Topic 7 was not tested directly by the Foundations Check, but three of its findings point straight at it.
| Code | Sub-topic | Expected difficulty for you |
|---|---|---|
| E7.1 | Transformations | Low — but the describing is where the marks go |
| E7.2 | Vectors in two dimensions | Medium — arithmetic you can already do, in a new shape |
| E7.3 | Magnitude of a vector | High — Pythagoras again, answers left as surds |
| E7.4 | Vector geometry | High — the reasoning step, and the most valuable part |
Paper 2 is non-calculator: 2 hours, 100 marks, 50% of your grade. This is one of the few topics where that changes almost nothing, because there is nothing here a machine would have helped with. It is small whole numbers, one square root, and clear reasoning. Magnitudes are left in surd form throughout.
Number sense and the four operations: solid on all seven rungs. No gap anywhere, from primary arithmetic right up to Extended. That is unusually good news for this topic in particular.
Vectors involve no new arithmetic at all. Adding two column vectors is adding two pairs of small numbers. Multiplying by a scalar is a times table. The magnitude is a square and a square root. Every calculation in topic 7 is one you can already do without thinking — which means the marks here are for saying what you did precisely, not for computing. That is a genuinely winnable topic, and it is worth roughly the same as any other on the paper.
Every idea appears four times, with less help each time.
Column vectors are typed on one line, as (3,-2), and surds as 2root5 or
2√5. Section 5 is a mixed set: fifteen questions, unlabelled and out of order.
There are four transformations on the syllabus. Performing them is straightforward and you will pick it up quickly. Describing them is where the marks actually live, and it is where most students quietly lose two marks a paper for years without ever finding out why.
Cambridge marks a description as a checklist. Each transformation has a required number of details. Give them all and you get the marks; give all but one and you usually get nothing, because a partial description does not identify the transformation. Learn the checklist first, before the geometry.
| Transformation | Details required | How many |
|---|---|---|
| Reflection | the word reflection, and the equation of the mirror line | 2 |
| Rotation | the word rotation, the angle, the direction, the centre | 4 |
| Enlargement | the word enlargement, the scale factor, the centre | 3 |
| Translation | the word translation, and the column vector | 2 |
Every point moves straight across the mirror line to the same distance on the other side, along a line perpendicular to the mirror. Points on the mirror line do not move at all.
The mirror line must be given as an equation: x = 3, y = −1,
y = x, y = −x, or “the y-axis” which is x = 0.
y = x and
y = −x, count diagonally, or use the shortcut: reflecting in y = x swaps the
coordinates, so (3, 1) becomes (1, 3).The whole shape turns about a fixed point. You must state the angle, the direction and the centre — three things, because leaving any one out leaves the transformation ambiguous.
(x, y) becomes (−y, x) anticlockwise and
(y, −x) clockwise, and a 180° turn gives (−x, −y) either way.Every point moves along the ray from the centre, and its distance from the centre is multiplied by the scale
factor k. The image is similar to the object: same shape, angles unchanged, all lengths
× k.
Two cases sound wrong and are not:
k = ½ makes the image smaller. It is still called an
enlargement — there is no such word as “reduction” in this syllabus. Say
“enlargement, scale factor ½, centre (0, 0)”.k = −2 sends every point to the opposite side of the
centre, twice as far out, and the image comes out upside down.k = −2 as if it were
k = 2 and drawing the image on the same side of the centre. The vector from the centre to each point
is multiplied by −2, so its direction reverses. A point 1 right and 1 up from the centre lands
2 left and 2 down. Your diagnostic showed sign errors recurring across topics; this is one of the places they cost
a whole question rather than a mark, because the entire image ends up in the wrong quadrant.k = image length ÷ object length, using a pair
of matching sides. Then check the direction: same side of the centre means k is positive, opposite
side means negative. Remember that lengths scale by k but areas scale by k².The shape slides. Nothing turns, nothing flips, nothing changes size. The movement is given as a column vector: the top number is the movement right, the bottom number the movement up.
(6, −5), not
(6, 5). And a translation description must include the vector — the word
“translation” on its own is one mark out of two.Do the first transformation, then apply the second one to the image, not to the original shape. To find the single transformation that does both, follow one or two points through, using these rules:
| transformation | the point (x, y) goes to |
|---|---|
| reflection in the y-axis | (−x, y) |
| reflection in the x-axis | (x, −y) |
| reflection in y = x | (y, x) |
| reflection in y = −x | (−y, −x) |
| rotation 90° anticlockwise about O | (−y, x) |
| rotation 90° clockwise about O | (y, −x) |
| rotation 180° about O | (−x, −y) |
A vector has two properties and only two: a size and a direction. It does not have a position. An arrow 3 right and 2 up is the same vector wherever you draw it on the grid — which is exactly why the same vector can describe a translation applied to any shape anywhere.
Written as a column vector, the top number is the movement to the right and the bottom number is the movement up. Left and down are negative. That is the whole notation.
Printed vectors are in bold; handwritten ones are underlined, like a with a line under it.
