HookWhy the plane pointed sideways at the runway
On 18 February 2022, Storm Eunice drove 120 mph gusts across southern England, and a live YouTube channel called Big Jet TV pointed a camera down the runway at Heathrow. Hundreds of thousands of people watched something that looks impossible until you understand it: airliners approaching the tarmac not head-on but visibly angled, nose pointed off to one side, drifting in almost sideways before straightening at the last second. Pilots call it 'crabbing', and it is a vector diagram made flesh. The aeroplane's engines push it through the air in the direction the nose points; the wind pushes the whole body of air sideways across the ground; and the path the plane actually traces over the runway — the thing that must stay straight — is the sum of those two velocities. The crab angle you can see with your own eyes is the corner of a triangle whose three sides are vectors.
That is the entire idea of this section: a quantity with both a size and a direction, drawn as an arrow, computed with numbers. You will write vectors in component form in two and three dimensions; find a vector's magnitude and direction and convert between the two descriptions; add vectors and multiply them by scalars, reading the geometry off a diagram; use position vectors to locate points and measure the distance between them; and finally deploy all of it on problems in context, including forces — where the crabbing aeroplane, a boat crossing a tide, and three ropes holding a sign steady are all the same piece of mathematics.
ModelWhat a vector is — components in two and three dimensions
A scalar is a plain number with a unit — a mass of 5 kg, a temperature of 20°C. A vector carries a direction as well as a size: a velocity of 250 km/h due north, a force of 30 N pulling up the slope. We draw it as an arrow, and we write it in component form: in two dimensions as a column \(\begin{pmatrix} x \\ y \end{pmatrix}\), or equivalently as \(x\mathbf{i}+y\mathbf{j}\), where \(\mathbf{i}\) and \(\mathbf{j}\) are the unit vectors of length one pointing along the positive \(x\)- and \(y\)-axes. The components are just instructions: 'go \(x\) across, then \(y\) up'.
Three dimensions add one more axis and one more unit vector \(\mathbf{k}\), so a 3-D vector is \(\begin{pmatrix} x \\ y \\ z \end{pmatrix}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\) — across, up, and out of the page. Almost everything you learn in 2-D carries over unchanged to 3-D; the third component just comes along for the ride. Two vectors are equal when all their components match, which means a vector is free to slide anywhere in the plane or in space — only its size and direction matter, not where the arrow is drawn. The zero vector \(\mathbf{0}\) has all components zero and no direction, and \(-\mathbf{a}\) is \(\mathbf{a}\) reversed, the same length pointing the opposite way.
ModelMagnitude and direction — Pythagoras meets trigonometry
A vector's magnitude is its length, written \(|\mathbf{a}|\), and it comes straight from Pythagoras on the components: \(|\mathbf{a}|=\sqrt{x^2+y^2}\) in two dimensions and \(|\mathbf{a}|=\sqrt{x^2+y^2+z^2}\) in three. The direction in 2-D is usually given as an angle \(\theta\) measured anticlockwise from the positive \(x\)-axis, found from \(\tan\theta=\dfrac{y}{x}\) — but you must look at the signs of \(x\) and \(y\) to place the arrow in the correct quadrant, because a calculator's \(\arctan\) only ever returns an angle between \(-90^\circ\) and \(90^\circ\).
Going the other way, from a magnitude \(r\) and a direction \(\theta\) back to components, you resolve: \(x=r\cos\theta\) and \(y=r\sin\theta\). This 'resolving into components' is the single most useful trick in the whole of applied mathematics, because it turns an awkward diagonal quantity into two clean perpendicular ones you can add separately. A unit vector in the direction of \(\mathbf{a}\) is \(\hat{\mathbf{a}}=\dfrac{1}{|\mathbf{a}|}\mathbf{a}\) — divide a vector by its own length and you keep the direction but scale it to length one.
Find the magnitude and direction of \(\mathbf{a}=3\mathbf{i}+4\mathbf{j}\), the magnitude of \(\mathbf{b}=\begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}\), and the components of a velocity of \(20\) m/s on a bearing that makes \(30^\circ\) above the horizontal.
For \(\mathbf{a}\): \(|\mathbf{a}|=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5\). The direction is \(\theta=\arctan\dfrac{4}{3}=53.1^\circ\) above the positive \(x\)-axis, and since both components are positive that first-quadrant answer is correct as it stands.
