Worked example · Easy
Move a point
Translate \(A(2,3)\) by \(\binom{4}{-2}\).
Add components: \(A\mapsto (2+4,3-2)=\boxed{(6,1)}\).
Chapter 43 · Completely rewritten
Vectors
Clear explanations, original worked examples and a full graded exercise set, mapped to the textbook.
What you will learn
Describe vector translations; use vector notation, column operations, scalars, position vectors and magnitudes; explain geometric relationships with vector arguments.
Begin with the easy examples, then attempt the medium and exam-style examples before using the hidden solutions.
1
Subtopic 1
Vectors and translations
A vector specifies both a direction and a distance. A translation moves every point of a shape by the same vector without turning it, reflecting it or changing its size. A vector \(\binom{a}{b}\) means move \(a\) units horizontally and \(b\) vertically; left and down movements are negative.
Essential method
\((x,y)\xrightarrow{\binom{a}{b}}(x+a,y+b)\). A vector from \(A(x_1,y_1)\) to \(B(x_2,y_2)\) is \(\binom{x_2-x_1}{y_2-y_1}\).
Exam tip: Check the sign of each movement. Translation affects every vertex in the same way.
Worked examples · Easy → Medium → Exam-style
Every solution explains the mathematical method and gives the result.
Worked example · Medium
Translation between points
Find the vector taking \(P(-2,5)\) to \(Q(4,1)\).
Horizontal \(=4-(-2)=6\). Vertical \(=1-5=-4\). Vector \(=\boxed{\binom6{-4}}\).
Worked example · Exam-style
Triangle translation
Triangle \(ABC\) has vertices \(A(1,-1),B(3,2),C(-2,4)\). Translate it by \(\binom{-3}{5}\).
Add \((-3,5)\) to each point: \(A\mapsto\boxed{(-2,4)},\ B\mapsto\boxed{(0,7)},\ C\mapsto\boxed{(-5,9)}\).
2
Subtopic 2
Vector notation
Vector arrows indicate direction. \(\overrightarrow{AB}\) is the displacement from point \(A\) to point \(B\), while \(\overrightarrow{BA}\) is its negative. Different arrows represent the same vector when they have equal magnitude and direction, regardless of where they start.
Bold lower-case letters such as \(\mathbf a\) name vectors, while a plain number is a scalar.
Essential method
\(\overrightarrow{BA}=-\overrightarrow{AB}\); \(\overrightarrow{AA}=\mathbf0\). Equal vectors have matching components, not merely equal length.
Exam tip: Do not mistake the vector from A to B for the coordinates of B.
Worked examples · Easy → Medium → Exam-style
Every solution explains the mathematical method and gives the result.
Worked example · Easy
Reverse direction
If \(\overrightarrow{AB}=\binom{-3}{5}\), find \(\overrightarrow{BA}\).
Reverse the vector: \(\overrightarrow{BA}=-\binom{-3}{5}=\boxed{\binom3{-5}}\).
Worked example · Medium
Equivalent vectors
\(\mathbf a=\binom{2}{-4}\) and \(\mathbf b=\binom{2}{-4}\) have different starting points. Are they equal?
Yes. Their horizontal and vertical components match, so \(\boxed{\mathbf a=\mathbf b}\).
Worked example · Exam-style
Link two displacements
If \(\overrightarrow{PQ}=\mathbf p\) and \(\overrightarrow{QR}=\mathbf q\), express \(\overrightarrow{RP}\).
Along \(P\to Q\to R\), \(\overrightarrow{PR}=\mathbf p+\mathbf q\). Reverse it: \(\boxed{\overrightarrow{RP}=-\mathbf p-\mathbf q}\).
3
Subtopic 3
Multiplying a vector by a scalar
Multiplying by a positive scalar changes the length of a vector while preserving direction. A negative scalar reverses its direction. A scalar of \(0\) produces the zero vector. Parallel vectors are scalar multiples of one another.
Essential method
\(k\binom ab=\binom{ka}{kb}\). Magnitudes scale by \(|k|\).
Exam tip: Scale both components. A factor of −2 doubles the length and reverses direction.
Worked examples · Easy → Medium → Exam-style
Every solution explains the mathematical method and gives the result.
Worked example · Easy
Double a vector
Find \(2\binom{-1}{3}\).
