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Module 6 · L7 of 20 ~35 min ⚡ +95 XP available

Component Form in 2D

A vector $\vec{v}$ pointing 3 units right and 4 units up can be written as $3\,\mathbf{i} + 4\,\mathbf{j}$ — but what exactly are $\mathbf{i}$ and $\mathbf{j}$, and why does writing vectors this way make every calculation faster? Component form is the algebraic language of vectors: once you master it, addition, subtraction and scalar multiplication all reduce to working with ordinary numbers.

Today's hook — A drone flies 5 units east and 12 units north. Before learning any formula, estimate the vector that describes this flight. How would you write it using two separate numbers? Jot your idea — you'll refine it in card 05.

0/5QUESTS

Recall — your gut answer first

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A drone travels 5 units east and 12 units north. Without any formula — how would you describe this displacement as a single mathematical object that captures both pieces of information? Write your idea below.

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The key idea

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Every 2D vector can be broken into two perpendicular parts: a horizontal part (in the $x$-direction) and a vertical part (in the $y$-direction). We write these parts using unit vectors $\mathbf{i}$ and $\mathbf{j}$ — each of magnitude 1, pointing along the positive $x$- and $y$-axes respectively.

Any vector $\vec{v}$ in the plane can be written as $\vec{v} = a\,\mathbf{i} + b\,\mathbf{j}$ where $a$ is the $x$-component (horizontal displacement) and $b$ is the $y$-component (vertical displacement).

$\vec{v} = a\,\mathbf{i} + b\,\mathbf{j}$ or equivalently $\vec{v} = \begin{pmatrix} a \\ b \end{pmatrix}$

$\vec{v} = a\,\mathbf{i} + b\,\mathbf{j}$

$\mathbf{i}$ and $\mathbf{j}$ are unit vectors

$|\mathbf{i}| = |\mathbf{j}| = 1$. They are perpendicular and give the direction; the scalars $a$ and $b$ give the size in each direction.

Column vector equivalent

$a\,\mathbf{i} + b\,\mathbf{j}$ and $\begin{pmatrix} a \\ b \end{pmatrix}$ mean exactly the same thing. The HSC accepts both notations.

Negative components

$a < 0$ means the vector points left; $b < 0$ means it points down. For example $-3\,\mathbf{i} + 2\,\mathbf{j}$ goes left 3 and up 2.

What you'll master

Know

Key facts

$\mathbf{i} = \begin{pmatrix}1\\0\end{pmatrix}$ and $\mathbf{j} = \begin{pmatrix}0\\1\end{pmatrix}$ are the standard unit vectors in 2D
Any vector $\vec{v}$ in the plane can be written $a\,\mathbf{i} + b\,\mathbf{j}$
Component form makes algebraic operations with vectors straightforward

Understand

Concepts

Why $\mathbf{i}$ and $\mathbf{j}$ are called basis vectors for 2D space
How to read the components of a vector from a diagram or a coordinate description
Why adding and subtracting vectors algebraically is equivalent to the triangle/parallelogram law

Can do

Skills

Write a given 2D vector in component form $a\,\mathbf{i} + b\,\mathbf{j}$
Add, subtract and scalar-multiply vectors algebraically using components
Find the component form of a position vector from one point to another

Key terms

Unit vector $\mathbf{i}$The vector of magnitude 1 pointing in the positive $x$-direction: $\mathbf{i} = \begin{pmatrix}1\\0\end{pmatrix}$.

Unit vector $\mathbf{j}$The vector of magnitude 1 pointing in the positive $y$-direction: $\mathbf{j} = \begin{pmatrix}0\\1\end{pmatrix}$.

Component formWriting a vector as $a\,\mathbf{i} + b\,\mathbf{j}$, where $a$ and $b$ are real scalars called the $x$- and $y$-components.

$x$-componentThe scalar coefficient of $\mathbf{i}$; represents horizontal displacement (positive = right, negative = left).

$y$-componentThe scalar coefficient of $\mathbf{j}$; represents vertical displacement (positive = up, negative = down).

Basis vectorsA set of vectors ($\mathbf{i}$ and $\mathbf{j}$ in 2D) from which every other vector in the space can be built by linear combination.