For \(\mathbf{b}\), the third dimension changes nothing about the method: \(|\mathbf{b}|=\sqrt{2^2+(-3)^2+6^2}=\sqrt{4+9+36}=\sqrt{49}=7\).
To resolve the \(20\) m/s velocity, \(x=20\cos 30^\circ=20\times 0.8660=17.3\) and \(y=20\sin 30^\circ=20\times 0.5=10\), so the velocity is \(\begin{pmatrix} 17.3 \\ 10 \end{pmatrix}\) m/s. A quick check confirms the arithmetic: \(\sqrt{17.3^2+10^2}=\sqrt{299.3+100}\approx 20\), the speed we started with.
MechanismAdding vectors and scaling them — the triangle law
To add two vectors you add their components: \(\begin{pmatrix} a \\ b \end{pmatrix}+\begin{pmatrix} c \\ d \end{pmatrix}=\begin{pmatrix} a+c \\ b+d \end{pmatrix}\). Geometrically this is the triangle law: draw the first arrow, start the second from where the first ends (tip to tail), and the single arrow from the very start to the very end is the resultant \(\mathbf{a}+\mathbf{b}\). Drawn from a common start instead, the two vectors and their sum make a parallelogram — same result, different picture. Subtraction \(\mathbf{a}-\mathbf{b}\) is just \(\mathbf{a}+(-\mathbf{b})\): reverse the second arrow, then add.
Multiplying a vector by a scalar \(k\) multiplies every component by \(k\): \(k\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} kx \\ ky \end{pmatrix}\). This stretches the arrow (if \(|k|>1\)) or shrinks it (if \(|k|<1\)) and reverses it if \(k\) is negative, but it never changes the line the arrow lies along. That gives the key geometric test: two non-zero vectors are parallel exactly when one is a scalar multiple of the other, \(\mathbf{b}=k\mathbf{a}\). Combine that with a shared point and you can prove three points are collinear — that they lie on one straight line.
Points \(A\), \(B\) and \(C\) have \(\overrightarrow{AB}=\begin{pmatrix} 2 \\ 3 \end{pmatrix}\) and \(\overrightarrow{BC}=\begin{pmatrix} 6 \\ 9 \end{pmatrix}\). Show that \(A\), \(B\) and \(C\) are collinear.
Test whether \(\overrightarrow{BC}\) is a scalar multiple of \(\overrightarrow{AB}\): \(\begin{pmatrix} 6 \\ 9 \end{pmatrix}=3\begin{pmatrix} 2 \\ 3 \end{pmatrix}\), so \(\overrightarrow{BC}=3\,\overrightarrow{AB}\). The two vectors are therefore parallel, and because they share the point \(B\) they lie on the same straight line — not merely parallel lines. Hence \(A\), \(B\) and \(C\) are collinear, and \(C\) is three times as far along as \(B\), so \(BC\) is twice the length of \(AB\). Naming the shared point is the step candidates forget: parallel plus a common point is what upgrades 'parallel' to 'collinear'.
ModelPosition vectors and the distance between two points
A position vector pins a point to a fixed origin \(O\): the position vector of \(A\) is \(\overrightarrow{OA}=\mathbf{a}\), the arrow from the origin to \(A\), whose components are simply the coordinates of \(A\). Position vectors are the bridge between coordinate geometry and vectors — a point and its position vector are two names for the same thing.
The move that unlocks nearly every vectors problem is the rule for the vector between two points: \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\), read as 'destination minus start'. It says: to get from \(A\) to \(B\), undo the trip out to \(A\) and take the trip out to \(B\). Once you have \(\overrightarrow{AB}\) in components, the distance between the points is just its magnitude, \(|\overrightarrow{AB}|=|\mathbf{b}-\mathbf{a}|\) — which is exactly the coordinate distance formula in disguise, and works identically in two or three dimensions. The midpoint of \(AB\) has position vector \(\tfrac{1}{2}(\mathbf{a}+\mathbf{b})\), the average of the two ends.
Points \(A\) and \(B\) have position vectors \(\mathbf{a}=\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix} 5 \\ 5 \\ 14 \end{pmatrix}\). Find \(\overrightarrow{AB}\) and the distance \(AB\).