\(2\binom{-1}{3}=\boxed{\binom{-2}{6}}\).
Worked example · Medium
Negative scalar
Find \(-3\binom{2}{-4}\).
\(\binom{-6}{12}\). Both components change sign and triple in size: \(\boxed{\binom{-6}{12}}\).
Worked example · Exam-style
Parallel test
Are \(\binom6{-9}\) and \(\binom{-2}{3}\) parallel?
\(\binom6{-9}=-3\binom{-2}{3}\), so \(\boxed{\text{yes, they are parallel}}\).
4
Subtopic 4
Addition and subtraction of column vectors
Adding successive movements gives the resultant displacement. Add the corresponding components. To subtract a vector, add its opposite. Draw arrows head-to-tail if a visual check is useful.
Essential method
\(\binom ab+\binom cd=\binom{a+c}{b+d}\); \(\binom ab-\binom cd=\binom{a-c}{b-d}\).
Exam tip: Subtract every component in the stated order. Vector addition is commutative; subtraction is not.
Worked examples · Easy → Medium → Exam-style
Every solution explains the mathematical method and gives the result.
Worked example · Easy
Add two vectors
\(\binom24+\binom{-3}1\).
\(\binom{2-3}{4+1}=\boxed{\binom{-1}{5}}\).
Worked example · Medium
Subtract vectors
Find \(\binom5{-2}-\binom{-1}4\).
\(\binom{5-(-1)}{-2-4}=\boxed{\binom6{-6}}\).
Worked example · Exam-style
Multi-stage travel
A journey has vectors \(\binom3{-1}\), \(\binom{-5}4\) and \(\binom23\). Find the overall displacement.
Add horizontal components \(3-5+2=0\); vertical \(-1+4+3=6\). Result \(=\boxed{\binom06}\).
5
Subtopic 5
Position vectors
A position vector starts at the origin \(O\). If point \(P\) has coordinates \((x,y)\), then \(\overrightarrow{OP}=\binom xy\). To move from one point to another, subtract their position vectors.
Essential method
\(\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}\), and \(\overrightarrow{OA}=\mathbf a\Rightarrow\overrightarrow{AO}=-\mathbf a\).
Exam tip: Keep the arrow direction fixed when subtracting endpoint vectors.
Worked examples · Easy → Medium → Exam-style
Every solution explains the mathematical method and gives the result.
Worked example · Easy
Read a position vector
Point \(P=(-2,5)\). Find \(\overrightarrow{OP}\).
\(\boxed{\overrightarrow{OP}=\binom{-2}{5}}\).
Worked example · Medium
Difference of positions
\(\overrightarrow{OA}=\binom13\) and \(\overrightarrow{OB}=\binom6{-2}\). Find \(\overrightarrow{AB}\).
\(\overrightarrow{AB}=\binom6{-2}-\binom13=\boxed{\binom5{-5}}\).
Worked example · Exam-style
Find an unknown position
\(\overrightarrow{OA}=\binom{-4}2\) and \(\overrightarrow{AB}=\binom75\). Find \(B\).
\(\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}=\binom{-4+7}{2+5}=\binom37\). Thus \(\boxed{B=(3,7)}\).
6
Subtopic 6
The magnitude of a vector
Magnitude means length, not direction. Horizontal and vertical components form a right triangle, so Pythagoras’ theorem gives the vector length. Magnitude cannot be negative.
Essential method
\(\left|\binom xy\right|=\sqrt{x^2+y^2}\). For a displacement from \(A\) to \(B\), use the components of \(\overrightarrow{AB}\).
Exam tip: Square negative components carefully; their squares are positive.
Worked examples · Easy → Medium → Exam-style
Every solution explains the mathematical method and gives the result.
Worked example · Easy
3–4–5 magnitude
Find \(\left|\binom34\right|\).
\(\sqrt{3^2+4^2}=\sqrt{25}=\boxed5\).
Worked example · Medium
Negative component
Find the length of \(\binom{-5}{12}\).
\(\sqrt{(-5)^2+12^2}=\sqrt{169}=\boxed{13}\).
Worked example · Exam-style
Distance between points
Find the distance from \(A(-2,1)\) to \(B(4,9)\).
\(\overrightarrow{AB}=\binom68\). Distance \(=\sqrt{6^2+8^2}=\boxed{10}\) units.
