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Module 14 · L01 of 12 ~40 min ⚡ +90 XP available

Vectors in 3D — Notation & Components

In Year 11 you treated vectors as arrows in a flat plane. Now lift them off the page. Every point in three-dimensional space can be addressed by three coordinates $(x, y, z)$ — and every directed quantity, from a flight path to a force, can be decomposed into three unit vectors $\mathbf{i}, \mathbf{j}, \mathbf{k}$. This lesson sets up the language and magnitude formula you'll use for the rest of Module 14.

Today's hook — A drone hovers at the point $P(3, 4, 12)$ relative to its launch pad (measured in metres east, north, up). Before reading on, sketch the position vector $\vec{OP}$ and predict its length. Then compute $|\vec{OP}|$ and check your guess. The number is much cleaner than you'd expect.

0/5QUESTS

Recall — your gut answer first

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Write the 2D vector from $A(1, 2)$ to $B(4, 6)$ in $\mathbf{i}, \mathbf{j}$ form and find its magnitude. Before checking — what extra component would you add if $B$ were lifted to $(4, 6, 12)$? Sketch your reasoning below.

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The two moves for writing a 3D vector

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Every 3D vector rewards two habits: pin down the three components (the signed displacements in $x$, $y$, $z$) and then choose a notation that suits the question (component form $\langle x, y, z\rangle$, column form, or $x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$). Mixing forms mid-working is the single biggest cause of arithmetic slips.

The components-axis-magnitude reading: (1) identify the displacement in each axis $x$, $y$, $z$, (2) attach the unit vectors $\mathbf{i}, \mathbf{j}, \mathbf{k}$, (3) compute the magnitude via Pythagoras in 3D.

Column: $\begin{pmatrix} x \\ y \\ z \end{pmatrix}$ · ijk: $x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$ · Magnitude: $\sqrt{x^2 + y^2 + z^2}$

$|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}$

i, j, k are perpendicular unit vectors

$\mathbf{i}, \mathbf{j}, \mathbf{k}$ each have length 1 and lie along the positive $x$-, $y$-, $z$-axes. Any 3D vector is a unique sum of scalar multiples of these three.

Position vs displacement

A position vector $\vec{OP}$ goes from the origin to $P$. A displacement vector $\vec{AB}$ goes from $A$ to $B$ and equals $\vec{OB} - \vec{OA}$ in components.

Equal vectors need equal components

Two vectors are equal iff all three components match: $a_1 = b_1$, $a_2 = b_2$, $a_3 = b_3$. Direction and magnitude alone aren't enough to check — compare component by component.

What you'll master

Know

Key facts

3D space is coordinatised by three perpendicular axes $x$, $y$, $z$
$\mathbf{i}, \mathbf{j}, \mathbf{k}$ are unit vectors along the positive $x$-, $y$-, $z$-axes
Component form, column form and ijk form are interchangeable
$|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}$ for $\mathbf{v} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$

Understand

Concepts

Why magnitude in 3D is a double application of Pythagoras
Why $\vec{AB} = \vec{OB} - \vec{OA}$ in any coordinate system
Why two vectors are equal only when all components match

Can do

Skills

Convert between column form, component form and ijk form
Compute the magnitude of any 3D vector
Write a position vector and the displacement between two points

Key terms

Unit vectors $\mathbf{i}, \mathbf{j}, \mathbf{k}$Mutually perpendicular vectors of length 1 along the positive $x$-, $y$- and $z$-axes respectively. They form an orthonormal basis for 3D space.

Component formA 3D vector written as $\mathbf{v} = \langle x, y, z\rangle$ where $x, y, z$ are the signed scalar projections onto each axis.

Column formA 3D vector written vertically as $\begin{pmatrix} x \\ y \\ z \end{pmatrix}$. Algebraically identical to component form; useful for matrix work later.

Position vectorThe vector $\vec{OP}$ from the origin $O$ to the point $P(x, y, z)$. Its components are exactly the coordinates of $P$.

MagnitudeThe length of a vector. For $\mathbf{v} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$, $|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}$. Magnitude is always non-negative.

Equal vectorsTwo vectors are equal iff they have the same magnitude AND the same direction — equivalently, identical components in every axis.

