}+7\mathbf{j}\) |
| 2 | \(|\vec{v}|=10\); \(\displaystyle\hat{v}= -\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}\) |
| 3 | \(\overrightarrow{PQ}=3\mathbf{i}+(k-2)\mathbf{j}\). \(|\overrightarrow{PQ}|=5\Rightarrow\sqrt{3^{2}+(k-2)^{2}}=5\Rightarrow(k-2)^{2}=16\). Hence \(k=6\) or \(k=-2\). |
| 4 | \(\vec{b}= \dfrac{5}{2}\vec{a}\) ⇒ parallel. \(|\vec{a}|=2\sqrt2\). Unit vector \(\displaystyle\hat{u}= \frac{1}{2\sqrt2}(2\mathbf{i}+2\mathbf{j})= \frac{1}{\sqrt2}\mathbf{i}+ \frac{1}{\sqrt2}\mathbf{j}\). |
| 5 | \(\vec{R}= (4-2)\mathbf{i}+(3+6)\mathbf{j}=2\mathbf{i}+9\mathbf{j}\). \(|\vec{R}|=\sqrt{2^{2}+9^{2}}=\sqrt{85}\). Unit vector \(\displaystyle\hat{R}= \frac{2}{\sqrt{85}}\mathbf{i}+ \frac{9}{\sqrt{85}}\mathbf{j}\). |
| 6 | Speed \(|\vec{v}|=25\;\text{m s}^{-1}\). Direction \(\theta=\tan^{-1}\!\left(\dfrac{24}{7}\right)\approx73^{\circ}\) anticlockwise from the +x‑axis. |
| 7 | Plane’s north‑west velocity \(\displaystyle\vec{v}_{\text{NW}}=200\bigl(\cos45^{\circ}\,\mathbf{j}+\sin45^{\circ}(-\mathbf{i})\bigr)= -141.4\mathbf{i}+141.4\mathbf{j}\). Add wind: \(\vec{v}_{\text{ground}}=(-141.4+50)\mathbf{i}+141.4\mathbf{j}= -91.4\mathbf{i}+141.4\mathbf{j}\). Ground speed \(|\vec{v}_{\text{ground}}|\approx166\;\text{km h}^{-1}\). Bearing \(\beta=\tan^{-1}\!\left(\dfrac{141.4}{91.4}\right)\approx57^{\circ}\) west of north ⇒ bearing \(360^{\circ}-57^{\circ}=303^{\circ}\) (or “N 57° W”). |
| 8 | North of west means \(180^{\circ}+20^{\circ}=200^{\circ}\) measured anticlockwise from +x‑axis. Components: \(\displaystyle\vec{W}=12\bigl(\cos200^{\circ}\,\mathbf{i}+\sin200^{\circ}\,\mathbf{j}\bigr)=12\bigl(-\cos20^{\circ}\,\mathbf{i}-\sin20^{\circ}\,\mathbf{j}\bigr)= -11.28\mathbf{i}-4.10\mathbf{j}\). Unit vector \(\displaystyle\hat{W}= \cos200^{\circ}\,\mathbf{i}+\sin200^{\circ}\,\mathbf{j}= -0.94\mathbf{i}-0.34\mathbf{j}\). |
Common Mistakes to Avoid
- Sign errors in \(\overrightarrow{AB}\): always subtract “tail – head” in the order \((x{\text{head}}-x{\text{tail}})\), \((y{\text{head}}-y{\text{tail}})\).
- Using the wrong magnitude: the divisor in the unit‑vector formula must be the magnitude of the *same* vector you are normalising.
- Forgetting that a unit vector has magnitude 1: a quick check \(|\hat{v}|=1\) catches algebra slips.
- Confusing \(\mathbf{i},\mathbf{j}\) with index symbols i, j: in component form \(\mathbf{i}\) and \(\mathbf{j}\) are vectors, not subscripts.
- Angle conventions: unless otherwise stated, angles are measured anticlockwise from the positive x‑axis. Bearings are measured clockwise from north.
- Neglecting direction when multiplying by a negative scalar: a negative scalar reverses the vector’s direction.
Summary Checklist
- Write any vector as \(a\mathbf{i}+b\mathbf{j}\) (or \((a,b)\)).
- Compute its magnitude with \(\sqrt{a^{2}+b^{2}}\) – always a positive number.
- Obtain a parallel unit vector by dividing each component by the magnitude; denote it \(\hat{v}\).
- For addition/subtraction, combine like components (i‑components with i‑components, j‑components with j‑components).
- When a direction (angle or bearing) is given, convert to components using \(\cos\) and \(\sin\). Remember the sign of the y‑component for south or west directions.
- Verify the final unit vector has magnitude 1 and points the required way.
Mastering these steps will enable you to answer every vector‑based question in the IGCSE/A‑Level Additional Mathematics examination, from simple geometric constructions to basic kinematics.