Transformations and Vectors
Cambridge O Level Mathematics (Syllabus D) 4024 Topic 7 revision chapter covering the whole of Transformations and Vectors for the 2025-2027 syllabus, version 2. It teaches all four official subtopics in order and in full. Transformations covers reflection of a shape in a straight line, rotation of a shape about a centre through multiples of ninety degrees, enlargement of a shape from a centre by a scale factor, and translation of a shape by a column vector, together with the combinations of transformations the syllabus allows. It sets out what a complete description of each transformation must contain, since a description that names the transformation but omits the mirror line, the centre, the angle and direction, the scale factor or the translation vector has not answered the instruction to describe it fully. It shows that the mirror line is the perpendicular bisector of every point to image segment, how corresponding points locate an unknown centre of rotation on a grid, how the relation CP prime equals k times CP places an image on the same ray for a positive scale factor and on the opposite ray for a negative one, that lengths multiply by the modulus of k while angles and shape are unchanged, and that area is multiplied by k squared. Vectors in two dimensions covers column vector, directed segment and lower case bold notation, addition, subtraction and multiplication by a scalar, the reversal rule that BA is the negative of AB, and the route rule that AB plus BC equals AC, from which AB equals b minus a follows when a and b are the position vectors of A and B. Magnitude of a vector covers the calculation of the length of a column vector as the square root of x squared plus y squared, its origin in Pythagoras, modulus notation, exact surd and integer answers, and the firm distinction between a vector and the non negative scalar that measures it. Vector geometry covers directed line segments, position vectors, expressing an unknown vector in terms of two coplanar vectors by choosing a route, midpoint vectors, division of a line in a given ratio derived rather than quoted, proofs that vectors are parallel, proofs that three points are collinear, parallelogram reasoning and problems involving ratio and similarity. A visible M-O-V-E protocol, a transformation description checklist, eleven mathematically checked coordinate diagrams, fully worked examples inside every subtopic, a negative enlargement clinic, a vector proof clinic, a mistake clinic, retrieval practice with answers held back until the whole question set has been attempted, an exam style mixed challenge and a spaced review plan support both first pass learning and last week revision.Show moreShow less
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What is Transformations and Vectors about?
1. \((-3,2)\) is in the second quadrant (left and up); \((4,-1)\) is in the fourth (right and down). 2. The \(x\)-axis is \(y=0\); the \(y\)-axis is \(x=0\); the vertical line through \((2,0)\) is \(x=2\); the \(45^\circ\) line through the origin is \(y=x\). 3. \(\sqrt{5^{2}+12^{2}}=\sqrt{169}=13\). 4. Length scale factor \(\tfrac{10}{4}=2.5\); area scale factor \(2.5^{2}=6.25\). 5. \(3+2=5\) parts, so the first part is \(\tfrac35\) of the whole, which is 12 cm; the second is 8 cm.
“Rotation of \(90^\circ\) about the origin.” No direction. There are two different \(90^\circ\) rotations about the origin and this does not say which. “Reflection.” No mirror line, so the transformation is not determined at all. “Enlargement, scale factor 3.” No centre. The scale factor fixes the size of the image but not where it is. “Translation 4 right and 3 down.” Acceptable in meaning, but the syllabus asks for the vector; write \(\begin{pmatrix}4\\-3\end{pmatrix}\). “Reflection in the line \(y=x\), then translation by \(\begin{pmatrix}0\\2\end{pmatrix}\).” Two transformations offered where one was asked for. Find the single equivalent transformation instead.
