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IGCSE · Practice questions · Vector geometry

Routes, ratios, proofs.

Extended. Six original Cambridge-style questions on vector geometry, from position vectors to a proof about a sign frame.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
The method

Follow a route of known vectors.

01
Recall
[2 marks]

\(A\) is the point \((-1, 4)\) and \(B\) is the point \((5, 2)\).

(a) Write down the position vector of \(A\). (b) Find \(\overrightarrow{AB}\).

  • (a) \(\begin{pmatrix}-1\\4\end{pmatrix}\) ✓
  • (b) \(\mathbf{b} - \mathbf{a} = \begin{pmatrix}5 - (-1)\\2 - 4\end{pmatrix} = \begin{pmatrix}6\\-2\end{pmatrix}\) ✓
02
Calculation
[2 marks]

In triangle \(OAB\), \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(M\) is the midpoint of \(OB\).

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

  • \(\overrightarrow{OM} = \dfrac{1}{2}\mathbf{b}\). ✓
  • \(\overrightarrow{AM} = \overrightarrow{AO} + \overrightarrow{OM} = -\mathbf{a} + \dfrac{1}{2}\mathbf{b}\). ✓
03
Calculation
[3 marks]

\(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(P\) lies on \(AB\) with \(AP : PB = 2 : 1\).

Find \(\overrightarrow{OP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\), in its simplest form.

  • \(\overrightarrow{AP} = \dfrac{2}{3}\overrightarrow{AB} = \dfrac{2}{3}(\mathbf{b} - \mathbf{a})\). ✓
  • \(\overrightarrow{OP} = \mathbf{a} + \dfrac{2}{3}\mathbf{b} - \dfrac{2}{3}\mathbf{a}\). ✓
  • \(\overrightarrow{OP} = \dfrac{1}{3}\mathbf{a} + \dfrac{2}{3}\mathbf{b}\). ✓
04
Proof
[3 marks]

\(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\), \(\overrightarrow{OC} = 2\mathbf{a}\) and \(\overrightarrow{OD} = 2\mathbf{b}\).

Show that \(CD\) is parallel to \(AB\) and twice as long.

  • \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\). ✓
  • \(\overrightarrow{CD} = -2\mathbf{a} + 2\mathbf{b} = 2(\mathbf{b} - \mathbf{a}) = 2\overrightarrow{AB}\). ✓
  • \(\overrightarrow{CD}\) is a multiple of \(\overrightarrow{AB}\), so \(CD\) is parallel to \(AB\), and the multiplier 2 means it is twice as long. ✓
05
Proof
[3 marks]

\(\overrightarrow{OP} = \mathbf{a} + \mathbf{b}\), \(\overrightarrow{OQ} = 2\mathbf{a} + 3\mathbf{b}\) and \(\overrightarrow{OR} = 4\mathbf{a} + 7\mathbf{b}\).

Show that \(P\), \(Q\) and \(R\) are collinear, and find \(PQ : QR\).

  • \(\overrightarrow{PQ} = (2\mathbf{a} + 3\mathbf{b}) - (\mathbf{a} + \mathbf{b}) = \mathbf{a} + 2\mathbf{b}\). ✓
  • \(\overrightarrow{QR} = (4\mathbf{a} + 7\mathbf{b}) - (2\mathbf{a} + 3\mathbf{b}) = 2\mathbf{a} + 4\mathbf{b} = 2\overrightarrow{PQ}\). ✓
  • \(PQ\) and \(QR\) are parallel and share the point \(Q\), so \(P\), \(Q\), \(R\) are collinear. \(PQ : QR = 1 : 2\). ✓
06
Analysis
[4 marks]

The frame of a sign is a triangle \(OAB\) with \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). A crossbar joins \(M\), the midpoint of \(OA\), to \(N\), the midpoint of \(OB\).

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

(b) Show that the crossbar is parallel to \(AB\).

(c) The side \(AB\) is 1.8 m long. How long is the crossbar?

  • (a) \(\overrightarrow{MN} = \overrightarrow{MO} + \overrightarrow{ON} = -\dfrac{1}{2}\mathbf{a} + \dfrac{1}{2}\mathbf{b}\). ✓
  • (b) \(\overrightarrow{MN} = \dfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \dfrac{1}{2}\overrightarrow{AB}\), a multiple of \(\overrightarrow{AB}\), so \(MN\) is parallel to \(AB\). ✓
  • (c) \(MN\) is half as long as \(AB\): \(1.8 \div 2 = 0.9\) m. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.