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Edexcel A-level Maths revision notes

Vectors

Section 10
Both years
Both years: this holds AS subject content and content the exam board adds beyond it for the full A-level.
5 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9MA0 section 10

Checked against Edexcel 9MA0 section 10. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

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10.1

Use vectors in two dimensions and in three dimensions.

Notes
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Explanation

  • A vector records magnitude and direction; in two or three dimensions it can be written in component form or with the unit vectors i\mathbf{i}, j\mathbf{j} and k\mathbf{k}.
  • Work component by component, keeping the xx, yy and zz entries aligned; subtract the initial point from the final point to form a displacement vector.
  • For example, from A(1,2,3)A(1,2,3) to B(4,0,5)B(4,0,5) the displacement is AB=(3,2,2)\overrightarrow{AB}=(3,-2,2).
  • A vector has no fixed location, whereas a point does; a common error is to confuse the coordinates of an endpoint with the components of the displacement leading to it.
Worked example

Given a=(2,1,3)\mathbf{a}=(2,-1,3) and b=(1,4,2)\mathbf{b}=(-1,4,2), find 2ab2\mathbf{a}-\mathbf{b}.

  1. 1.First 2a=(4,2,6)2\mathbf{a}=(4,-2,6).
  2. 2.Subtract corresponding components of b\mathbf{b} to get (4(1),24,62)=(5,6,4)(4-(-1),-2-4,6-2)=(5,-6,4).

Answer: 2ab=(5,6,4)2\mathbf{a}-\mathbf{b}=(5,-6,4)

Common mistakes

  • Don't change the coordinate order between vectors, so components from different axes are combined.
  • Don't combine vector components inconsistently, applying a scalar to only one coordinate.

Exam tip

Perform each vector operation component by component and preserve the coordinate order throughout.

Tier 1 · Easy

ORIGINAL

1.

The point AA has coordinates (2,5)(-2,5) and the point BB has coordinates (4,1)(4,1). Find AB\overrightarrow{AB}.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

Given a=(1,2,1)\mathbf{a}=(1,2,-1), b=(2,1,3)\mathbf{b}=(2,-1,3) and c=(5,0,5)\mathbf{c}=(5,0,5), find scalars λ\lambda and μ\mu such that λa+μb=c\lambda\mathbf{a}+\mu\mathbf{b}=\mathbf{c}.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

Let p=(1,1,2)\mathbf{p}=(1,1,2) and q=(2,1,1)\mathbf{q}=(2,-1,1). By writing (u,v,w)=λp+μq(u,v,w)=\lambda\mathbf{p}+\mu\mathbf{q}, show that (u,v,w)(u,v,w) lies in the span of p\mathbf{p} and q\mathbf{q} if and only if u+vw=0u+v-w=0. Hence find the value of kk for which (7,1,k)(7,1,k) lies in this span and the value of kk for which (5,k,7)(5,k,7) lies in this span. Can the same value of kk satisfy both conditions?

(4)

(Total for Question 1 is 4 marks)

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Answer conventions

Follow the wording on the question and its mark scheme. awrt means an appropriately rounded value is accepted; an exact answer must stay as a fraction, surd, logarithm or multiple of π when required, and a rounded decimal may be disallowed. Include requested units and forms. A cso tag protects that accuracy mark, while earlier method marks follow the question-specific dependencies.

10.2

Calculate the magnitude and direction of a vector and convert between component form and magnitude/direction form.

Notes
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Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For v=(a,b)\mathbf{v}=(a,b), its magnitude is v=a2+b2|\mathbf{v}|=\sqrt{a^2+b^2} and a direction angle must be stated relative to a specified axis or bearing convention.
  • A vector of magnitude rr at angle θ\theta anticlockwise from the positive xx-axis has components (rcosθ,rsinθ)(r\cos\theta,r\sin\theta); use signs or a quadrant-aware angle calculation when reversing the process.
  • For example, magnitude 1010 at 3030^\circ above the positive xx-axis gives (10cos30,10sin30)=(53,5)(10\cos30^\circ,10\sin30^\circ)=(5\sqrt3,5).
  • The value from tan1(b/a)\tan^{-1}(b/a) alone can select the wrong quadrant; a common error is to report an acute reference angle without checking the component signs.
  • A unit vector in the direction of non-zero a\mathbf a is a/a\mathbf a/|\mathbf a|; divide every component by the vector's magnitude.
Worked example

A vector has magnitude 1414 and direction 120120^\circ anticlockwise from the positive xx-axis. Write it in exact component form.

  1. 1.Use (14cos120,14sin120)(14\cos120^\circ,14\sin120^\circ).
  2. 2.Since cos120=12\cos120^\circ=-\frac12 and sin120=32\sin120^\circ=\frac{\sqrt3}{2}, the vector is (7,73)(-7,7\sqrt3).

