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7b · E7.2–E7.4

Vectors

Topic 7 Transformations and vectors · Extended only · Papers 2 and 4

EXTENDED Everything on this page is Extended content only. Core uses column vectors only to describe translations (7a).

🎯What you need to be able to do

  • Use vectors written as \( \begin{pmatrix} x \\ y \end{pmatrix} \), \( \overrightarrow{AB} \) or \( \mathbf{a} \); add, subtract and multiply them by a scalar.
  • Find the magnitude \( \sqrt{x^{2} + y^{2}} \) of a vector.
  • Use position vectors, and express vectors in terms of two given vectors.
  • Use vectors to show that lines are parallel, that three points are collinear, and to solve problems with ratio and similarity.

📚The mathematics

Column vectors

Add and subtract component by component; multiply every component by a scalar:

\( \begin{pmatrix} 3 \\ -4 \end{pmatrix} + \begin{pmatrix} -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \end{pmatrix} \)
\( 2\begin{pmatrix} 3 \\ -4 \end{pmatrix} = \begin{pmatrix} 6 \\ -8 \end{pmatrix} \)
\( \left|\begin{pmatrix} 3 \\ -4 \end{pmatrix}\right| = \sqrt{3^{2} + 4^{2}} = 5 \)

The vector from \(A\) to \(B\) is \( \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} \): “end minus start”. The position vector of a point is its vector from the origin.

Vector geometry

To find a vector in terms of \( \mathbf{a} \) and \( \mathbf{c} \), walk along known edges from start to finish, adding a vector for each step (and subtracting when you walk against an arrow). For a point dividing \(AB\) in the ratio \( m : n \), go from \(A\) a fraction \( \tfrac{m}{m + n} \) of the way along \( \overrightarrow{AB} \).

  • Parallel: one vector is a scalar multiple of the other, e.g. \( 4\mathbf{a} - 2\mathbf{c} = 2(2\mathbf{a} - \mathbf{c}) \).
  • Collinear (on one straight line): \( \overrightarrow{PQ} \) and \( \overrightarrow{QR} \) are parallel and share the point \(Q\). Say both in the answer.
Parallelogram OABC with OA = a along the bottom and OC = c up the left side. M is the midpoint of AB, and P lies on the diagonal OB with OP to PB equal to 2 to 1. The points C, P and M lie on one straight dashed line, with CP two thirds of CM.
The worked example: \( \overrightarrow{CM} = \tfrac{3}{2}\overrightarrow{CP} \), so \(C\), \(P\) and \(M\) are collinear.

✏️Worked example (non-calculator)

(a) \( \mathbf{a} = \begin{pmatrix} 3 \\ -4 \end{pmatrix} \) and \( \mathbf{b} = \begin{pmatrix} -1 \\ 2 \end{pmatrix} \). Find \( 2\mathbf{a} - 3\mathbf{b} \) and \( |\mathbf{a}| \). [3] (b) \(OABC\) is a parallelogram with \( \overrightarrow{OA} = \mathbf{a} \) and \( \overrightarrow{OC} = \mathbf{c} \). \(M\) is the midpoint of \(AB\), and \(P\) is on \(OB\) with \( OP : PB = 2 : 1 \). Find \( \overrightarrow{CP} \) and \( \overrightarrow{CM} \) in terms of \( \mathbf{a} \) and \( \mathbf{c} \), and show that \(C\), \(P\) and \(M\) are collinear. [5]

(a) \( \begin{pmatrix} 6 \\ -8 \end{pmatrix} - \begin{pmatrix} -3 \\ 6 \end{pmatrix} = \begin{pmatrix} 9 \\ -14 \end{pmatrix} \); \( |\mathbf{a}| = 5 \).

(b) \( \overrightarrow{OB} = \mathbf{a} + \mathbf{c} \), so \( \overrightarrow{OP} = \tfrac{2}{3}(\mathbf{a} + \mathbf{c}) \) and

\[ \overrightarrow{CP} = \overrightarrow{CO} + \overrightarrow{OP} = -\mathbf{c} + \tfrac{2}{3}\mathbf{a} + \tfrac{2}{3}\mathbf{c} = \tfrac{2}{3}\mathbf{a} - \tfrac{1}{3}\mathbf{c} \]

\( \overrightarrow{AB} = \mathbf{c} \), so \( \overrightarrow{OM} = \mathbf{a} + \tfrac{1}{2}\mathbf{c} \) and \( \overrightarrow{CM} = -\mathbf{c} + \mathbf{a} + \tfrac{1}{2}\mathbf{c} = \mathbf{a} - \tfrac{1}{2}\mathbf{c} \). Then \( \overrightarrow{CM} = \tfrac{3}{2}\left(\tfrac{2}{3}\mathbf{a} - \tfrac{1}{3}\mathbf{c}\right) = \tfrac{3}{2}\overrightarrow{CP} \). The vectors are parallel and share the point \(C\), so \(C\), \(P\) and \(M\) are collinear.