Retrieval check 7.1. Triangle \(A\) has vertices \((1,1)\), \((4,1)\) and \((1,3)\). Write down the vertices of its image under each of the following. (a) Reflection in \(y=-x\). The rule is \((x,y)\mapsto(-y,-x)\), so the image is \((-1,-1)\), \((-1,-4)\), \((-3,-1)\). (b) Rotation through \(90^\circ\) clockwise about the origin. The rule is \((x,y)\mapsto(y,-x)\), so the image is \((1,-1)\), \((1,-4)\), \((3,-1)\). (c) Enlargement, centre \((1,1)\), scale factor \(-2\). Use \(P'=C+k(P-C)\). The vertex \((1,1)\) is the centre, so it does not move. \((4,1)\): \(C+(-2)\begin{pmatrix}3\\0\end{pmatrix} =(1,1)+(-6,0)=(-5,1)\). \((1,3)\): \(C+(-2)\begin{pmatrix}0\\2\end{pmatrix} =(1,1)+(0,-4)=(1,-3)\). The image is \((1,1)\), \((-5,1)\), \((1,-3)\) — twice the size and on the far side of the centre.
Clinic check. A shape of area \(6\,\mathrm{cm^2}\) is enlarged from a centre \(C\) by scale factor \(-\tfrac32\). (a) Is the image bigger or smaller than the object? (b) On which side of \(C\) does it lie? (c) What is its area? (a) Bigger. The size depends on \(\left|-\tfrac32\right|=\tfrac32\), so every length is one and a half times as long. (b) On the opposite side of \(C\) from the object, because \(k\) is negative. (c) Area scales by \(k^{2}=\left(-\tfrac32\right)^{2}=\tfrac94\), so the image has area \(6\times\tfrac94=13.5\,\mathrm{cm^2}\). Note that \(k^2\) is positive even though \(k\) is not — an area can never come out negative.
Retrieval check 7.2. \(\mathbf p=\begin{pmatrix}-2\\5\end{pmatrix}\) and \(\mathbf q=\begin{pmatrix}3\\-1\end{pmatrix}\), with \(\overrightarrow{OP}=\mathbf p\) and \(\overrightarrow{OQ}=\mathbf q\). Find (a) \(\mathbf p+\mathbf q\), (b) \(2\mathbf p-3\mathbf q\), (c) \(\overrightarrow{PQ}\), (d) \(\overrightarrow{QP}\). (a) \(\begin{pmatrix}-2+3\\5+(-1)\end{pmatrix}=\begin{pmatrix}1\\4\end{pmatrix}\). (b) \(2\mathbf p=\begin{pmatrix}-4\\10\end{pmatrix}\) and \(3\mathbf q=\begin{pmatrix}9\\-3\end{pmatrix}\), so \(2\mathbf p-3\mathbf q=\begin{pmatrix}-4-9\\10-(-3)\end{pmatrix} =\begin{pmatrix}-13\\13\end{pmatrix}\). (c) \(\overrightarrow{PQ}=\mathbf q-\mathbf p =\begin{pmatrix}3-(-2)\\-1-5\end{pmatrix}=\begin{pmatrix}5\\-6\end{pmatrix}\). Check against the points: \(P(-2,5)\) to \(Q(3,-1)\) is 5 right and 6 down. (d) \(\overrightarrow{QP}=-\overrightarrow{PQ} =\begin{pmatrix}-5\\6\end{pmatrix}\).
Route \(O\to P\to Q\). Since \(PQ\) is opposite and equal to \(OR\), \(\overrightarrow{PQ}=\mathbf r\), so \[\overrightarrow{OQ}=\overrightarrow{OP}+\overrightarrow{PQ}=\mathbf p+\mathbf r.\] The route \(O\to R\to Q\) gives \(\mathbf r+\mathbf p\), the same answer — a useful reassurance that the choice of route does not matter.
Key ideas to remember
- The one-line version. Name the rule, give every parameter it needs, and for vectors, pick a route and say what it proves.
- Derive it, do not memorise it. Two lines of route rule get you \(\overrightarrow{AB}=\mathbf b-\mathbf a\) in about four seconds, and they get it right every time. Memorising the result alone is exactly how it ends up reversed under pressure. End point minus start point is the same rule you already use for the gradient and the distance between two points.