Writing vectors in component form

core concept

Given a vector from point $A(x_1, y_1)$ to point $B(x_2, y_2)$, the component form is:

$$\overrightarrow{AB} = (x_2 - x_1)\,\mathbf{i} + (y_2 - y_1)\,\mathbf{j}$$

This is simply end minus start for each coordinate. The $x$-component tells you how far right (or left, if negative), and the $y$-component tells you how far up (or down).

Reading from a diagram: Count the horizontal squares for the $x$-component and the vertical squares for the $y$-component.

Example: The drone from the hook: 5 units east (positive $x$-direction) and 12 units north (positive $y$-direction) gives:

$\vec{v} = 5\,\mathbf{i} + 12\,\mathbf{j}$

From $A(1, 3)$ to $B(4, -2)$: $\overrightarrow{AB} = (4-1)\,\mathbf{i} + (-2-3)\,\mathbf{j} = 3\,\mathbf{i} - 5\,\mathbf{j}$

Compass directions. East = positive $\mathbf{i}$, West = negative $\mathbf{i}$, North = positive $\mathbf{j}$, South = negative $\mathbf{j}$. A bearing problem that says "4 km east and 3 km south" becomes $4\,\mathbf{i} - 3\,\mathbf{j}$ immediately.

Component form: $\overrightarrow{AB}=\begin{pmatrix}x_2-x_1\\y_2-y_1\end{pmatrix}$. Operations: $\vec{a}\pm\vec{b}=\begin{pmatrix}a_1\pm b_1\\a_2\pm b_2\end{pmatrix}$; $k\vec{a}=\begin{pmatrix}ka_1\\ka_2\end{pmatrix}$.

Pause — copy the end-minus-start rule: $\overrightarrow{AB}=B-A=\begin{pmatrix}x_2-x_1\\y_2-y_1\end{pmatrix}$ with a worked example into your book.

Quick check: Point $A$ is at $(2, 5)$ and point $B$ is at $(6, 1)$. What is $\overrightarrow{AB}$ in component form?

Algebraic operations in component form

core concept

We just saw that $\overrightarrow{AB}=\begin{pmatrix}x_2-x_1\\y_2-y_1\end{pmatrix}$ (end minus start). That raises a question: once all vectors are in component form, what are the exact component-by-component rules for addition, subtraction, and scalar multiplication? This card answers it → $\vec{a}+\vec{b}=\begin{pmatrix}a_1+b_1\\a_2+b_2\end{pmatrix}$; $\vec{a}-\vec{b}=\begin{pmatrix}a_1-b_1\\a_2-b_2\end{pmatrix}$; $k\vec{a}=\begin{pmatrix}ka_1\\ka_2\end{pmatrix}$.

Once vectors are in component form, all operations reduce to component-by-component arithmetic. There is no geometry needed — just algebra.

$$\text{Addition: } (a\,\mathbf{i}+b\,\mathbf{j})+(c\,\mathbf{i}+d\,\mathbf{j}) = (a+c)\,\mathbf{i}+(b+d)\,\mathbf{j}$$

$$\text{Subtraction: } (a\,\mathbf{i}+b\,\mathbf{j})-(c\,\mathbf{i}+d\,\mathbf{j}) = (a-c)\,\mathbf{i}+(b-d)\,\mathbf{j}$$

$$\text{Scalar multiplication: } k(a\,\mathbf{i}+b\,\mathbf{j}) = ka\,\mathbf{i}+kb\,\mathbf{j}$$

Why this works: $\mathbf{i}$-terms and $\mathbf{j}$-terms are perpendicular, so they behave like separate "channels" that never mix. You can treat each one independently, just like collecting like terms in algebra.

Example: Let $\vec{u} = 3\,\mathbf{i} - 2\,\mathbf{j}$ and $\vec{v} = -\mathbf{i} + 5\,\mathbf{j}$.

