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Module 6 · L8 of 20 ~35 min ⚡ +95 XP available

Component Form in 3D

You already know how to write $\vec{v} = a\,\mathbf{i} + b\,\mathbf{j}$ in 2D. Now add one more dimension — a $z$-axis — and a third unit vector $\mathbf{k}$. The jump from flat geometry to three-dimensional space sounds daunting, but the algebra barely changes: you just carry one extra component. This lesson builds your intuition for 3D space and makes the extension feel natural.

Today's hook — A submarine travels 4 km east, 3 km north, and then dives 2 km underwater. You already know how to handle the first two parts. Before studying this lesson — how do you think you'd extend the notation to include depth? Jot your idea.

0/5QUESTS

Recall — your gut answer first

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A submarine travels 4 km east, 3 km north, and 2 km downward (into the sea). Without any formula — how would you write this as a single vector that captures all three directions? What do you think changes when you move from 2D to 3D notation?

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The key idea — adding a third axis

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In 2D we had two perpendicular unit vectors $\mathbf{i}$ and $\mathbf{j}$. In 3D we add a third unit vector $\mathbf{k}$ pointing in the positive $z$-direction (often interpreted as upward or out-of-the-page), perpendicular to both $\mathbf{i}$ and $\mathbf{j}$.

Every vector in 3D space can be written as $\vec{v} = a\,\mathbf{i} + b\,\mathbf{j} + c\,\mathbf{k}$, where $a$, $b$, $c$ are the $x$-, $y$- and $z$-components. The column vector equivalent is $\begin{pmatrix}a\\b\\c\end{pmatrix}$.

$\mathbf{i}=\begin{pmatrix}1\\0\\0\end{pmatrix}$, $\mathbf{j}=\begin{pmatrix}0\\1\\0\end{pmatrix}$, $\mathbf{k}=\begin{pmatrix}0\\0\\1\end{pmatrix}$

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

$\mathbf{k}$ is a unit vector

$|\mathbf{k}| = 1$, and $\mathbf{k}$ is perpendicular to both $\mathbf{i}$ and $\mathbf{j}$. The three vectors $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ form a right-handed coordinate system.

Column vector equivalent

$a\,\mathbf{i}+b\,\mathbf{j}+c\,\mathbf{k}$ and $\begin{pmatrix}a\\b\\c\end{pmatrix}$ are interchangeable. The HSC accepts both.

2D is a special case

A 2D vector $a\,\mathbf{i}+b\,\mathbf{j}$ is the same as the 3D vector $a\,\mathbf{i}+b\,\mathbf{j}+0\,\mathbf{k}$. The $z$-component is simply zero.

What you'll master

Know

Key facts

$\mathbf{k} = \begin{pmatrix}0\\0\\1\end{pmatrix}$ is the unit vector in the positive $z$-direction
Any 3D vector is $a\,\mathbf{i}+b\,\mathbf{j}+c\,\mathbf{k}$ or equivalently $\begin{pmatrix}a\\b\\c\end{pmatrix}$
3D operations use the same rules as 2D — just three components instead of two

Understand

Concepts

Why $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ are mutually perpendicular and form a basis for 3D space
How to read 3D components from a coordinate description or a diagram
How the end-minus-start rule extends from 2D to 3D points

Can do

Skills

Write any 3D vector in component form $a\,\mathbf{i}+b\,\mathbf{j}+c\,\mathbf{k}$
Add, subtract and scalar-multiply 3D vectors by working component-by-component
Find $\overrightarrow{AB}$ in 3D given the coordinates of $A$ and $B$

Key terms

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

3D component formWriting a vector as $a\,\mathbf{i}+b\,\mathbf{j}+c\,\mathbf{k}$ where $a$, $b$, $c$ are the $x$-, $y$- and $z$-components.

$z$-componentThe scalar coefficient of $\mathbf{k}$; represents displacement in the $z$-direction (often height or depth).

