GCSE · Vectors

Vectors on a shapeExam-style questions

Expressing any journey around a figure in terms of two given vectors — why every route gives the same answer, why going against an arrow flips the sign, and why a point splitting a side in the ratio 2:3 sits two fifths along, not two thirds.

12 questions · GCSE (Higher) level · mark scheme and worked solution on each · no sign-up to try

DifficultyAllLevel 2Level 3Level 4Level 5
  1. 1
    PQRSPQRS is a parallelogram.

    PQ→=a\overrightarrow{PQ} = \mathbf{a} and PS→=b\overrightarrow{PS} = \mathbf{b}.abPQRSThe diagram is a sketch and is not accurately drawn.

    (a) Write down, in terms of a\mathbf{a} and/or b\mathbf{b}, the vector SR→\overrightarrow{SR}.

    (b) Give a reason for your answer to part (a).

    (c) Find QS→\overrightarrow{QS} in terms of a\mathbf{a} and b\mathbf{b}.
    Level 2 · 4 marksTry it
  2. 2
    OABOAB is a triangle. OA→=a\overrightarrow{OA} = \mathbf{a} and OB→=b\overrightarrow{OB} = \mathbf{b}.

    MM is the midpoint of ABAB.abOABMThe diagram is a sketch and is not accurately drawn.

    (a) Find AB→\overrightarrow{AB} in terms of a\mathbf{a} and b\mathbf{b}.

    (b) Find OM→\overrightarrow{OM} in terms of a\mathbf{a} and b\mathbf{b}, giving your answer in its simplest form.
    Level 3 · 4 marksTry it
  3. 3
    ABCDABCD is a trapezium. AB→=a\overrightarrow{AB} = \mathbf{a} and BC→=b\overrightarrow{BC} = \mathbf{b}.

    ADAD is parallel to BCBC, and AD=3×BCAD = 3 \times BC.abABCDThe diagram is a sketch and is not accurately drawn.

    (a) Write down AD→\overrightarrow{AD} in terms of b\mathbf{b}.

    (b) Find DC→\overrightarrow{DC} in terms of a\mathbf{a} and b\mathbf{b}, giving your answer in its simplest form.
    Level 3 · 4 marksTry it
  4. 4
    ABCDABCD is a quadrilateral.

    AB→=a\overrightarrow{AB} = \mathbf{a}, BC→=b\overrightarrow{BC} = \mathbf{b} and CD→=2b−a\overrightarrow{CD} = 2\mathbf{b} - \mathbf{a}.ab2b − aABCDThe diagram is a sketch and is not accurately drawn.

    (a) Find AD→\overrightarrow{AD} in terms of a\mathbf{a} and b\mathbf{b}, giving your answer in its simplest form.

    (b) Find DB→\overrightarrow{DB} in terms of a\mathbf{a} and b\mathbf{b}, giving your answer in its simplest form.
    Level 3 · 4 marksTry it
  5. 5
    OABOAB is a triangle. OA→=a\overrightarrow{OA} = \mathbf{a} and OB→=b\overrightarrow{OB} = \mathbf{b}.

    MM is the midpoint of ABAB.

    (a) Write AB→\overrightarrow{AB} in terms of a\mathbf{a} and b\mathbf{b}. (1 mark)

    (b) Write OM→\overrightarrow{OM} in terms of a\mathbf{a} and b\mathbf{b}. Give your answer in its simplest form. (2 marks)
    Level 3 · 3 marksTry it

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