Answer: (7,73)(-7,7\sqrt3)

Common mistakes

  • Don't normalise a vector by dividing by the sum of its components rather than by its magnitude.
  • Don't use sine for the horizontal component and cosine for the vertical component despite the stated reference axis.

Exam tip

Draw the direction from the positive x-axis, then use cosine horizontally and sine vertically with correct signs.

Tier 1 · Easy

ORIGINAL

1.

Find the magnitude and direction of the vector (3,4)(3,4), giving the direction anticlockwise from the positive xx-axis to 11 decimal place.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Find the magnitude and direction of the vector (6,8)(-6,8), measuring the direction anticlockwise from the positive xx-axis and giving the direction to 33 significant figures.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1.

A two-dimensional vector has magnitude 1313, and the cosine of the angle it makes with the positive xx-axis is 513\frac5{13}. Find all possible component forms and explain the ambiguity.

(4)

(Total for Question 1 is 4 marks)

10.3

Add vectors diagrammatically and perform the algebraic operations of vector addition and multiplication by scalars, and understand their geometrical interpretations.

Notes
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Explanation

  • Vector addition combines successive displacements: placing the tail of b\mathbf{b} at the head of a\mathbf{a} makes the resultant from the first tail to the final head equal to a+b\mathbf{a}+\mathbf{b}.
  • Add corresponding components and multiply every component by a scalar; subtraction is addition of the opposite vector.
  • The vector kak\mathbf{a} is parallel to a\mathbf{a}, has magnitude ka|k||\mathbf{a}|, and points in the reverse direction when k<0k<0.
  • A common error is to multiply only one component by a scalar or to draw vectors head-to-head instead of using a head-to-tail or parallelogram construction.
The head-to-tail construction places v\mathbf v at the head of 2u2\mathbf u; the direct arrow is their resultant.
Worked example

Let u=(4,1)\mathbf{u}=(4,1) and v=(2,5)\mathbf{v}=(-2,5). Find 2u+v2\mathbf{u}+\mathbf{v} and describe a head-to-tail construction for this resultant.

  1. 1.Calculate 2u=(8,2)2\mathbf{u}=(8,2), then add v\mathbf{v} to obtain (82,2+5)=(6,7)(8-2,2+5)=(6,7).
  2. 2.Diagrammatically, translate the vectors without rotating them and place the second u\mathbf{u} after the first, followed by v\mathbf{v}; the direct closing vector is the sum.

Answer: 2u+v=(6,7)2\mathbf{u}+\mathbf{v}=(6,7).; Place two copies of u\mathbf{u} and then one copy of v\mathbf{v} head-to-tail; the resultant joins the initial tail to the final head.

Common mistakes

  • Don't reverse a displacement vector in the diagram but leave its algebraic sign unchanged.
  • Don't add vector diagrams tail-to-tail without completing the parallelogram or head-to-tail construction.

Exam tip

For a resultant, scale each vector first and join them head-to-tail in the order represented algebraically.

Tier 1 · Easy

ORIGINAL

1.

Given u=(2,3)\mathbf{u}=(2,-3) and v=(5,4)\mathbf{v}=(-5,4), find u+v\mathbf{u}+\mathbf{v}.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1.

In triangle ABCABC, AB=u\overrightarrow{AB}=\mathbf{u} and AC=v\overrightarrow{AC}=\mathbf{v}. The point MM is the midpoint of BCBC. Express AM\overrightarrow{AM} in terms of u\mathbf{u} and v\mathbf{v}.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1.

The vectors a=(1,2)\mathbf{a}=(1,2) and b=(3,1)\mathbf{b}=(3,-1) combine to give w=(11,1)\mathbf{w}=(11,1). Find α\alpha and β\beta such that w=αa+βb\mathbf{w}=\alpha\mathbf{a}+\beta\mathbf{b}, and interpret the result geometrically.

(4)

(Total for Question 1 is 4 marks)

10.4

Understand and use position vectors; calculate the distance between two points represented by position vectors.

Notes
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Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • The position vector of a point AA is OA\overrightarrow{OA} from a fixed origin OO; if these vectors are a\mathbf{a} and b\mathbf{b}, then AB=ba\overrightarrow{AB}=\mathbf{b}-\mathbf{a}.
  • Find a displacement by subtracting position vectors in final-minus-initial order, then find distance by taking the magnitude of that displacement.
  • A point dividing ABAB internally in the fraction tt from AA to BB has position vector a+t(ba)\mathbf{a}+t(\mathbf{b}-\mathbf{a}).
  • Distance is a non-negative scalar, not a vector; a common error is to quote ba\mathbf{b}-\mathbf{a} as the distance without calculating its magnitude.
Worked example

The position vectors of AA and BB are (1,2,4)(-1,2,4) and (3,4,7)(3,-4,7). Calculate the exact distance ABAB.