$\vec{u} + \vec{v} = (3-1)\,\mathbf{i} + (-2+5)\,\mathbf{j} = 2\,\mathbf{i} + 3\,\mathbf{j}$
$\vec{u} - \vec{v} = (3-(-1))\,\mathbf{i} + (-2-5)\,\mathbf{j} = 4\,\mathbf{i} - 7\,\mathbf{j}$
$3\vec{u} = 9\,\mathbf{i} - 6\,\mathbf{j}$

Like terms in disguise. $3\,\mathbf{i} + 2\,\mathbf{i} = 5\,\mathbf{i}$ for the same reason that $3x + 2x = 5x$. You cannot add $\mathbf{i}$-terms to $\mathbf{j}$-terms, just as you cannot add $x$-terms to $y$-terms.

Once vectors are in component form, all operations reduce to component-by-component arithmetic . There is no geometry needed — just algebra.

Pause — copy all three component operations: addition, subtraction, and scalar multiplication, each with a short example into your book.

Did you get this? True or false: $(2\,\mathbf{i} - 3\,\mathbf{j}) + (-5\,\mathbf{i} + \mathbf{j}) = -3\,\mathbf{i} - 2\,\mathbf{j}$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · WRITING COMPONENT FORM

Write the vector from $P(1, -3)$ to $Q(7, 2)$ in component form.

$\overrightarrow{PQ} = (x_Q - x_P)\,\mathbf{i} + (y_Q - y_P)\,\mathbf{j}$

Use the end-minus-start rule. Identify the start point $P(1,-3)$ and end point $Q(7,2)$.

PROBLEM 2 · VECTOR ADDITION

Given $\vec{a} = 4\,\mathbf{i} - \mathbf{j}$ and $\vec{b} = -2\,\mathbf{i} + 6\,\mathbf{j}$, find $2\vec{a} + \vec{b}$.

$2\vec{a} = 2(4\,\mathbf{i} - \mathbf{j}) = 8\,\mathbf{i} - 2\,\mathbf{j}$

Apply scalar multiplication: multiply each component by 2.

PROBLEM 3 · FIND AN UNKNOWN COMPONENT

Vectors $\vec{p} = 3\,\mathbf{i} + k\,\mathbf{j}$ and $\vec{q} = -\mathbf{i} + 2\,\mathbf{j}$. If $\vec{p} + \vec{q} = 2\,\mathbf{i} + 5\,\mathbf{j}$, find $k$.

$\vec{p}+\vec{q} = (3-1)\,\mathbf{i} + (k+2)\,\mathbf{j} = 2\,\mathbf{i} + (k+2)\,\mathbf{j}$

Add the components: $x$-component $= 3+(-1) = 2$ ✓; $y$-component $= k+2$, which we still need to determine.

Fill the gap: If $\vec{u} = 5\,\mathbf{i} - 2\,\mathbf{j}$ and $\vec{v} = -3\,\mathbf{i} + 7\,\mathbf{j}$, then $\vec{u} + \vec{v} =$ $\,\mathbf{i} +$ $\,\mathbf{j}$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Start minus end instead of end minus start

$\overrightarrow{AB}$ is end ($B$) minus start ($A$). Writing $A - B$ gives the reverse vector $\overrightarrow{BA}$. Always ask: where am I going? That's the end point — subtract the start from it.

Trap 02

Mixing $\mathbf{i}$ and $\mathbf{j}$ components

$(2\,\mathbf{i}+3\,\mathbf{j}) + (4\,\mathbf{i}+\mathbf{j}) \neq 9\,\mathbf{i}$ or $9\,\mathbf{j}$. Add $x$-to-$x$ and $y$-to-$y$ separately. $\mathbf{i}$ and $\mathbf{j}$ terms cannot combine because the directions are perpendicular.

Trap 03

Forgetting the scalar multiplies all components

$3(2\,\mathbf{i}+5\,\mathbf{j}) = 6\,\mathbf{i}+15\,\mathbf{j}$, not $6\,\mathbf{i}+5\,\mathbf{j}$. The scalar distributes over both components. This is the same distributive law as $3(2x+5y)$.

Did you get this? True or false: $4(3\,\mathbf{i} - 2\,\mathbf{j}) = 12\,\mathbf{i} - 8\,\mathbf{j}$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Write the vector from $A(-1, 4)$ to $B(5, -2)$ in component form.

Given $\vec{u} = 2\,\mathbf{i} + 7\,\mathbf{j}$ and $\vec{v} = 4\,\mathbf{i} - 3\,\mathbf{j}$, find $\vec{u} - 2\vec{v}$.