7
Subtopic 7
Vector geometry
Use routes between vertices to prove properties without measuring a diagram. Opposite sides of a parallelogram are equal vectors when taken in corresponding directions. A midpoint divides a displacement into equal halves; a point in a given ratio divides its segment into proportional vector parts.
Essential method
\(\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}\); midpoint \(M\) of \(AB\): \(\overrightarrow{OM}=\tfrac12(\mathbf a+\mathbf b)\). If \(AP:PB=m:n\), then \(\overrightarrow{OP}=\frac{n\mathbf a+m\mathbf b}{m+n}\).
Exam tip: When proving parallel lines, show one vector is a scalar multiple of the other; explain why the ratios match.
Worked examples · Easy → Medium → Exam-style
Every solution explains the mathematical method and gives the result.
Worked example · Easy
Parallelogram diagonal
In a parallelogram \(OABC\), \(\overrightarrow{OA}=\mathbf a\) and \(\overrightarrow{OC}=\mathbf c\). Find \(\overrightarrow{OB}\).
Travel \(O\to A\to B\); \(\overrightarrow{AB}=\mathbf c\). Thus \(\boxed{\overrightarrow{OB}=\mathbf a+\mathbf c}\).
Worked example · Medium
Midpoint
Position vectors of \(A\) and \(B\) are \(\binom24\) and \(\binom{8}{-2}\). Find midpoint position.
\(\tfrac12\left(\binom24+\binom8{-2}\right)=\tfrac12\binom{10}{2}=\boxed{\binom51}\).
Worked example · Exam-style
Divide a line in ratio
\(\overrightarrow{OA}=\mathbf a\), \(\overrightarrow{OB}=\mathbf b\). Point \(P\) divides \(AB\) in the ratio \(AP:PB=2:1\). Express \(\overrightarrow{OP}\).
\(\overrightarrow{AP}=\tfrac23(\mathbf b-\mathbf a)\). Therefore \(\overrightarrow{OP}=\mathbf a+\tfrac23(\mathbf b-\mathbf a)=\boxed{(\mathbf a+2\mathbf b)/3}\).
Chapter 43 — Student exercises
Questions are arranged by textbook subtopic. Within each section they start with confidence-building tasks and finish with extended examination-style reasoning. Try each question before opening its solution.
Vectors and translations
1Question 1Foundation
Move \((3,4)\) by \(\binom{2}{-1}\).
Show Answer
Step-by-step solution: \((3+2,4-1)=\boxed{(5,3)}\).
2Question 2Foundation
Find the vector from \((0,2)\) to \((5,7)\).
Show Answer
Step-by-step solution: \(\binom{5-0}{7-2}=\boxed{\binom55}\).
3Question 3Intermediate
Find the vector taking \((-4,1)\) to \((1,-6)\).
Show Answer
Step-by-step solution: \(\binom{1-(-4)}{-6-1}=\boxed{\binom5{-7}}\).
4Question 4Exam-style
Point \(X(8,-1)\) is the image of \(Y(2,3)\) under a translation. Find the translation vector.
Show Answer
Step-by-step solution: Image minus original: \(\binom{8-2}{-1-3}=\boxed{\binom6{-4}}\).
5Question 5Exam-style
A triangle \(A(0,0),B(4,0),C(1,3)\) moves by \(\binom{-2}{-5}\). State all image vertices.
Show Answer
Step-by-step solution: \(A^\prime=(-2,-5),\ B^\prime=(2,-5),\ C^\prime=(-1,-2)\). Every point receives the same movement.
Vector notation
6Question 6Foundation
If \(\overrightarrow{CD}=\binom41\), find \(\overrightarrow{DC}\).
Show Answer
Step-by-step solution: Negate both components: \(\boxed{\binom{-4}{-1}}\).
7Question 7Foundation
Which equals \(\mathbf v=\binom{-2}{3}\): \(\binom{2}{-3}\) or \(\binom{-2}{3}\)?
Show Answer
Step-by-step solution: The second has the same components: \(\boxed{\binom{-2}{3}}\).
8Question 8Intermediate
If \(\overrightarrow{AB}=\mathbf t\), write \(\overrightarrow{BA}\) using \(\mathbf t\).
