Vectors are the modelling backbone of Topic 3 in IB Mathematics Applications & Interpretation HL. AI treats a vector less as an abstract arrow and more as a way to carry two facts at once — a size and a direction — which is exactly what you need to describe a position, a displacement or a velocity in the plane. Because a calculator is always allowed, the marks live in setting the problem up as vectors and reading the GDC's answer correctly, not in grinding the arithmetic by hand.
This cheatsheet condenses AHL 3.10–3.13 — column-vector notation and magnitude, unit vectors, the algebra of vectors, the scalar (dot) product and the angle between two vectors, the perpendicularity test, the vector equation of a line, and the AI applications of velocity vectors and particles that meet — onto one page, and flags the traps that quietly cost method marks. The print-ready PDF is at the bottom, free to download.
§1 — Vectors, column notation & magnitude AHL 3.13
A vector carries both magnitude and direction. Write it as a column of components — one per axis — and its magnitude (length) comes straight from Pythagoras.
Column vector:$\vec{a}=\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$ lists the components along each axis (drop $a_3$ in 2-D)
Magnitude (2-D):$|\vec{a}|=\sqrt{a_1^{2}+a_2^{2}}$
Magnitude (3-D):$|\vec{a}|=\sqrt{a_1^{2}+a_2^{2}+a_3^{2}}$ — the length of the arrow
Displacement:$\overrightarrow{AB}=\vec{b}-\vec{a}$ — the position vector of $B$ minus that of $A$
TrickSquare each component, add, then square-root — never just add the components. In AI the magnitude is the speed of a velocity vector or the distance of a displacement.
TrapThe signs of the components never affect $|\vec{a}|$ because squaring removes them — but they entirely decide the direction. Keep a point's position vector and a line's direction vector as separate objects.
§2 — Unit vectors & direction AHL 3.13
A unit vector has length exactly $1$ and points the same way as the original. You build one by dividing every component by the magnitude — "normalising".
Unit vector:$\hat{a}=\dfrac{1}{|\vec{a}|}\,\vec{a}$ — divide every component by the magnitude
Check:$|\hat{a}|=1$ for any non-zero $\vec{a}$
Rescale:a vector of length $L$ in the direction of $\vec{a}$ is $L\hat{a}$
TrickFind $|\vec{a}|$ first, then divide. A velocity of speed $s$ heading in direction $\vec{d}$ is simply $s\hat{d}$ — normalise the direction, then scale by the speed.
§3 — Adding, subtracting & scaling AHL 3.13
Addition, subtraction and scalar multiplication all act component by component. Geometrically, adding two vectors places them tip-to-tail; the sum is the diagonal of the parallelogram they span.
Add / subtract:$\vec{a}\pm\vec{b}$ adds or subtracts matching components
Scalar multiple:$k\vec{a}$ multiplies every component by $k$; its length becomes $|k|\,|\vec{a}|$, and $k<0$ reverses the direction
Resultant:$\vec{a}+\vec{b}$ is the diagonal of the parallelogram built on $\vec{a}$ and $\vec{b}$
Two vectors from a common point O: complete the parallelogram and the resultant $\vec{a}+\vec{b}$ is its diagonal. The angle θ between $\vec{a}$ and $\vec{b}$ is the one the dot product returns (§5).
TrickLine the two columns up and work one row at a time, top to bottom, so no component is dropped or mismatched.
NoteTwo vectors are parallel exactly when one is a scalar multiple of the other: $\vec{a}\parallel\vec{b}\iff\vec{a}=k\vec{b}$ for some scalar $k$. Same $k$ sign means same direction; opposite sign means anti-parallel.
§4 — The scalar (dot) product AHL 3.10
The scalar product multiplies matching components and adds them. Crucially, the result is a single number (a scalar) — not a vector — and it is the bridge from components to the angle between two vectors.
Definition:$\vec{a}\cdot\vec{b}=a_1b_1+a_2b_2+a_3b_3$
Geometric form:$\vec{a}\cdot\vec{b}=|\vec{a}|\,|\vec{b}|\cos\theta$, where $\theta$ is the angle between them
Self product:$\vec{a}\cdot\vec{a}=|\vec{a}|^{2}$
TrickLine the vectors up component by component — $a_1$ with $b_1$, $a_2$ with $b_2$ — so you never drop or misalign a term.