A ship travels $3\,\mathbf{i} + 4\,\mathbf{j}$ km then $-\mathbf{i} + 2\,\mathbf{j}$ km. Write the total displacement as a single vector.

If $\vec{a} = m\,\mathbf{i} + 3\,\mathbf{j}$ and $\vec{b} = 2\,\mathbf{i} - \mathbf{j}$, and $\vec{a} + \vec{b} = 5\,\mathbf{i} + 2\,\mathbf{j}$, find $m$.

Show that $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$ for $A(1,2)$, $B(4,5)$, $C(0,3)$ using component form.

Odd one out: Three of these are equal to $2\,\mathbf{i} + 6\,\mathbf{j}$. Which one is NOT?

Revisit your thinking

Earlier you described the drone's displacement of 5 units east and 12 units north.

The component form is $\vec{v} = 5\,\mathbf{i} + 12\,\mathbf{j}$. The power of this notation is that it packs two pieces of information into one compact expression, and all operations — addition, subtraction, scalar multiplication — become pure arithmetic on those two numbers. Did your original description capture both pieces of information? What did you get right, and what would you change?

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 31 mark

Q1. Write the vector from $A(2, -1)$ to $B(-3, 4)$ in component form. (1 mark)

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ApplyBand 42 marks

Q2. Given $\vec{p} = 3\,\mathbf{i} - 4\,\mathbf{j}$ and $\vec{q} = -\mathbf{i} + 2\,\mathbf{j}$, find $2\vec{p} - 3\vec{q}$. (2 marks)

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AnalyseBand 53 marks

Q3. Points $A$, $B$, $C$ have position vectors $\vec{a} = 2\,\mathbf{i}+\mathbf{j}$, $\vec{b} = 5\,\mathbf{i}-2\,\mathbf{j}$, $\vec{c} = -\mathbf{i}+4\,\mathbf{j}$ respectively. Find $\overrightarrow{BC}$ and hence write $\overrightarrow{BC}$ in terms of $\vec{a}$, $\vec{b}$ and $\vec{c}$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers: 1. $\overrightarrow{AB} = 6\,\mathbf{i}-6\,\mathbf{j}$ · 2. $\vec{u}-2\vec{v} = (2-8)\,\mathbf{i}+(7+6)\,\mathbf{j} = -6\,\mathbf{i}+13\,\mathbf{j}$ · 3. $2\,\mathbf{i}+6\,\mathbf{j}$ · 4. $m=3$ · 5. $\overrightarrow{AB}=3\,\mathbf{i}+3\,\mathbf{j}$, $\overrightarrow{BC}=-4\,\mathbf{i}-2\,\mathbf{j}$, $\overrightarrow{AC}=-\mathbf{i}+\mathbf{j}$; sum $= -\mathbf{i}+\mathbf{j} = \overrightarrow{AC}$ ✓

Q1 (1 mark): $\overrightarrow{AB} = (-3-2)\,\mathbf{i}+(4-(-1))\,\mathbf{j} = -5\,\mathbf{i}+5\,\mathbf{j}$ [1].

Q2 (2 marks): $2\vec{p} = 6\,\mathbf{i}-8\,\mathbf{j}$ [1]; $3\vec{q} = -3\,\mathbf{i}+6\,\mathbf{j}$; $2\vec{p}-3\vec{q} = (6+3)\,\mathbf{i}+(-8-6)\,\mathbf{j} = \mathbf{9\,i-14\,j}$ [1].

Q3 (3 marks): $\overrightarrow{BC} = \vec{c}-\vec{b} = (-1-5)\,\mathbf{i}+(4-(-2))\,\mathbf{j} = -6\,\mathbf{i}+6\,\mathbf{j}$ [1]; in terms of position vectors $\overrightarrow{BC} = \vec{c}-\vec{b}$ [1]; confirm: $-6\,\mathbf{i}+6\,\mathbf{j}$ [1].

Boss battle · The Component Master

earn bronze · silver · gold

Five timed questions on 2D component form. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering 2D vector component questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 6 · Scalar Multiplication Lesson 8 · Component Form in 3D →

Module overview · Extension 1 · Checkpoint 2