Show Answer
Step-by-step solution: Reverse direction: \(\boxed{-\mathbf t}\).
9Question 9Exam-style
If \(\overrightarrow{AB}=\mathbf a\), \(\overrightarrow{BC}=\mathbf b\), write \(\overrightarrow{AC}\).
Show Answer
Step-by-step solution: Follow the route A–B–C: \(\boxed{\mathbf a+\mathbf b}\).
10Question 10Exam-style
For a closed triangle, simplify \(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}\).
Show Answer
Step-by-step solution: The route returns to its start, so the resultant is \(\boxed{\mathbf0}\).
Multiplying a vector by a scalar
11Question 11Foundation
Calculate \(3\binom21\).
Show Answer
Step-by-step solution: \(\boxed{\binom63}\).
12Question 12Foundation
Calculate \(-2\binom{-4}{3}\).
Show Answer
Step-by-step solution: \(\boxed{\binom8{-6}}\).
13Question 13Intermediate
Find \(\tfrac12\binom8{-6}\).
Show Answer
Step-by-step solution: \(\boxed{\binom4{-3}}\).
14Question 14Exam-style
If \(\mathbf a=\binom{-3}{5}\), find \(4\mathbf a\).
Show Answer
Step-by-step solution: \(4\mathbf a=\boxed{\binom{-12}{20}}\).
15Question 15Exam-style
Are \(\binom{10}{-15}\) and \(\binom{-2}{3}\) parallel? Explain.
Show Answer
Step-by-step solution: The first is \(-5\) times the second, so \(\boxed{\text{parallel}}\).
Addition and subtraction of column vectors
16Question 16Foundation
Add \(\binom13+\binom24\).
Show Answer
Step-by-step solution: \(\boxed{\binom37}\).
17Question 17Foundation
Subtract \(\binom52-\binom11\).
Show Answer
Step-by-step solution: \(\boxed{\binom41}\).
18Question 18Intermediate
Find \(\binom{-3}5+\binom4{-2}\).
Show Answer
Step-by-step solution: \(\boxed{\binom13}\).
19Question 19Exam-style
Simplify \(2\binom12-\binom{-3}5\).
Show Answer
Step-by-step solution: \(\binom24-\binom{-3}5=\boxed{\binom5{-1}}\).
20Question 20Exam-style
A hiker moves \(\binom{6}{-1}\) km, then \(\binom{-2}{4}\) km. What vector takes the hiker directly back to the start?
Show Answer
Step-by-step solution: Total \(=\binom43\). Return vector is its negative: \(\boxed{\binom{-4}{-3}}\text{ km}\).
Position vectors
21Question 21Foundation
What is the position vector of \(Q(4,-1)\)?
Show Answer
Step-by-step solution: \(\boxed{\binom4{-1}}\).
22Question 22Foundation
\(A=(1,2)\), \(B=(6,3)\). Find \(\overrightarrow{AB}\).
Show Answer
Step-by-step solution: \(\binom{6-1}{3-2}=\boxed{\binom51}\).
23Question 23Intermediate
If \(\overrightarrow{OA}=\binom{3}{-4}\), what is \(\overrightarrow{AO}\)?
Show Answer
Step-by-step solution: \(\boxed{\binom{-3}{4}}\).
24Question 24Exam-style
If \(\overrightarrow{OA}=\binom14\) and \(\overrightarrow{AB}=\binom{-3}2\), find \(B\).
Show Answer
Step-by-step solution: \(\binom14+\binom{-3}2=\binom{-2}6\), so \(\boxed{B=(-2,6)}\).
25Question 25Exam-style
\(A=(-3,2)\) and \(B=(5,-4)\). Express \(\overrightarrow{AB}\) and \(\overrightarrow{BA}\).
Show Answer
Step-by-step solution: \(\overrightarrow{AB}=\binom{8}{-6}\); reversing gives \(\overrightarrow{BA}=\boxed{\binom{-8}{6}}\).
The magnitude of a vector
26Question 26Foundation
Find the magnitude of \(\binom{6}{8}\).
Show Answer
Step-by-step solution: \(\sqrt{36+64}=\boxed{10}\).
27Question 27Foundation
Find \(\left|\binom{-3}{-4}\right|\).
