HookThe night a 747 crossed the Atlantic at 825 mph
Just before midnight on 8 February 2020, a British Airways Boeing 747 — flight BA112 — lifted off from New York's JFK bound for London. Storm Ciara was driving a jet stream across the North Atlantic at more than 250 mph, and the aircraft climbed straight into the back of it. Its ground speed, the rate at which its shadow crossed the ocean, briefly touched 825 mph, and it landed at Heathrow 4 hours and 56 minutes later — the fastest subsonic New York to London crossing on record, arriving around eighty minutes early.
The aeroplane never flew at 825 mph through the air around it; its airspeed was the usual 570 mph or so. The extra speed came from the wind. Ground velocity is the sum of two arrows: where the aircraft points and how fast it moves through the air, plus how fast and in what direction the air itself is moving. Add the arrows tip to tail and the single arrow from start to finish is the real journey over the ground. That addition of arrows is a vector, and this section teaches you to write vectors as columns, to add, subtract and scale them, to find the length of a resultant, and — the part that carries the most marks at Higher — to use vectors to prove geometric facts about parallel lines, midpoints and points that lie in a straight line.
ModelThe vector as a journey — translations and column vectors
A vector is a quantity with both size and direction — a journey, not just a distance. On a grid the neatest way to write one is a column vector \(\begin{pmatrix} x \\ y \end{pmatrix}\), where the top number is the movement in the \(x\)-direction (right if positive, left if negative) and the bottom number is the movement in the \(y\)-direction (up if positive, down if negative). It describes a translation exactly: to translate a shape by \(\begin{pmatrix} 4 \\ -3 \end{pmatrix}\) you slide every point \(4\) right and \(3\) down, and the shape keeps its size, its orientation and the way it faces. Translation is the one transformation that changes nothing but position.
Notation matters in the exam. The vector from \(A\) to \(B\) is written \(\overrightarrow{AB}\), or as a single bold or underlined letter such as \(\mathbf{a}\). Reversing the journey reverses the sign: \(\overrightarrow{BA}=-\overrightarrow{AB}\), because walking back covers the same distance the opposite way. Getting that minus sign right is the foundation of everything that follows, from resultants to full geometric proofs.
A triangle is translated by \(\begin{pmatrix} 5 \\ -2 \end{pmatrix}\), moving one vertex from \((1,\ 4)\). Where does that vertex land, and what single vector translates it back? Add the column vector to the point component by component: \((1+5,\ 4-2)=(6,\ 2)\). The return translation is the negative of the original, \(\begin{pmatrix} -5 \\ 2 \end{pmatrix}\), which undoes the move exactly. A translation and its reverse are always negatives of each other — the same fact as \(\overrightarrow{BA}=-\overrightarrow{AB}\).
MechanismAdding, subtracting and scaling — and the resultant
Vectors are added and subtracted component by component: \(\begin{pmatrix} 3 \\ 1 \end{pmatrix}+\begin{pmatrix} 2 \\ 4 \end{pmatrix}=\begin{pmatrix} 5 \\ 5 \end{pmatrix}\), and subtraction works the same way with minus signs. Multiplying by a scalar — an ordinary number — stretches a vector without turning it: \(3\begin{pmatrix} 2 \\ -1 \end{pmatrix}=\begin{pmatrix} 6 \\ -3 \end{pmatrix}\) points the same way but is three times as long. On a diagram, adding vectors means drawing them nose-to-tail and then the single arrow from the very start to the very end — the resultant.
That resultant is exactly the jet-stream story: the aircraft's velocity through the air plus the wind's velocity gives its velocity over the ground. When the two combine at right angles, the length of the resultant comes straight from Pythagoras, linking this section back to the mensuration work. The magnitude (length) of \(\begin{pmatrix} x \\ y \end{pmatrix}\) is \(\sqrt{x^2+y^2}\), the straight-line size of the whole journey.
A light aircraft flies due east at \(\begin{pmatrix} 500 \\ 0 \end{pmatrix}\) km/h through the air, while a wind from the south adds \(\begin{pmatrix} 0 \\ 120 \end{pmatrix}\) km/h. Find its velocity and its speed over the ground. Add the vectors component by component: \(\begin{pmatrix} 500 \\ 0 \end{pmatrix}+\begin{pmatrix} 0 \\ 120 \end{pmatrix}=\begin{pmatrix} 500 \\ 120 \end{pmatrix}\). The ground speed is the magnitude of this resultant: \(\sqrt{500^2+120^2}=\sqrt{250{,}000+14{,}400}=\sqrt{264{,}400}\approx 514\) km/h. The plane is pushed sideways as well as forwards, and its speed over the ground exceeds its airspeed — the everyday reason a tailwind shortens a flight.