TrapThe dot product is a scalar. If your answer still carries components or an $\vec{i},\vec{j}$, you have made a slip — collapse it to one number.
§5 — The angle between two vectors AHL 3.10
Rearranging the geometric form gives the angle directly. This is the single most-tested vector skill in AI HL.
Angle:$\cos\theta=\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}$, then $\theta=\cos^{-1}\!\left(\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}\right)$
For $\vec{a}=\begin{pmatrix}3\\4\end{pmatrix}$ and $\vec{b}=\begin{pmatrix}4\\1\end{pmatrix}$: $\vec{a}\cdot\vec{b}=12+4=16$, $|\vec{a}|=5$, $|\vec{b}|=\sqrt{17}$, so $\cos\theta=\dfrac{16}{5\sqrt{17}}=0.776$ and $\theta=39.1^\circ$ (3 s.f.).
TrickCompute the dot product and both magnitudes separately, then combine — and keep your GDC in degree mode unless radians are asked for.
TrapA negative dot product means an obtuse angle ($\theta>90^\circ$). Don't discard the sign — $\cos^{-1}$ returns the correct obtuse angle automatically only if you feed it the signed value.
NoteThe dot product gives $\cos\theta$. HL also has the vector (cross) product, whose magnitude $|\vec{a}\times\vec{b}|=|\vec{a}|\,|\vec{b}|\sin\theta$ gives areas and a perpendicular direction in 3-D — reach for it only when a question asks for an area or a normal vector, not for a plain angle.
§6 — Perpendicular vectors: dot product = 0 AHL 3.10
Since $\cos 90^\circ=0$, two non-zero vectors are perpendicular exactly when their scalar product is zero. Set the dot product to $0$ and solve for any unknown.
Perpendicular test:$\vec{a}\perp\vec{b}\iff\vec{a}\cdot\vec{b}=0$ (for non-zero $\vec{a},\vec{b}$)
Find $k$ so that $\begin{pmatrix}2\\k\end{pmatrix}$ and $\begin{pmatrix}6\\-3\end{pmatrix}$ are perpendicular: $\ (2)(6)+(k)(-3)=12-3k=0\Rightarrow k=4$.
TrickWrite the dot product out in full, set it equal to $0$, collect the $k$-terms and solve — it is always a single linear equation.
TrapMind the signs when a component is negative — a dropped minus flips the answer. And $\vec{a}\cdot\vec{b}=0$ only signals perpendicularity when both vectors are non-zero.
§7 — The vector equation of a line AHL 3.11
A line is a base point plus every scalar multiple of a fixed direction. Each value of the parameter $t$ picks out one point on the line.
Vector line:$\vec{r}=\vec{a}+t\vec{d}$ — $\vec{a}$ is a point on the line, $\vec{d}$ its direction, $t\in\mathbb{R}$
Point on line:$R$ lies on the line only if one value of $t$ satisfies every coordinate
Is $\begin{pmatrix}7\\10\end{pmatrix}$ on $\vec{r}=\begin{pmatrix}1\\2\end{pmatrix}+t\begin{pmatrix}3\\4\end{pmatrix}$? Row 1: $1+3t=7\Rightarrow t=2$; check row 2: $2+4(2)=10$ ✓ — yes, at $t=2$.
TrickSolve for $t$ from one row, then verify the remaining row(s) give the identical $t$.
TrapA point is on the line only if a single $t$ fits all coordinates — matching just one coordinate is not enough. Two lines are parallel when their direction vectors are scalar multiples of each other.
§8 — Velocity vectors & motion AHL 3.12
Constant-velocity motion is a vector line where the parameter is time: start point plus velocity times $t$. This is where AI puts vectors to work.