Show Answer
Step-by-step solution: \(\sqrt9+\sqrt{16}\) is not the method; use \(\sqrt{9+16}=\boxed5\).
28Question 28Intermediate
Find the magnitude of \(\binom{1}{2}\) to 2 d.p.
Show Answer
Step-by-step solution: \(\sqrt{1+4}=\sqrt5\approx\boxed{2.24}\).
29Question 29Exam-style
How far apart are \(P(2,-3)\) and \(Q(5,1)\)?
Show Answer
Step-by-step solution: Displacement \(\binom34\), magnitude \(\sqrt{3^2+4^2}=\boxed5\) units.
30Question 30Exam-style
A vector \(\mathbf v=\binom{x}{12}\) has magnitude 13. Find all possible values of \(x\).
Show Answer
Step-by-step solution: \(x^2+12^2=13^2\), so \(x^2=25\), giving \(\boxed{x=5\text{ or }-5}\).
Vector geometry
31Question 31Foundation
If \(\overrightarrow{AB}=\mathbf a\) and \(\overrightarrow{BC}=\mathbf b\), write \(\overrightarrow{AC}\).
Show Answer
Step-by-step solution: Travel from A via B to C: \(\boxed{\mathbf a+\mathbf b}\).
32Question 32Foundation
If \(\overrightarrow{OA}=\mathbf a\) and \(\overrightarrow{OB}=\mathbf b\), find \(\overrightarrow{AB}\).
Show Answer
Step-by-step solution: \(\overrightarrow{AO}+\overrightarrow{OB}=-\mathbf a+\mathbf b=\boxed{\mathbf b-\mathbf a}\).
33Question 33Intermediate
Find the midpoint of \(A(2,3)\) and \(B(8,7)\).
Show Answer
Step-by-step solution: \(\left(\frac{2+8}{2},\frac{3+7}{2}\right)=\boxed{(5,5)}\).
34Question 34Exam-style
Show \(\binom{4}{6}\) and \(\binom{10}{15}\) are parallel.
Show Answer
Step-by-step solution: \(\binom{10}{15}=\tfrac52\binom46\). Since one is a scalar multiple of the other, they are parallel.
35Question 35Exam-style
\(\overrightarrow{OA}=\binom10\), \(\overrightarrow{OB}=\binom{7}{12}\), and \(P\) divides \(AB\) in the ratio 1:2. Find \(P\).
Show Answer
Step-by-step solution: \(AP=\tfrac13AB=\tfrac13\binom6{12}=\binom24\). Add to \(A\): \(OP=\binom10+\binom24=\binom34\). Thus \(\boxed{P=(3,4)}\).
36Question 36Exam-style challenge
In a parallelogram \(OACB\), \(\overrightarrow{OA}=\mathbf a\), \(\overrightarrow{OB}=\mathbf b\) and \(M\) is the midpoint of \(AB\). Using vectors, prove that \(O\), \(M\) and \(C\) lie on a straight line. Give the ratio \(OM:MC\).
Show Answer
Step 1 — Find \(\overrightarrow{OC}\): The diagonal of the parallelogram is \(\overrightarrow{OC}=\mathbf a+\mathbf b\).
Step 2 — Use the midpoint rule: \(\overrightarrow{OM}=\frac12(\mathbf a+\mathbf b)\).
Step 3 — Prove collinearity: Because \(\overrightarrow{OM}=\frac12\overrightarrow{OC}\), the two vectors are parallel and have the same starting point \(O\). Hence \(O\), \(M\) and \(C\) are collinear.
Step 4 — State the ratio: \(M\) is halfway from \(O\) to \(C\), so \(\boxed{OM:MC=1:1}\).
High-grade revision checklist
- Can I solve unfamiliar questions using Vectors and translations and explain my method?
- Can I solve unfamiliar questions using Vector notation and explain my method?
- Can I solve unfamiliar questions using Multiplying a vector by a scalar and explain my method?
- Can I solve unfamiliar questions using Addition and subtraction of column vectors and explain my method?
- Can I solve unfamiliar questions using Position vectors and explain my method?
- Can I solve unfamiliar questions using The magnitude of a vector and explain my method?
- Can I solve unfamiliar questions using Vector geometry and explain my method?
Before the exam, practise without looking at the answers, check units and rounding, and write geometric or statistical reasons where they are needed.