MEX-V1NESA outcome (Further Work with Vectors): introduces three-dimensional vectors and operations on them, including notation $\mathbf{i}, \mathbf{j}, \mathbf{k}$, magnitude and the geometry of 3D space.

Notation: column, component and ijk form

core concept

A 3D vector $\mathbf{v}$ with $x$-component $a$, $y$-component $b$, $z$-component $c$ can be written in three interchangeable ways:

Column form $\mathbf{v} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$ — stacks the components; ideal for hand-working and for matrix-vector products later.
Component (angle bracket) form $\mathbf{v} = \langle a, b, c\rangle$ — compact, used in calculus and physics.
ijk form $\mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}$ — explicit; emphasises that any 3D vector is a sum of axis-aligned pieces.

Worked through the hook: The drone position vector $\vec{OP}$ with $P(3, 4, 12)$:

Column: $\vec{OP} = \begin{pmatrix} 3 \\ 4 \\ 12 \end{pmatrix}$
Component: $\vec{OP} = \langle 3, 4, 12 \rangle$
ijk: $\vec{OP} = 3\mathbf{i} + 4\mathbf{j} + 12\mathbf{k}$
Magnitude: $|\vec{OP}| = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$ metres.

Connecting to magnitude. The formula $|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}$ is Pythagoras applied twice: first in the $xy$-plane (distance $\sqrt{x^2+y^2}$ from the origin to the foot of the perpendicular), then up the $z$-axis using that distance and $z$ as the two legs of a new right triangle.

Three forms: column $\binom{a}{b}\!\!\binom{c}{}$, component $\langle a,b,c\rangle$, ijk $a\mathbf{i}+b\mathbf{j}+c\mathbf{k}$ · Position vector $\vec{OP}$ has components = coordinates of $P$ · $|\mathbf{v}| = \sqrt{x^2+y^2+z^2}$ — Pythagoras applied twice · Drone example: $P(3,4,12)$ gives $|\vec{OP}| = 13$

Pause — copy the three notational forms (column, component, $\mathbf{i}\mathbf{j}\mathbf{k}$), the magnitude formula $|\mathbf{v}| = \sqrt{x^2+y^2+z^2}$, and the position vector of a point into your book.

Quick check: Which expression is the magnitude of $\mathbf{v} = 2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}$?

Position vectors and displacement between points

core concept

We just saw that a 3D vector is written in column, component $\langle a,b,c \rangle$, or $\mathbf{i}\mathbf{j}\mathbf{k}$ form with $|\mathbf{v}| = \sqrt{x^2+y^2+z^2}$ (Pythagoras applied twice). That raises a question: how do position vectors and point coordinates relate, and how do we find the displacement vector between two points? This card answers it → $\vec{AB} = \vec{OB} - \vec{OA}$ (head minus tail) and equal vectors require all three components to match.

Every point $P(x, y, z)$ in 3D space has a unique position vector $\vec{OP} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$ pointing from the origin. To get the vector from $A$ to $B$:

$\vec{AB} = \vec{OB} - \vec{OA}$ — subtract the position vector of the tail from the position vector of the head.
In components: if $A(a_1, a_2, a_3)$ and $B(b_1, b_2, b_3)$, then $\vec{AB} = (b_1 - a_1)\mathbf{i} + (b_2 - a_2)\mathbf{j} + (b_3 - a_3)\mathbf{k}$.

Why equal vectors require equal components:

$\mathbf{u} = u_1\mathbf{i} + u_2\mathbf{j} + u_3\mathbf{k}$ and $\mathbf{v} = v_1\mathbf{i} + v_2\mathbf{j} + v_3\mathbf{k}$ are equal iff $u_1 = v_1$, $u_2 = v_2$ and $u_3 = v_3$ — direction and magnitude alone could agree but lie along different axes.

$$\vec{AB} = \vec{OB} - \vec{OA} = (b_1 - a_1)\mathbf{i} + (b_2 - a_2)\mathbf{j} + (b_3 - a_3)\mathbf{k}$$

Common mistake. Students sometimes compute $\vec{AB}$ as $\vec{OA} - \vec{OB}$, getting the vector pointing the wrong way. Mnemonic: "head minus tail" — the head of the arrow is the second-named point, so $\vec{AB} = B - A$.