Position at time $t$:$\vec{r}(t)=\vec{r}_0+t\vec{v}$ — initial position plus velocity $\times$ time
Speed:$\text{speed}=|\vec{v}|$ — the magnitude of the velocity vector
Distance from origin:$|\vec{r}(t)|$ — find the position vector first, then its magnitude
With $\vec{r}(t)=\begin{pmatrix}1\\2\end{pmatrix}+t\begin{pmatrix}3\\4\end{pmatrix}$: speed $=\left|\begin{pmatrix}3\\4\end{pmatrix}\right|=5$; at $t=2$, $\vec{r}=\begin{pmatrix}7\\10\end{pmatrix}$, so the distance from the origin is $\sqrt{7^{2}+10^{2}}=\sqrt{149}=12.2$ (3 s.f.).
TrickThe coefficient of $t$ is the velocity vector; its magnitude is the (constant) speed. You never differentiate for constant-velocity motion.
TrapDistance from the origin is $|\vec{r}(t)|$ — evaluate the position at that time first, then take the magnitude. Taking the magnitude of the velocity by mistake gives the speed, not the distance.
§9 — Application: when two particles meet AHL 3.12
Two particles meet only when their position vectors are equal at the same time. Equate the position vectors, solve one coordinate for $t$, and the other coordinate must confirm that same $t$.
Meeting condition:$\vec{r}_A(t)=\vec{r}_B(t)$ for a single value of $t$
Method:equate $x$-components and solve for $t$; the $y$-components must give the same $t$
TrapCrossing paths is not the same as meeting. If the two coordinate equations give different values of $t$, the paths cross in space but the particles pass that point at different moments — so they never actually meet.
NoteIf instead a common time $t$ is given and you want how far apart the particles are, form the connecting vector $\vec{r}_B(t)-\vec{r}_A(t)$ and take its magnitude — that is the distance between them.
§10 — Exam attack plan All sections
| Question cue | What to do | Watch for |
|---|
| "Find $|\vec{a}|$ / the speed" | $\sqrt{a_1^{2}+a_2^{2}(+a_3^{2})}$ | Square-add-root; signs don't matter |
| "Find the unit vector" | $\dfrac{1}{|\vec{a}|}\vec{a}$ — divide each component by $|\vec{a}|$ | Find $|\vec{a}|$ first; the result has length 1 |
| "Angle between two vectors" | $\cos\theta=\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}$, then $\cos^{-1}$ | Keep the sign; negative $\Rightarrow$ obtuse |
| "Show perpendicular / find $k$" | Set $\vec{a}\cdot\vec{b}=0$ and solve | One linear equation; mind negative signs |
| "Is the point on the line?" | Solve $t$ from one row, check the rest | A single $t$ must fit all rows |
| "Speed / position of a particle" | Speed $=|\vec{v}|$; position $=\vec{r}_0+t\vec{v}$ | Coefficient of $t$ is velocity, not position |
| "Do the two particles meet?" | Equate position vectors; solve; check $t$ agrees | Different $t$ $\Rightarrow$ paths cross, no meeting |
| "Distance between them at time $t$" | $|\vec{r}_B(t)-\vec{r}_A(t)|$ | Subtract first, then take the magnitude |
Worked Example — Two Drones on Constant-Velocity Paths
Question (AI HL Paper 2 style — 7 marks)
Two delivery drones fly over a zone. Their positions (in km) at time $t$ hours after 09:00 are
$\vec{r}_A=\begin{pmatrix}1\\2\end{pmatrix}+t\begin{pmatrix}3\\4\end{pmatrix}$ and $\vec{r}_B=\begin{pmatrix}-1\\8\end{pmatrix}+t\begin{pmatrix}4\\1\end{pmatrix}$.
(a) Find the speed of drone A.
(b) Find the distance between the two drones at $t=0$.
(c) Show that the drones meet, and state the time and the point of meeting.