Right-handed systemThe standard orientation of $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$: if you point your right-hand fingers in the $\mathbf{i}$ direction and curl them toward $\mathbf{j}$, your thumb points in the $\mathbf{k}$ direction.

3D position vectorThe vector from the origin $O$ to a point $P(x,y,z)$: $\overrightarrow{OP} = x\,\mathbf{i}+y\,\mathbf{j}+z\,\mathbf{k}$.

Zero vector in 3D$\mathbf{0} = 0\,\mathbf{i}+0\,\mathbf{j}+0\,\mathbf{k} = \begin{pmatrix}0\\0\\0\end{pmatrix}$ — has no direction and magnitude zero.

Writing vectors in 3D component form

core concept

The end-minus-start rule from 2D extends directly to 3D. For points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$:

$$\overrightarrow{AB} = (x_2-x_1)\,\mathbf{i} + (y_2-y_1)\,\mathbf{j} + (z_2-z_1)\,\mathbf{k}$$

The position vector of a point $P(x,y,z)$ from the origin is $\overrightarrow{OP} = x\,\mathbf{i}+y\,\mathbf{j}+z\,\mathbf{k}$.

Example — the submarine: Starting at the origin, it travels 4 km east ($+x$), 3 km north ($+y$), 2 km down ($-z$, since downward is the negative $z$-direction):

$\vec{d} = 4\,\mathbf{i} + 3\,\mathbf{j} - 2\,\mathbf{k}$

Example — from coordinates: From $A(1, -2, 3)$ to $B(4, 1, -1)$:

$\overrightarrow{AB} = (4-1)\,\mathbf{i}+(1-(-2))\,\mathbf{j}+(-1-3)\,\mathbf{k} = 3\,\mathbf{i}+3\,\mathbf{j}-4\,\mathbf{k}$

Positive $z$ direction matters. Always check the context: in many HSC problems, $\mathbf{k}$ points upward (so $+\mathbf{k}$ = going up, $-\mathbf{k}$ = going down). The submarine dives, so its $z$-component is negative.

The end-minus-start rule from 2D extends directly to 3D. For points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$:

Pause — copy the 3D component form: $\overrightarrow{AB}=\begin{pmatrix}x_2-x_1\\y_2-y_1\\z_2-z_1\end{pmatrix}$ with a worked example of two 3D points into your book.

Quick check: What is $\overrightarrow{AB}$ for $A(2, -1, 4)$ and $B(5, 3, -2)$?

Algebraic operations in 3D

core concept

We just saw that $\overrightarrow{AB}=\begin{pmatrix}x_2-x_1\\y_2-y_1\\z_2-z_1\end{pmatrix}$ in 3D — the end-minus-start rule extends directly by adding a $z$-component. That raises a question: do all vector algebraic operations in 3D simply gain a third component, or are there operations that work differently in 3D? This card answers it → addition, subtraction, and scalar multiplication all extend by adding the $z$-component; magnitude uses $\sqrt{a_1^2+a_2^2+a_3^2}$.

All operations in 3D follow exactly the same rules as 2D — just with a third component. Work component by component across all three axes.

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

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

Example: Let $\vec{u} = 2\,\mathbf{i} - \mathbf{j} + 3\,\mathbf{k}$ and $\vec{v} = -\mathbf{i} + 4\,\mathbf{j} - 2\,\mathbf{k}$.

$\vec{u}+\vec{v} = (2-1)\,\mathbf{i}+(-1+4)\,\mathbf{j}+(3-2)\,\mathbf{k} = \mathbf{i}+3\,\mathbf{j}+\mathbf{k}$
$\vec{u}-\vec{v} = (2+1)\,\mathbf{i}+(-1-4)\,\mathbf{j}+(3+2)\,\mathbf{k} = 3\,\mathbf{i}-5\,\mathbf{j}+5\,\mathbf{k}$
$-2\vec{v} = 2\,\mathbf{i}-8\,\mathbf{j}+4\,\mathbf{k}$

Nothing new to learn. Every rule you used in 2D works in 3D. The only difference is keeping track of a third component. In an exam, set out your working with $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ clearly labelled to avoid arithmetic slips.