Do underline them — a mark scheme can distinguish a vector a from a length a, and
so should you.
To add two vectors, add the tops and add the bottoms. Geometrically, you draw the second arrow starting from the tip of the first, and the sum is the single arrow from the very start to the very end.
The top row and the bottom row never mix. Two completely separate additions, side by side.
Subtract the tops and subtract the bottoms. The difficulty is never the idea, it is the arithmetic when a negative number is involved.
3 − (−1) written as 2. Subtracting a negative adds,
so it is 4. Your check flagged sign errors recurring across strands, and vector subtraction is where they appear
most often in this topic, because every question has two chances to make one — the top row and the bottom
row. Write the brackets in. 2 − (−5) looks like an addition once the brackets are on the
page, and it is one.a − b as a + (−b).
Flip both signs of b first, then add. One rule applied twice beats two rules applied once each, and
addition is where you make no errors.a − b is the arrow running from the tip of b
to the tip of a. If your answer points the other way, you have computed b − a instead.
Those two are always negatives of each other.A scalar is an ordinary number. Multiplying a vector by it multiplies both components:
2a points the same way as a and is twice as long.−a is the same length as a and points exactly the opposite way.½a points the same way and is half as long.(6, −3) is parallel to (2, −1), and so is
(−4, 2) — pointing backwards still counts as parallel. This single test does most of the
work in vector geometry proofs.The magnitude of a vector is its length: how far the arrow reaches, ignoring which way it points. It is written with vertical bars, called modulus bars, exactly as with the modulus of a number.
There is nothing new to learn here. The two components are the short sides of a right-angled triangle and the vector itself is the hypotenuse, so this is the formula from topic 6 wearing different notation.
| AB | means the length of AB, a plain number.
AB without bars means the vector, which carries a direction as well. A mark scheme distinguishes them
and so should your working — it is a free way to show the examiner you know which object you are handling.√20 is not a finished answer; 2√5 is. To simplify,
find the largest square factor and take its root outside: 20 = 4 × 5, so
√20 = √4 × √5 = 2√5. The square factors worth spotting are 4, 9, 16, 25,
36, 49, 64, 81 and 100.(−5)² is
25, not −25. A magnitude is a length, so it can never come out negative — if yours has, you
squared a negative wrongly. This is the friendliest place in the whole topic for your sign habit, because both
components get squared and every sign disappears.| a | and | −a | are equal:
reversing a vector does not change its length. Second, | 3a | = 3 | a | — scaling a vector
scales its length by the same factor, so you can often take a common factor out before squaring anything.This is the part of the topic that carries real marks, and it is the part that looks hardest. It is not hard. It rests on one idea and one rule, and once you have both, every question in it becomes the same question.
A position vector is the vector from the origin O to a point. If OA = a, then
a is the position vector of A. It fixes where A is, rather than merely describing a movement.
To travel from A to B when you know both position vectors, go backwards to the origin and then forwards:
Read that carefully, because the order is the single most common error in the whole topic.
AB = b − a, not a − b. Finish minus start. The letter you are heading
towards comes first.
AB = a − b. That is
BA, the same arrow pointing the other way, and it will make every subsequent line of the proof come
out negative. Say “finish minus start” under your breath every time. AB = b − a,
BA = a − b, PQ = q − p.XY = XO + OA + AY, or whatever the
picture allows. All routes between the same two points give the same answer, which is itself a free check.If M is the midpoint of AB, travel there: from O to A, then half of the way from A to B.
The tidy form ½(a + b) is worth remembering, but derive it the first few times so that the
same method works when the point is not a midpoint. If P divides AB in the ratio 1 : 2, then P is a third
of the way along, so OP = a + 1⁄3(b − a).
That is the whole test. Work out both vectors in terms of a and b, then show one is a
multiple of the other. A negative k is fine — opposite directions are still parallel.
Collinear means three or more points lie on one straight line. It needs two things, and a proof that gives only the first is incomplete:
So to prove P, Q and R collinear: show QR = k × PQ, then write the sentence “PQ and QR
are parallel and share the point Q, therefore P, Q and R are collinear.” That sentence is a mark.
Nothing here is labelled, and the four sub-topics are shuffled together. In an exam nobody tells you whether a question wants a description, a column-vector calculation, a magnitude or a piece of reasoning, and knowing which is which is the skill that blocked practice never builds.
Two habits to keep switched on throughout: write brackets round every negative before you add, and say “finish minus start” before every directed line segment.
Six things. Between them they cover nearly everything an examiner can ask in topic 7.
a − b becomes a + (−b), with both of b flipped first. That one habit removes most of the sign errors your diagnostic picked up.√20 is not finished, 2√5 is.AB = b − a. Then to prove things parallel, factorise until one vector is visibly a multiple of the other — and for collinear, add the sentence naming the shared point.Come back to section 5 in a week without reading anything above it. This topic rewards a cold run more than most, because almost nothing in it needs to be understood twice — it needs to be written down precisely once.