- Interleave, do not block. Topic 7 shares its machinery with three other chapters: straight-line equations from Chapter 3 give you the mirror line and the perpendicular bisector, Pythagoras from Chapter 4 gives you the magnitude, and similarity from Chapter 4 is what an enlargement produces. A revision session that mixes Topic 7 with those is worth more than one that drills Topic 7 alone.
What you need to be able to do
- I can reflect a shape in a given straight line and state that the mirror line is the perpendicular bisector of every point-to-image segment.
- I can rotate a shape about a given centre through \(90^\circ\), \(180^\circ\) or \(270^\circ\) in a stated direction.
- I can use corresponding points to locate an unknown centre of rotation on a grid.
- I can enlarge a shape from a centre using a scale factor that is greater than 1, a positive fraction, or negative, and I know which side of the centre the image lands on.
- I can translate a shape by a column vector and read a translation off a diagram.
- I can apply two transformations in a stated order and record the intermediate image.
- I can write a complete description of any of the four transformations, with every parameter it requires.
- I can read and write a vector as a column, as \(\overrightarrow{AB}\), and as \(\mathbf a\).
- I can add and subtract column vectors and multiply one by a scalar.
- I can use \(\overrightarrow{BA}=-\overrightarrow{AB}\) to reverse a directed segment.
- I can build a route with \(\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\) and find an equivalent route between the same two points.
- I can derive \(\overrightarrow{AB}=\mathbf b-\mathbf a\) rather than memorising it, and I get the subtraction the right way round.
- I can calculate \(\left|\begin{pmatrix}x\\y\end{pmatrix}\right|=\sqrt{x^{2}+y^{2}}\) and explain why it is Pythagoras.
- I can leave a magnitude as an exact surd, or give it to 3 significant figures when asked for a decimal.
- I can use modulus notation correctly and I never write a magnitude as a column.
- I know a magnitude is a non-negative scalar, so a negative answer is a signal to check.
- I can represent a vector by a directed line segment and use position vectors from an origin \(O\).
- I can express an unknown vector in terms of two coplanar vectors by choosing a route.
- I can find the position vector of a midpoint, and of a point dividing a line in a given ratio, by deriving it rather than quoting a formula.
- I can prove two vectors are parallel by showing one is a scalar multiple of the other.
- I can prove three points are collinear, and I finish the proof with an explicit sentence.
- I can use vectors to settle a ratio or a similarity claim in a geometric figure.
Key terms in Transformations and Vectors
- Vectors in Two Dimensions
- A vector in two dimensions is a quantity with both magnitude and direction, recorded as a column vector whose top entry is the horizontal displacement and whose bottom entry is the vertical displacement. Cambridge prints vectors either as a directed line segment such as AB with an arrow above it, or as a lower case bold letter such as a. A vector fixes a movement but not a starting point, so two arrows of the same length and direction anywhere in the plane represent the same vector. Vectors are added and subtracted entry by entry and multiplied by a scalar entry by entry, so k times the column x, y is the column kx, ky. Reversing a directed segment reverses its vector, giving BA equal to minus AB, and following one directed segment by another gives the route rule AB plus BC equals AC. When the position vectors of A and B from an origin O are a and b, the route from A through O to B gives AB equal to b minus a, with the subtraction always in that order. A column vector is also exactly what describes a translation, which is why the same notation serves both halves of this topic.
- Transformations
- A transformation is a rule that assigns to every point of the plane a single image point, so that a whole shape is carried to an image shape. Cambridge O Level Mathematics D 4024 examines four: reflection of a shape in a straight line, rotation of a shape about a centre through a multiple of 90 degrees, enlargement of a shape from a centre by a scale factor that may be greater than one, a positive fraction or negative, and translation of a shape by a column vector. Reflection, rotation and translation preserve every length and angle, so the image is congruent to the object; reflection alone reverses orientation. Enlargement preserves angles and shape but multiplies every length by the modulus of the scale factor and every area by its square, so the image is similar to the object. Describing a transformation completely means stating its name together with every parameter it requires: the mirror line for a reflection, the angle, direction and centre for a rotation, the centre and scale factor for an enlargement, and the column vector for a translation. Questions may combine two transformations, and the order in which they are applied usually changes the result.