Check it. With \( \mathbf{a} = \begin{pmatrix} 3 \\ 0 \end{pmatrix} \) and \( \mathbf{c} = \begin{pmatrix} 0 \\ 3 \end{pmatrix} \) (a square), \(C = (0, 3)\), \(P = (2, 2)\), \(M = (3, 1.5)\): the slopes \(C\to P\) and \(P\to M\) are both \( -\tfrac{1}{2} \) ✓.
Half an argument. “\( \overrightarrow{CM} \) is a multiple of \( \overrightarrow{CP} \)” only shows the lines are parallel. For collinear you must also say they share a common point.

📝Practise

All Extended, non-calculator style.

1. \( \mathbf{a} = \begin{pmatrix} 2 \\ 5 \end{pmatrix} \) and \( \mathbf{b} = \begin{pmatrix} -3 \\ 1 \end{pmatrix} \). Find \( \mathbf{a} + \mathbf{b} \) and \( 3\mathbf{a} - 2\mathbf{b} \). [3]
\( \begin{pmatrix} -1 \\ 6 \end{pmatrix} \); \( \begin{pmatrix} 6 \\ 15 \end{pmatrix} - \begin{pmatrix} -6 \\ 2 \end{pmatrix} = \begin{pmatrix} 12 \\ 13 \end{pmatrix} \).
2. Find the magnitude of \( \begin{pmatrix} -5 \\ 12 \end{pmatrix} \). [2]
\( \sqrt{25 + 144} = 13 \).
3. \(A\) is \( (1, 2) \) and \(B\) is \( (7, -6) \). Find \( \overrightarrow{AB} \) and \( |\overrightarrow{AB}| \). [3]
\( \begin{pmatrix} 6 \\ -8 \end{pmatrix} \); magnitude \( \sqrt{36 + 64} = 10 \).
4. The vectors \( \begin{pmatrix} k \\ 6 \end{pmatrix} \) and \( \begin{pmatrix} 2 \\ 3 \end{pmatrix} \) are parallel. Find \(k\). [2]
\( \begin{pmatrix} k \\ 6 \end{pmatrix} = 2\begin{pmatrix} 2 \\ 3 \end{pmatrix} \), so \( k = 4 \).
5. \( \overrightarrow{OA} = \mathbf{a} \) and \( \overrightarrow{OB} = \mathbf{b} \). \(X\) lies on \(AB\) with \( AX : XB = 1 : 3 \). Find \( \overrightarrow{OX} \) in terms of \( \mathbf{a} \) and \( \mathbf{b} \), in its simplest form. [3]
\( \overrightarrow{OX} = \mathbf{a} + \tfrac{1}{4}(\mathbf{b} - \mathbf{a}) = \tfrac{3}{4}\mathbf{a} + \tfrac{1}{4}\mathbf{b} \).
6. \( \overrightarrow{OP} = \mathbf{a} + 2\mathbf{b} \), \( \overrightarrow{OQ} = 3\mathbf{a} + 4\mathbf{b} \) and \( \overrightarrow{OR} = 7\mathbf{a} + 8\mathbf{b} \). Show that \(P\), \(Q\) and \(R\) lie on a straight line. [3]
\( \overrightarrow{PQ} = 2\mathbf{a} + 2\mathbf{b} \) and \( \overrightarrow{QR} = 4\mathbf{a} + 4\mathbf{b} = 2\overrightarrow{PQ} \). They are parallel and share the point \(Q\), so \(P\), \(Q\) and \(R\) are collinear.
7. In triangle \(OAB\), \( \overrightarrow{OA} = \mathbf{a} \) and \( \overrightarrow{OB} = \mathbf{b} \). \(N\) lies on \(OB\) with \( ON : NB = 1 : 2 \). Find \( \overrightarrow{AN} \). [2]
\( \overrightarrow{AN} = \overrightarrow{AO} + \overrightarrow{ON} = -\mathbf{a} + \tfrac{1}{3}\mathbf{b} \).

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • GeoGebra — drag the vertices of a parallelogram and watch vector expressions stay true
  • Corbettmaths — vector proof, parallel vectors and collinear points