  1. 1.AB=(3(1),42,74)=(4,6,3)\overrightarrow{AB}=(3-(-1),-4-2,7-4)=(4,-6,3).
  2. 2.Hence AB=AB=42+(6)2+32=61AB=|\overrightarrow{AB}|=\sqrt{4^2+(-6)^2+3^2}=\sqrt{61}.

Answer: AB=61AB=\sqrt{61}

Common mistakes

  • Don't square the coordinates of the two points separately instead of squaring their component differences.
  • Don't find the difference of position vectors but forget to take its Euclidean magnitude.

Exam tip

Subtract the endpoint position vectors in a consistent order, then square and sum every component for the distance.

Tier 1 · Easy

ORIGINAL

1.

Points AA and BB have position vectors (2,1)(2,-1) and (7,3)(7,3). Find AB\overrightarrow{AB} and the exact distance ABAB.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

The position vectors of AA and BB are (2,1,5)(-2,1,5) and (6,5,3)(6,5,-3). Find the midpoint of ABAB and the point PP that divides ABAB internally in the ratio AP:PB=3:1AP:PB=3:1.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

Points AA and BB have position vectors (1,2,3)(1,-2,3) and (9,6,1)(9,6,-1). The point MM divides ABAB internally in the ratio AM:MB=3:1AM:MB=3:1. Find the position vector of MM and the exact distance OMOM.

(5)

(Total for Question 1 is 5 marks)

10.5

Use vectors to solve problems in pure mathematics and in context (including forces).

Notes
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Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Vector models turn geometrical displacements or forces into component equations; equilibrium means that the vector sum of all forces is zero.
  • Choose and state positive coordinate directions, resolve every vector consistently, then equate components or use position-vector relationships.
  • In a parallelogram with adjacent position vectors a\mathbf{a} and b\mathbf{b}, the opposite vertex has position vector a+b\mathbf{a}+\mathbf{b} and both diagonals share midpoint 12(a+b)\frac12(\mathbf{a}+\mathbf{b}).
  • A common error is to balance force magnitudes without balancing directions; equal numerical magnitudes do not guarantee equilibrium unless the vector sum is zero.
For the parallelogram with position vectors 0,a,b,a+b\mathbf 0,\mathbf a,\mathbf b,\mathbf a+\mathbf b, both diagonals share midpoint (a+b)/2(\mathbf a+\mathbf b)/2.
Worked example

The points O,A,B,CO,A,B,C have position vectors 0,a,b,a+b\mathbf{0},\mathbf{a},\mathbf{b},\mathbf{a}+\mathbf{b} respectively. Use vectors to prove that the diagonals OCOC and ABAB bisect each other.

  1. 1.The midpoint of OCOC has position vector 12[0+(a+b)]=12(a+b)\frac12[\mathbf{0}+(\mathbf{a}+\mathbf{b})]=\frac12(\mathbf{a}+\mathbf{b}).
  2. 2.The midpoint of ABAB has position vector 12(a+b)\frac12(\mathbf{a}+\mathbf{b}) as well.
  3. 3.Since the two diagonals have the same midpoint, each bisects the other.

Answer: Both diagonals have midpoint position vector 12(a+b)\frac12(\mathbf{a}+\mathbf{b}), so they bisect each other.

Common mistakes

  • Don't assume collinearity from proportional-looking coordinates without finding one common scalar multiplier.
  • Don't use a diagram alone as proof and never shows that the two candidate midpoints have the same position vector.

Exam tip

In a vector proof, calculate both relevant position vectors and use their equality to justify the geometric conclusion.

Tier 1 · Easy

ORIGINAL

1.

Two forces acting on a particle are F1=(4,1)\mathbf{F}_1=(4,-1) N and F2=(2,5)\mathbf{F}_2=(-2,5) N. Find the resultant force and its magnitude.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Two forces acting on a particle are F1=(4,1)N\mathbf{F}_1=(4,-1)\,\text{N} and F2=(2,5)N\mathbf{F}_2=(-2,5)\,\text{N}. Find the third force required for equilibrium and its magnitude.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1.

Three vectors sum to the zero vector. Two of them are a=(3,4,2)\mathbf{a}=(3,4,-2) and b=(5,1,6)\mathbf{b}=(-5,1,6). Find the third vector c\mathbf{c}, its exact magnitude and a unit vector in its direction.

(5)

(Total for Question 1 is 5 marks)

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