Solution
- (a) The velocity of A is $\begin{pmatrix}3\\4\end{pmatrix}$, so speed $=\sqrt{3^{2}+4^{2}}=\sqrt{25}=5$ km h$^{-1}$. (M1)(A1)
- (b) At $t=0$, A is at $\begin{pmatrix}1\\2\end{pmatrix}$ and B is at $\begin{pmatrix}-1\\8\end{pmatrix}$, so $\overrightarrow{AB}=\begin{pmatrix}-1-1\\8-2\end{pmatrix}=\begin{pmatrix}-2\\6\end{pmatrix}$. Distance $=\sqrt{(-2)^{2}+6^{2}}=\sqrt{40}=6.32$ km (3 s.f.). (M1)(A1)
- (c) Set $\vec{r}_A=\vec{r}_B$. Row 1: $1+3t=-1+4t\Rightarrow t=2$. Row 2: $2+4t=8+t\Rightarrow 3t=6\Rightarrow t=2$. (M1) Both rows give the same $t=2$, so the drones do meet. (A1) At $t=2$ the position is $\begin{pmatrix}1+3(2)\\2+4(2)\end{pmatrix}=\begin{pmatrix}7\\10\end{pmatrix}$, so they meet at 11:00 at the point $(7,10)$. (A1)
Examiner's note: the whole of part (c) turns on one idea — a single value of $t$ must satisfy both coordinate equations. Had the two rows given different times, the flight paths would still cross, but the drones would pass that crossing point at different moments and never meet. Keep the velocity (the coefficient of $t$) distinct from the position vector, and give units and a 3 s.f. answer where the numbers aren't exact.
Common Student Questions
What is the difference between a position vector and a velocity vector?
The position vector $\vec{r}(t)$ tells you where a particle is at time $t$; the velocity vector $\vec{v}$ — the coefficient of $t$ in $\vec{r}(t)=\vec{r}_0+t\vec{v}$ — tells you how fast and in which direction it moves. The particle's speed is the magnitude $|\vec{v}|$. Confusing the two is the most common vectors slip: a "distance" question wants $|\vec{r}(t)|$, a "speed" question wants $|\vec{v}|$.
How do I find the angle between two vectors in the exam?
Use $\cos\theta=\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}$, then take $\cos^{-1}$. Compute the dot product and the two magnitudes separately, divide, and apply $\cos^{-1}$ with your GDC in degree mode. Keep the sign of the dot product: a negative value gives an obtuse angle, which $\cos^{-1}$ returns correctly on its own.
Two vectors are perpendicular — what equation do I write?
Set their scalar product equal to zero: $\vec{a}\cdot\vec{b}=0$. Because $\cos 90^\circ=0$, that is precisely the perpendicularity condition for two non-zero vectors. If one component is an unknown $k$, this hands you a single linear equation — for example $\begin{pmatrix}2\\k\end{pmatrix}\cdot\begin{pmatrix}6\\-3\end{pmatrix}=12-3k=0$ gives $k=4$. Just watch the signs on negative components.
My two coordinate equations give different values of t — is that a mistake?
Not necessarily. When you test whether two particles meet, each coordinate gives its own equation for $t$. If they give the same $t$, the particles are at the same place at the same time — they meet. If they give different values, the paths cross in space but the particles reach that crossing point at different moments, so they never meet. Different $t$ values are a valid (and common) "they do not meet" answer, not an arithmetic error.
Do I ever need the cross product in AI HL Vectors?
Occasionally — for areas and for a direction perpendicular to two vectors in 3-D, using $|\vec{a}\times\vec{b}|=|\vec{a}|\,|\vec{b}|\sin\theta$ (so the area of a triangle is $\tfrac12|\overrightarrow{AB}\times\overrightarrow{AC}|$). But for the bread-and-butter AI HL tasks on this sheet — magnitudes, angles, perpendicularity, points on a line and particles that meet — the dot product and the magnitude are all you need. Reach for the cross product only when the word "area" or "perpendicular vector" appears.
What's NOT in this cheatsheet
This page gives you the formulas and traps for free. The full Photon Academy IB Math AI HL library (enrolled students, or the lifetime resource library) adds:
- Notes PDF — every vector rule, method and worked edge case in full, from magnitude to meeting particles.
- Tutorial booklet — 25+ AI HL-style questions from fluency to full exam applications (velocity vectors, particles that meet, perpendicularity).
- Tutorial Solutions — mark-scheme-style solutions with M1/A1 annotations.
- Predicted-paper questions — the exact vector question types most likely in the next AI HL session.
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