- Magnitude of a Vector
- The magnitude of a vector is its length, written between modulus signs, so the magnitude of a is written mod a and the magnitude of the directed segment AB is written mod AB. For a column vector with entries x and y the magnitude is the square root of x squared plus y squared, because the two components are the perpendicular legs of a right-angled triangle whose hypotenuse is the vector itself, so the result is Pythagoras theorem applied to a displacement. A magnitude is a scalar, not a vector: it is a single non-negative number that measures how long the vector is and says nothing about which way it points. Squaring removes any minus signs, so a vector and its reverse always have the same magnitude, and the answer can never come out negative. Magnitudes may be left exact as an integer or a surd, or given as a decimal to three significant figures when a decimal is asked for. Multiplying a vector by a scalar k multiplies its magnitude by the modulus of k. The magnitude formula is not printed in the list of formulas on the examination paper, so it has to be recalled.
- Vector Geometry
- Vector geometry is the use of directed line segments and position vectors to settle geometric questions by calculation instead of by measurement. Fixing an origin O gives every point A a position vector OA, usually written a, and any vector between two points is then the difference of their position vectors, so AB equals b minus a. An unknown vector is found by choosing a route from its start point to its end point built only from vectors that are already known, since every valid route gives the same answer. The midpoint M of AB has position vector one half of a plus b, and a point P dividing AB internally in the ratio m to n has position vector n a plus m b all over m plus n, a result derived by travelling to A and then m over m plus n of the way along AB rather than quoted. Two non-zero vectors are parallel exactly when one is a scalar multiple of the other, and three points A, B and C are collinear exactly when AB is a scalar multiple of AC, because the two vectors are then parallel and share the point A. A parallelogram is established by showing one pair of opposite sides is equal as vectors, which fixes both length and direction, since equal magnitudes alone would not. The size and sign of the scalar also give relative positions and length scale factors, which is how vectors settle questions about ratio and similarity.
Common mistakes to avoid
- A sixth, quieter one. Order matters when transformations are combined. Two translations are the safe exception: they always give the same result in either order, because adding column vectors is commutative. A handful of other pairs commute too — two rotations about the same centre, two enlargements from the same centre — but every one of those shares a fixed point. Where no fixed point is shared, assume reversing the order changes the image.
- The \(180^\circ\) exception, and the \(k=-1\) overlap. A half-turn needs no direction, because clockwise and anticlockwise land in the same place. And a rotation of \(180^\circ\) about \(C\) puts every point exactly where an enlargement from \(C\) with \(k=-1\) puts it, so when a question shows only that mapping, both descriptions are correct and either is accepted. That is the one genuine overlap between the four; everywhere else exactly one description fits.
- “The mirror line” is not an answer. Nor is drawing it on the diagram without naming it. The mark is for the equation. If the mirror is the vertical line through \((3,0)\), write \(x=3\) — not \(y=3\), which is a different line entirely.
- Size and sign are different questions. The sign of \(k\) decides which side of the centre the image is on; \(|k|\) decides how big it is. So \(k=-3\) makes the image three times as large and puts it on the opposite side, while \(k=\tfrac12\) makes it half as large on the same side. Reading “negative” as “smaller” conflates the two and answers the wrong question twice over. The negative enlargement clinic works through all four cases from one centre.
- Top row across, bottom row up. Writing \(\begin{pmatrix}3\\-2\end{pmatrix}\) when you meant 3 down and 2 left is not a small slip; it is a different transformation. Say the vector aloud as “three across, two down” before you write it, and check the sign of each component against the diagram.