All operations in 3D follow exactly the same rules as 2D — just with a third component. Work component by component across all three axes.

Pause — copy the 3D operation rules: $\vec{a}+\vec{b}=\begin{pmatrix}a_1+b_1\\a_2+b_2\\a_3+b_3\end{pmatrix}$ and $|\vec{a}|=\sqrt{a_1^2+a_2^2+a_3^2}$ into your book.

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

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

PROBLEM 1 · WRITING 3D COMPONENT FORM

Write the vector from $A(3, 0, -2)$ to $B(-1, 5, 4)$ in component form.

$\overrightarrow{AB} = (x_B-x_A)\,\mathbf{i}+(y_B-y_A)\,\mathbf{j}+(z_B-z_A)\,\mathbf{k}$

Apply end-minus-start in all three coordinates. Start at $A(3,0,-2)$, end at $B(-1,5,4)$.

PROBLEM 2 · 3D VECTOR ARITHMETIC

Given $\vec{a} = 2\,\mathbf{i}-3\,\mathbf{j}+\mathbf{k}$ and $\vec{b} = -\mathbf{i}+2\,\mathbf{j}-4\,\mathbf{k}$, find $3\vec{a}+2\vec{b}$.

$3\vec{a} = 3(2\,\mathbf{i}-3\,\mathbf{j}+\mathbf{k}) = 6\,\mathbf{i}-9\,\mathbf{j}+3\,\mathbf{k}$

Multiply every component of $\vec{a}$ by 3: $2\to6$, $-3\to-9$, $1\to3$.

PROBLEM 3 · FIND UNKNOWN COMPONENT IN 3D

Vectors $\vec{p} = 2\,\mathbf{i}+m\,\mathbf{j}-\mathbf{k}$ and $\vec{q} = -3\,\mathbf{i}+\mathbf{j}+4\,\mathbf{k}$. If $\vec{p}-\vec{q} = 5\,\mathbf{i}+2\,\mathbf{j}-5\,\mathbf{k}$, find $m$.

$\vec{p}-\vec{q} = (2-(-3))\,\mathbf{i}+(m-1)\,\mathbf{j}+(-1-4)\,\mathbf{k} = 5\,\mathbf{i}+(m-1)\,\mathbf{j}-5\,\mathbf{k}$

Compute the subtraction component by component. The $x$- and $z$-components are fully determined; only the $y$-component involves $m$.

Fill the gap: For $\vec{u} = \mathbf{i}-2\,\mathbf{j}+3\,\mathbf{k}$ and $\vec{v} = 4\,\mathbf{i}+\mathbf{j}-\mathbf{k}$, the $z$-component of $\vec{u}+\vec{v}$ is .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting the third component

When adding or subtracting vectors, students sometimes forget to operate on the $z$-component. Write all three components in every step, even if the $z$-component is zero. Missing one component gives the wrong vector.

Trap 02

Double-negative errors in the $z$-component

$\overrightarrow{AB}$ with $z_A = -2$ and $z_B = 4$ gives $z_B - z_A = 4 - (-2) = 6$, not $4-2=2$. In 3D, negative coordinates for the $z$-axis are very common — slow down and write out each subtraction explicitly.

Trap 03

Confusing $\mathbf{j}$ and $\mathbf{k}$ directions

$\mathbf{j}$ is the $y$-direction (often north/south or left/right), while $\mathbf{k}$ is the $z$-direction (often up/down or in/out). Always match the unit vector to the correct physical direction stated in the problem.

Did you get this? True or false: the $z$-component of $\overrightarrow{AB}$ where $A(1,2,-3)$ and $B(4,0,1)$ is $4$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Write the position vector of $P(3, -1, 5)$ from the origin in component form.

Find $\overrightarrow{PQ}$ for $P(2, 4, -1)$ and $Q(-3, 1, 5)$.