$\vec{OP}$ = position vector with components matching coordinates of $P$ · $\vec{AB} = \vec{OB} - \vec{OA}$ (head minus tail) · Equal vectors $\Leftrightarrow$ all three components match · Magnitude is always non-negative, even if components are negative

Pause — copy $\vec{AB} = \vec{OB} - \vec{OA}$ (head minus tail), the equal-vector condition (all three components match), and the non-negativity of magnitude into your book.

Did you get this? True or false: if $A(1, -2, 5)$ and $B(4, 0, -1)$, then $\vec{AB} = 3\mathbf{i} + 2\mathbf{j} - 6\mathbf{k}$.

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

PROBLEM 1 · CONVERT BETWEEN FORMS

Write $\mathbf{v} = -2\mathbf{i} + 5\mathbf{j} - \mathbf{k}$ in (a) column form and (b) component (angle-bracket) form, and compute $|\mathbf{v}|$.

Identify components: $x = -2$, $y = 5$, $z = -1$. Note the implicit "$-1$" coefficient on $\mathbf{k}$ — easy to miss.

Always isolate the three scalar coefficients before changing notation. Sign errors here propagate everywhere.

PROBLEM 2 · DISPLACEMENT BETWEEN POINTS

Given $A(2, -1, 4)$ and $B(5, 3, -2)$, find $\vec{AB}$ in ijk form and the distance $|AB|$.

Apply "head minus tail": $\vec{AB} = \vec{OB} - \vec{OA} = (5 - 2)\mathbf{i} + (3 - (-1))\mathbf{j} + (-2 - 4)\mathbf{k}$.

Subtract component by component, watching signs on the negatives. Write the differences before simplifying.

PROBLEM 3 · EQUAL VECTORS

Find $a, b, c$ such that $\mathbf{u} = (a+1)\mathbf{i} + (2b - 3)\mathbf{j} + (c^2)\mathbf{k}$ equals $\mathbf{v} = 4\mathbf{i} + \mathbf{j} + 9\mathbf{k}$. Give all solutions.

Equal vectors require equal components. Match each pair: $a + 1 = 4$, $2b - 3 = 1$, $c^2 = 9$.

Equality of vectors in $\mathbb{R}^3$ is component-wise — you get a separate scalar equation for each axis. Treat them independently.

Fill the gap: For $\mathbf{v} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$, the magnitude is $|\mathbf{v}| = \sqrt{x^2 +$ $+$ $}$, which is Pythagoras applied twice in 3D space.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting the implicit $\pm 1$ coefficient

A vector written $\mathbf{v} = \mathbf{i} - \mathbf{k}$ has $x = 1$, $y = 0$, $z = -1$. Students often record $z = 1$ (missing the sign) or $y = 1$ (forgetting that no $\mathbf{j}$ term means $y = 0$). Always rewrite with every coefficient explicit before computing magnitude.

Trap 02

Computing $\vec{AB}$ as $A - B$

The displacement vector $\vec{AB}$ goes FROM $A$ TO $B$, so it equals $\vec{OB} - \vec{OA}$ — head minus tail. Reversing gives $\vec{BA}$ (the opposite direction). Always say the mnemonic "head minus tail" before subtracting.

Trap 03

Adding signed components inside the square root

$|2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}| \neq \sqrt{2 - 3 + 6}$. You must square each component first (which kills the signs), then sum, then take the square root: $\sqrt{4 + 9 + 36} = 7$. The squaring step is what makes 3D Pythagoras work.

Did you get this? True or false: the magnitude of $\mathbf{v} = \mathbf{i} - \mathbf{k}$ is $\sqrt{2}$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Write $\mathbf{v} = 4\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ in column form and component form, and find $|\mathbf{v}|$.

Find $\vec{PQ}$ and $|PQ|$ for $P(0, 2, -3)$ and $Q(6, -1, 5)$.

Find the value(s) of $t$ such that $|\langle t, 2, -2\rangle| = 3$.

