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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 formulahow 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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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

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

Module overview · Extension 1 · Checkpoint 2