Given $\vec{r} = \mathbf{i}+2\,\mathbf{j}-\mathbf{k}$ and $\vec{s} = 3\,\mathbf{i}-\mathbf{j}+4\,\mathbf{k}$, find $2\vec{r}-\vec{s}$.

If $\vec{m} = 2\,\mathbf{i}+\mathbf{j}+n\,\mathbf{k}$ and $\vec{m}+3\,\mathbf{k} = 2\,\mathbf{i}+\mathbf{j}+\mathbf{k}$, find $n$.

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

Odd one out: Three of these are correct statements about 3D unit vectors. Which one is NOT?

Revisit your thinking

Earlier you tried to describe the submarine's displacement of 4 km east, 3 km north, 2 km down using vector notation.

The answer is $\vec{d} = 4\,\mathbf{i}+3\,\mathbf{j}-2\,\mathbf{k}$. The only thing that changed from 2D is the addition of a third component with the unit vector $\mathbf{k}$. Every operation you already know — addition, subtraction, scalar multiplication — works in exactly the same way. Did your initial idea come close? What would you change now?

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Multiple choice

+5 XP per correct · +25 XP all-correct

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 $P(4, -1, 3)$ to $Q(1, 2, -2)$ in component form. (1 mark)

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ApplyBand 42 marks

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

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AnalyseBand 53 marks

Q3. Points $A(2, -1, 3)$, $B(5, 2, k)$, $C(8, 5, -1)$ are collinear (lie on a straight line). Show that $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel and find $k$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers: 1. $3\,\mathbf{i}-\mathbf{j}+5\,\mathbf{k}$ · 2. $\overrightarrow{PQ} = -5\,\mathbf{i}+3\,\mathbf{j}-3\,\mathbf{k}$ · 3. $2\vec{r} = 2\,\mathbf{i}+4\,\mathbf{j}-2\,\mathbf{k}$; $2\vec{r}-\vec{s} = (2-3)\,\mathbf{i}+(4+1)\,\mathbf{j}+(-2-4)\,\mathbf{k} = -\mathbf{i}+5\,\mathbf{j}-6\,\mathbf{k}$ · 4. $n = -2$ · 5. $\overrightarrow{AB}=2\,\mathbf{i}+\mathbf{j}-3\,\mathbf{k}$, $\overrightarrow{BC}=-3\,\mathbf{i}+3\,\mathbf{j}+4\,\mathbf{k}$, $\overrightarrow{AC}=-\mathbf{i}+4\,\mathbf{j}+\mathbf{k}$; sum $=-\mathbf{i}+4\,\mathbf{j}+\mathbf{k}=\overrightarrow{AC}$ ✓

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

Q2 (2 marks): $3\vec{u} = 3\,\mathbf{i}-9\,\mathbf{j}+6\,\mathbf{k}$ [1]; $2\vec{v} = 8\,\mathbf{i}+4\,\mathbf{j}-2\,\mathbf{k}$; $3\vec{u}-2\vec{v} = (3-8)\,\mathbf{i}+(-9-4)\,\mathbf{j}+(6+2)\,\mathbf{k} = \mathbf{-5\,i-13\,j+8\,k}$ [1].

Q3 (3 marks): $\overrightarrow{AB} = 3\,\mathbf{i}+3\,\mathbf{j}+(k-3)\,\mathbf{k}$; $\overrightarrow{AC} = 6\,\mathbf{i}+6\,\mathbf{j}-4\,\mathbf{k}$ [1]. From $x$- and $y$-components: $\overrightarrow{AC} = 2\,\overrightarrow{AB}$ [1], so $z$-component: $-4 = 2(k-3) \Rightarrow k-3 = -2 \Rightarrow k = 1$ [1].

Boss battle · The 3D Vector Master

earn bronze · silver · gold

Five timed questions on 3D 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 3D vector component questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 7 · Component Form in 2D Lesson 9 · Magnitude of a Vector →

Module overview · Extension 1 · Checkpoint 2