Edexcel · GCSE Maths · 1MA1 · Foundation and Higher

M35 · Vectors

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Revision notes, worked examples and methods for vectors.

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Vector notation and arithmetic

  • A vector has magnitude and direction. A column vector (4−3) represents 4 units right and 3 down. Its components are signed displacements, not a coordinate pair identifying one fixed point.
  • Equal vectors have equal components even when drawn in different positions. The negative vector reverses direction: −(4−3) = (−43).
  • Add component by component: (23) + (4−1) = (62). Geometrically, place the second arrow's tail at the first arrow's head.
    Head-to-tail vector additionA vector two right and three up is followed by four right and one down. The resultant is six right and two up.a = (2, 3)b = (4, −1)a + b = (6, 2)Place the second vector head-to-tail
    Head-to-tail vector addition
  • Subtract using the reverse vector: (52) − (14) = (4−2). Order matters.
  • Multiplying by a scalar changes length and may reverse direction. 3(2−1) = (6−3), while −2(2−1) = (−42).
  • If A = (1, 2) and B = (5, −1), vector AB is B − A = (4−3). Vector BA is its negative.
  • A vector's magnitude follows Pythagoras: |(34)| = √(32 + 42) = 5. A translation's vector is measured in the coordinate system's units.

Higher — vector geometry and proofs

  • Use bold letters such as a or directed labels such as AB→ for vectors. If OA = a and OB = b, then AB = b − a.
  • The midpoint M of AB has OM = (a + b). If M divides AB in ratio 1 : 2 from A, OM = a + (b − a).
  • Vectors that are non-zero scalar multiples are parallel. To prove three points are collinear, show vectors along two joining segments are scalar multiples and share a point.
  • Worked example: In triangle OAB, midpoints M of OA and N of OB have OM = a and ON = b. Therefore MN = (b − a) = AB, proving MN is parallel to AB and half its length.
    Midpoint vector routes. Triangle OAB with midpoint and labelled vector routes
  • For a parallelogram OABC in order, if OA = a and OC = b, then OB = a + b. Diagonals can be expressed in more than one route to locate their intersection.
  • Direction matters in each route: AB + BC = AC, but AB + CB is different. Draw arrows and keep start/end labels consistent.
  • For an intersection problem, express its position along each line with parameters, equate coefficients of independent vectors and solve the resulting simultaneous equations.

Test yourself

50 questions · Sets of 10 from the selected tier. For fractions, use / when typing; for powers, use superscripts or ^. Follow each question's answer format. These quick checks support revision; practise full written solutions and proofs too.

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M35 M35 mind map: Vectors, Arithmetic, Positions, Proof / ratios. A text version follows.
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Vectors

  • Components: Magnitude and direction; equal components, different positions
  • Length: Magnitude √(x²+y²); (3,4) has length 5

Arithmetic

  • Add / subtract: Component by component; head-to-tail; subtraction reverses
  • Scalar: Scales length; negative reverses direction

Positions

  • Direction: Higher: AB=B−A; if OA=a,OB=b then AB=b−a
  • Routes: Higher: AB+BC=AC; keep arrow start/end labels consistent

Proof / ratios

  • Divide: Higher: Midpoint (a+b)/2; ratio1:2 from A → a+(b−a)/3
  • Parallel / collinear: Higher: Non-zero scalar multiples; collinear shares common point
  • Intersection: Higher: Use two routes; equate independent vector coefficients

Connections

  • Arithmetic → Positions: Component arithmetic agrees with directed routes
  • Positions → Proof / ratios: Scalar multiples prove parallel relationships