- Why the lines through \(P,P'\) all meet at the centre. Because \(\overrightarrow{CP'}=k\overrightarrow{CP}\), the points \(C\), \(P\) and \(P'\) are always collinear — whatever the value of \(k\). So every line joining an object vertex to its image passes through the centre, and two of them are enough to find it. This is the same fact as the dashed rays in Figures 7.3 and 7.4, used in reverse.
- A point is not a vector. \(A\) is a place. \(\overrightarrow{OA}\) is the journey from the origin to that place, and \(\mathbf a\) is a name for that journey. They happen to carry the same pair of numbers when the journey starts at \(O\), which is why the two are easily confused — but \(\overrightarrow{AB}\) does not start at \(O\), and its components are differences, not coordinates. In handwriting, underline a vector letter (\(\underline{a}\)); print uses bold.
- \(\mathbf a-\mathbf b\) and \(\overrightarrow{AB}\) are not the same thing. In the example above \(\mathbf a-\mathbf b=\begin{pmatrix}6\\-2\end{pmatrix}\) but \(\overrightarrow{AB}=\begin{pmatrix}-6\\2\end{pmatrix}\). They differ by a sign, because \(\overrightarrow{AB}=\mathbf b-\mathbf a\) subtracts the start from the end. Whenever the letters in a directed segment run \(A\) to \(B\), the vector runs \(\mathbf b\) minus \(\mathbf a\) — the reverse of the reading order, which is exactly why it catches people out.
- Three things a magnitude can never be. It can never be a column — \(\left|\begin{pmatrix}-5\\12\end{pmatrix}\right|\) is \(13\), not \(\begin{pmatrix}5\\12\end{pmatrix}\). It can never be negative, so a minus sign in an answer means a slip in the squaring. And it can never be the sum of the components: \(-5+12=7\) is not the length of anything here. If your answer to a magnitude question is not a single non-negative number, stop and re-read the working.
- Both vectors must start at the same point. Showing that \(\overrightarrow{MN}\) is a multiple of \(\overrightarrow{NP}\) also works, because those two share \(N\). But showing \(\overrightarrow{MN}\) is a multiple of some unrelated vector proves only that two lines are parallel — parallel lines need never meet, so no collinearity follows. The shared point is doing half the work in every one of these proofs, and it must be named in the conclusion.
- Equal magnitudes are not enough. \(\left|\begin{pmatrix}4\\1\end{pmatrix}\right| =\left|\begin{pmatrix}1\\4\end{pmatrix}\right|=\sqrt{17}\), but those two vectors point in completely different directions. A quadrilateral with one pair of opposite sides merely equal in length could be an isosceles trapezium, not a parallelogram. The vector statement \(\overrightarrow{AB}=\overrightarrow{DC}\) is stronger than the length statement \(|AB|=|DC|\), and it is the stronger one you need.
- Mind the letter order. For the quadrilateral \(ABCD\), the side opposite \(AB\) is \(DC\), not \(CD\) — because going round the quadrilateral is \(A\to B\to C\to D\), so \(AB\) and \(CD\) run in opposite senses. Writing \(\overrightarrow{AB}=\overrightarrow{CD}\) claims something false; the correct statement is \(\overrightarrow{AB}=\overrightarrow{DC}\), or equivalently \(\overrightarrow{AB}=-\overrightarrow{CD}\).
- These rules are for the origin only. For any other centre, work with vectors from that centre instead: find \(\overrightarrow{CP}\), transform it, then add it back to \(C\). Applying an origin rule to a non-origin centre is a common and expensive slip.
Examiner tips
- The objective to check first is the last one in 7.1: writing a complete description. Whenever a question says “describe fully”, it is asking for a fixed list of parameters, and supplying that list costs nothing beyond the work you have already done. It is the one objective on this page that can be secure in the mathematics and still missing on the page.
- Read the notes column, not just the skill. Three of the most useful facts in Topic 7 live in the syllabus notes rather than in the numbered statements: rotations are through multiples of \(90^\circ\); scale factors may be positive, fractional or negative; and questions may combine transformations. Each of those is a place candidates are surprised.