Apply 'destination minus start': \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix} 5-1 \\ 5-2 \\ 14-2 \end{pmatrix}=\begin{pmatrix} 4 \\ 3 \\ 12 \end{pmatrix}\). The distance is the magnitude of that vector:
\[AB=\sqrt{4^2+3^2+12^2}=\sqrt{16+9+144}=\sqrt{169}=13.\]
So \(A\) and \(B\) are exactly \(13\) units apart. Notice the order does not matter for distance — \(\overrightarrow{BA}=\mathbf{a}-\mathbf{b}\) has components \((-4,-3,-12)\), the same length \(13\) pointing the other way — but it matters enormously for direction, so always subtract in the order the question's arrow demands.
CaseVectors in the real world — crosswinds and forces
The final spec point turns vectors loose on physical problems, and two contexts dominate: velocities and forces. For velocities, the resultant of an object's own motion and the motion of the medium it travels through is a vector sum — which is precisely the crabbing aeroplane from the top of this page. The plane's velocity through the air, plus the wind's velocity, equals its velocity over the ground; the pilot angles the nose so that this ground velocity lines up with the runway. A boat crossing a tidal river, a swimmer in a current and a drone in a breeze are all the identical triangle.
For forces, vectors describe pushes and pulls, and the governing idea is resultant force: add all the forces on an object as vectors, and the single vector you get determines what happens next through Newton's second law, \(\mathbf{F}=m\mathbf{a}\). A body is in equilibrium — at rest or moving at steady velocity — exactly when the resultant force is the zero vector, \(\sum \mathbf{F}=\mathbf{0}\), so the forces cancel in both the \(\mathbf{i}\) and the \(\mathbf{j}\) directions independently. That single condition solves the sign held by three cables, the lamp hung from two wires, and the box resting on a slope. The examiner's habit is to hand you a real context and expect you to translate it into arrows, resolve, and add.
An aircraft points due north and flies at \(250\) km/h relative to the air. A wind blows from the west at \(40\) km/h (i.e. towards the east). Taking \(\mathbf{i}\) east and \(\mathbf{j}\) north, find the aircraft's velocity over the ground, its ground speed, and its bearing.
Write each velocity in components. The aircraft's air velocity is \(\begin{pmatrix} 0 \\ 250 \end{pmatrix}\) and the wind is \(\begin{pmatrix} 40 \\ 0 \end{pmatrix}\). Add them for the ground velocity:
\[\mathbf{v}=\begin{pmatrix} 0 \\ 250 \end{pmatrix}+\begin{pmatrix} 40 \\ 0 \end{pmatrix}=\begin{pmatrix} 40 \\ 250 \end{pmatrix}.\]
The ground speed is the magnitude, \(|\mathbf{v}|=\sqrt{40^2+250^2}=\sqrt{1600+62500}=\sqrt{64100}\approx 253\) km/h — slightly faster than the airspeed, because the wind has a component along the flight. The direction east of north is \(\arctan\dfrac{40}{250}=9.1^\circ\), so the true track is a bearing of \(009^\circ\): the plane is swept nine degrees off the way its nose points, which is exactly the crab you can see on the livestream.
The same arithmetic runs backwards for forces. If a sign is held by two known forces \(\mathbf{F}_1=\begin{pmatrix} 5 \\ 2 \end{pmatrix}\) N and \(\mathbf{F}_2=\begin{pmatrix} -3 \\ 4 \end{pmatrix}\) N, then for equilibrium the third force must make the total zero: \(\mathbf{F}_3=-(\mathbf{F}_1+\mathbf{F}_2)=-\begin{pmatrix} 2 \\ 6 \end{pmatrix}=\begin{pmatrix} -2 \\ -6 \end{pmatrix}\) N. Add all three and you get \(\mathbf{0}\), confirming the sign stays put.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Draw the diagram. Almost every vectors problem is transparent once the arrows are on paper and impossible to see without them — mark the resultant, label the origin, and put the wind or the second force tip-to-tail. When you find a direction, resist quoting the raw \(\arctan\): check the quadrant against the signs of the components and, in mechanics contexts, convert to a proper bearing measured clockwise from north.
For 'show that' collinearity or parallel questions, state the scalar multiple explicitly (\(\overrightarrow{BC}=3\,\overrightarrow{AB}\)) and, for collinearity, name the point the two vectors share — the mark scheme wants both halves. In distance work, write \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\) before taking the magnitude, and keep three-dimensional components lined up in a column so you do not lose the \(z\)-term. On a forces or velocity problem, resolve everything into \(\mathbf{i}\) and \(\mathbf{j}\) components first, add those separately, and only convert back to a magnitude and direction at the very end; for equilibrium, set each component of the resultant to zero and solve — never set the magnitudes to zero, set the components. Units belong on every physical answer.