Given $\mathbf{a} = (k-1)\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}$ and $\mathbf{b} = 4\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}$, find $k$ so $\mathbf{a} = \mathbf{b}$.

A particle at $A(1, 1, 1)$ travels to $B(4, 5, 13)$. Find the position vector of $B$, the displacement $\vec{AB}$ and the distance travelled.

Odd one out: Three of these expressions represent the same 3D vector. Which one is the odd one out?

Revisit your thinking

Earlier you sketched $\vec{OP}$ for the drone at $P(3, 4, 12)$ and predicted its length.

Most students guess somewhere between 15 and 20, but the answer is exactly 13 — a clean Pythagorean triple (the $3$-$4$-$12$-$13$ quadruple). The position vector is $\vec{OP} = 3\mathbf{i} + 4\mathbf{j} + 12\mathbf{k}$ and its magnitude $\sqrt{9 + 16 + 144} = \sqrt{169} = 13$ metres. Spotting these clean triples ($3$-$4$-$12$-$13$, $1$-$2$-$2$-$3$, $2$-$3$-$6$-$7$) will save you arithmetic time throughout Module 14.

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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 32 marks

Q1. Write the position vector of $P(-1, 4, 8)$ in ijk form and find its magnitude. (2 marks)

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ApplyBand 43 marks

Q2. The points $A$, $B$ and $C$ have coordinates $(1, 2, 3)$, $(4, 6, 3)$ and $(1, 2, 8)$. Find $\vec{AB}$, $\vec{AC}$ and the lengths $|AB|$, $|AC|$. (3 marks)

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AnalyseBand 53 marks

Q3. Find all values of $\lambda$ such that the vector $\mathbf{v} = \lambda\mathbf{i} + (\lambda + 1)\mathbf{j} + 2\mathbf{k}$ has magnitude $\sqrt{13}$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. Column: $\begin{pmatrix} 4 \\ -1 \\ 2 \end{pmatrix}$. Component: $\langle 4, -1, 2\rangle$. $|\mathbf{v}| = \sqrt{16 + 1 + 4} = \sqrt{21}$.

2. $\vec{PQ} = (6-0)\mathbf{i} + (-1-2)\mathbf{j} + (5-(-3))\mathbf{k} = 6\mathbf{i} - 3\mathbf{j} + 8\mathbf{k}$. $|PQ| = \sqrt{36 + 9 + 64} = \sqrt{109}$.

3. $t^2 + 4 + 4 = 9 \Rightarrow t^2 = 1 \Rightarrow t = \pm 1$.

4. Component match: $k - 1 = 4 \Rightarrow k = 5$ (the $\mathbf{j}$ and $\mathbf{k}$ components already agree).

5. $\vec{OB} = 4\mathbf{i} + 5\mathbf{j} + 13\mathbf{k}$. $\vec{AB} = 3\mathbf{i} + 4\mathbf{j} + 12\mathbf{k}$. $|AB| = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$.

Q1 (2 marks): $\vec{OP} = -\mathbf{i} + 4\mathbf{j} + 8\mathbf{k}$ [1]. $|\vec{OP}| = \sqrt{1 + 16 + 64} = \sqrt{81} = 9$ [1].

Q2 (3 marks): $\vec{AB} = 3\mathbf{i} + 4\mathbf{j}$, $|AB| = \sqrt{9 + 16} = 5$ [1]. $\vec{AC} = 5\mathbf{k}$, $|AC| = 5$ [1]. Both displacements have length 5 — note that $AB$ lies in a horizontal plane while $AC$ is purely vertical [1].

Q3 (3 marks): Set $|\mathbf{v}|^2 = 13$: $\lambda^2 + (\lambda+1)^2 + 4 = 13$ [1]. Expand: $2\lambda^2 + 2\lambda + 5 = 13 \Rightarrow \lambda^2 + \lambda - 4 = 0$ [1]. Solve via the quadratic formula: $\lambda = \dfrac{-1 \pm \sqrt{17}}{2}$ [1].

Boss battle · The 3D Cartographer

earn bronze · silver · gold

Five timed questions on 3D notation, position vectors, displacement and magnitude. 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 quick 3D vector questions. Lighter alternative to the boss.

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Module overview · Extension 2 · Checkpoint 1