- On rounding. The general 4024 convention applies here as everywhere: on Paper 2, give non-exact numerical answers to 3 significant figures, do not round intermediate values, and use the unrounded value if a later part depends on it. Most of Topic 7 produces exact answers anyway, so the convention bites mainly on magnitudes.
- M-O-V-E is a specialisation of R-I-S-E, the whole-course protocol — read and represent, identify connections, solve visibly, evaluate. If you already use R-I-S-E, keep it; M-O-V-E just names what “represent” and “evaluate” mean when the objects are shapes and directed segments.
- Tracing paper is allowed, and you may ask for it in both papers. Trace the object, put your pencil point on a candidate centre, and turn the paper through the angle: if the tracing lands on the image, the centre and angle are right. Use it to confirm a centre you found by the method above, not to replace the working — the description still has to be written out in full.
- The pair you can rely on. Two translations always give the same result in either order, because \(\mathbf u+\mathbf v=\mathbf v+\mathbf u\). A few other pairs happen to commute as well — two rotations about the same centre, or two enlargements from the same centre — but they all share a fixed point, and questions rarely hand you that. Every pairing that does not, such as reflection then rotation or rotation then enlargement, should be assumed to depend on the order unless you have checked otherwise.
- Simplify to the standard form. Leave an answer as a tidy combination such as \(\tfrac12\mathbf p+\tfrac12\mathbf r\) or \(\mathbf r-\mathbf p\), with like terms collected. An unsimplified expression such as \(\mathbf p+\tfrac12\mathbf r-\tfrac12\mathbf p\) is not yet an answer in the form the question asked for, and it hides the scalar multiple you may need in the next part.
- Not on the formula sheet. The list of formulas printed on the paper covers areas, volumes, the quadratic formula and the sine and cosine rules. Vector magnitude is not on it, so this is a recall item. Since it is just Pythagoras, the safest way to “remember” it is to sketch the right triangle.
- The last column is not optional prose. The syllabus asks you to show that vectors are parallel and to show that three points are collinear, and an equation on its own has not yet shown anything — it is the working that leads to the statement. Reaching the correct scalar multiple and stopping leaves the question unanswered. One sentence, every time.
- Read the question before choosing. If it says “describe fully the single rotation”, give the rotation. If it says “describe fully the single enlargement”, give \(k=-1\). If it just says “describe fully”, either is fine — but give one of them completely rather than hedging with both.
- Six of these eight cost no mathematics at all. Errors 1, 2, 4, 6, 7 and 8 are failures of recording, not of understanding — the work was done and the answer was written down badly. That is good news: they are the cheapest marks in Topic 7 to recover, and rereading this list the night before is a genuine use of ten minutes.
- None of these is on the formula sheet. The printed list covers areas, volumes, the quadratic formula and the sine and cosine rules — nothing from Topic 7. Every one of the eight above is a recall item. The good news is that six of them can be rebuilt in a line or two from the route rule and Pythagoras, so what you really need to hold is the route rule, the enlargement relation, and the habit of deriving the rest.
- Mark yourself against the descriptions, not just the numbers. In questions 1, 5 and 6 the coordinates are worth about half the marks and the complete description is worth the rest. If you got every image right and still scored under half, the problem is not your mathematics — go back to the description checklist.
How Transformations and Vectors is examined
- Both papers are two hours, both carry 100 marks, and both may assess any part of Topic 7. Paper 1 is non-calculator; Paper 2 allows a scientific calculator. Every question on both papers is compulsory, and a ruler is needed for both — the syllabus states explicitly that a ruler must be used for all straight edges in transformation work.
- Completeness before elegance. “Describe fully” asks for a fixed list of parameters. A description missing one of them has not answered the instruction, however good the drawing is.
- Exact components stay exact. A vector answer of \(\begin{pmatrix}7\\-9\end{pmatrix}\) is the answer; there is nothing to round.