MechanismPosition vectors and reading a route off a diagram
In geometry problems the vectors are usually given as letters — \(\overrightarrow{OA}=\mathbf{a}\), \(\overrightarrow{OB}=\mathbf{b}\) — and the skill is to express any other vector in the picture as a route built from them. To travel from \(A\) to \(B\) you go back down \(\mathbf{a}\) to the origin and out along \(\mathbf{b}\): \(\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=-\mathbf{a}+\mathbf{b}=\mathbf{b}-\mathbf{a}\). A position vector is just the route from the origin \(O\) to a point, so \(\mathbf{a}\) is the position vector of \(A\).
The move that unlocks these problems is finding a fraction of a line. If \(M\) is the midpoint of \(AB\) then \(\overrightarrow{AM}=\tfrac{1}{2}\overrightarrow{AB}\); if a point divides \(AB\) in the ratio \(1:2\) then it lies \(\tfrac{1}{3}\) of the way along, so \(\overrightarrow{AP}=\tfrac{1}{3}\overrightarrow{AB}\). Build the route to the point you want out of these fractions of known vectors, then simplify by collecting the \(\mathbf{a}\) terms and the \(\mathbf{b}\) terms separately — exactly as you collect like terms in algebra.
CaseVector proofs — parallel, collinear and midpoints
The hardest and best-rewarded vector questions ask you to prove something geometric, and two facts do almost all the work. First, two vectors are parallel exactly when one is a scalar multiple of the other: \(\overrightarrow{QR}=k\,\overrightarrow{PQ}\) means \(QR\) is parallel to \(PQ\), and \(k\) tells you how many times longer it is. Second, three points are collinear — in one straight line — when two vectors joining them are parallel and share a common point; parallel alone is not enough, because parallel lines can sit apart.
So a proof runs to a fixed rhythm: express the two vectors you care about as routes in \(\mathbf{a}\) and \(\mathbf{b}\), simplify each, then spot that one is a multiple of the other. If \(\overrightarrow{XY}=2(\mathbf{a}+\mathbf{b})\) and \(\overrightarrow{XZ}=3(\mathbf{a}+\mathbf{b})\), then \(\overrightarrow{XZ}=\tfrac{3}{2}\overrightarrow{XY}\); the two are parallel and both start at \(X\), so \(X\), \(Y\) and \(Z\) are collinear. Finish with a sentence stating the conclusion in words — that final claim is where the reasoning mark is awarded.
In triangle \(OAB\), \(\overrightarrow{OA}=\mathbf{a}\) and \(\overrightarrow{OB}=\mathbf{b}\), and \(M\) is the midpoint of \(AB\). Prove that \(\overrightarrow{OM}=\tfrac{1}{2}(\mathbf{a}+\mathbf{b})\). Start with the side: \(\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=-\mathbf{a}+\mathbf{b}=\mathbf{b}-\mathbf{a}\). Because \(M\) is the midpoint, \(\overrightarrow{AM}=\tfrac{1}{2}\overrightarrow{AB}=\tfrac{1}{2}(\mathbf{b}-\mathbf{a})\). Now build the route from \(O\) to \(M\): \(\overrightarrow{OM}=\overrightarrow{OA}+\overrightarrow{AM}=\mathbf{a}+\tfrac{1}{2}(\mathbf{b}-\mathbf{a})=\mathbf{a}-\tfrac{1}{2}\mathbf{a}+\tfrac{1}{2}\mathbf{b}=\tfrac{1}{2}\mathbf{a}+\tfrac{1}{2}\mathbf{b}=\tfrac{1}{2}(\mathbf{a}+\mathbf{b})\), as required. Every step is a route in \(\mathbf{a}\) and \(\mathbf{b}\) followed by collecting like terms.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Vector questions split into two kinds, and they reward different habits. The short ones are pure arithmetic — add, subtract or scale column vectors component by component, and find a magnitude with \(\sqrt{x^2+y^2}\); here the marks are simply for accurate signs, so line the components up vertically and watch the negatives.
The longer 3-to-5 mark questions are the geometric proofs, and they are AO2/AO3 reasoning marks that many candidates leave blank. Do not: they follow a fixed method. Draw and label the diagram, express each vector you need as a route through the given letters (\(\mathbf{a}\), \(\mathbf{b}\)), and simplify by collecting like terms. To prove two lines parallel, show one vector is a scalar multiple of the other and say so. To prove three points collinear, show the vectors are parallel and name the point they share. Convert a ratio to a fraction carefully — a ratio of \(1:2\) means \(\tfrac{1}{3}\) of the way along, not \(\tfrac{1}{2}\). Above all, finish with a sentence that states the conclusion in words: 'therefore \(\overrightarrow{XZ}\) is a multiple of \(\overrightarrow{XY}\) and they share \(X\), so \(X,Y,Z\) are collinear.' That closing statement is the reasoning mark, and it is the one most often dropped.