- An accuracy instruction wins. If Paper 2 asks for an exact magnitude, \(\sqrt{53}\) is right and \(7.28\) is not. If it asks for 3 significant figures, the reverse holds.
- A proof needs a conclusion. Correct algebra that stops at \(\overrightarrow{AB}=3\overrightarrow{BC}\) has not yet said that the points are collinear.
- The diagram is not evidence. Points that look collinear, or lines that look parallel, prove nothing. The syllabus asks you to show it.
Frequently asked questions
What must a full description of a transformation contain?
Every parameter the transformation needs. A reflection needs the mirror line as an equation, such as \(y=x\) or \(x=-2\). A rotation needs the angle, the direction and the centre — “rotation through \(90^\circ\) clockwise about \((2,1)\)”. An enlargement needs the centre and the scale factor \(k\). A translation needs the column vector \(\begin{pmatrix}a\b\end{pmatrix}\). Count the parameters before writing: three for a rotation, two for an enlargement, one each for a reflection and a translation.
What does a negative scale factor do in an enlargement?
It sends the image through the centre to the opposite ray; it does not make the image smaller. Answer two questions separately: how big uses \(|k|\), so lengths are multiplied by \(|k|\) and area by \(k^{2}\); which side uses the sign of \(k\). A scale factor of \(-3\) makes the image three times as large and places it on the far side of the centre. An image is smaller only when \(|k|<1\).
How do you write a translation as a column vector?
The top entry is always the horizontal movement and the bottom entry is always the vertical movement: \(\begin{pmatrix}a\b\end{pmatrix}\) means \(a\) across, then \(b\) up. A shift of two left and three down is \(\begin{pmatrix}-2\-3\end{pmatrix}\), not \(\begin{pmatrix}3\-2\end{pmatrix}\). Read the vector back as a sentence — “two left, three down” — and check it against the diagram.
How do you find the vector \(\overrightarrow{AB}\) from position vectors?
Use the route rule rather than a memorised formula: \(\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=-\mathbf a+\mathbf b\), so \(\overrightarrow{AB}=\mathbf b-\mathbf a\), end point minus start point. Deriving it in two lines takes about four seconds and gets the subtraction the right way round every time; memorising the result alone is how it ends up reversed under pressure. Reversing a segment uses \(\overrightarrow{BA}=-\overrightarrow{AB}\).
How do you find the magnitude of a vector?
By Pythagoras, because the two components are perpendicular: \(\left|\begin{pmatrix}x\y\end{pmatrix}\right|=\sqrt{x^{2}+y^{2}}\). For \(\begin{pmatrix}-6\8\end{pmatrix}\) this is \(\sqrt{36+64}=10\). A magnitude is a non-negative scalar, never a column, so a negative answer is a signal to check. Leave it as an exact surd unless a decimal is asked for, in which case give 3 significant figures.
How do you prove three points are collinear using vectors?
Find two vectors joining the points, such as \(\overrightarrow{CD}\) and \(\overrightarrow{DE}\), by choosing routes through known segments. Show that one is a scalar multiple of the other, which proves they are parallel. Because they also share the point \(D\), the three points lie on one straight line. Finish with an explicit sentence stating that conclusion; the scalar multiple also gives the ratio \(CD:DE\).
Why does a rotation through \(180^\circ\) not need a direction?
Because a half-turn clockwise and a half-turn anticlockwise land in the same place. About the origin the rule is \((x,y)\mapsto(-x,-y)\), which sends \((2,-1)\) to \((-2,1)\) whichever way you turn. For \(90^\circ\) or \(270^\circ\) there are two different rotations about the same centre, so the direction must be stated. This syllabus rotates through multiples of \(90^\circ\) only.
Syllabus reference and sources
Written against: Cambridge O Level Mathematics – Syllabus D (4024) 2025–2027 Syllabus (Subject Content, Topic 7: Transformations and Vectors).
Written by: Academiq Edu Instructor Panel
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