Home / IGCSE Maths 0580 / Transformations and vectors / Vector geometry
Cambridge IGCSE · Topic 7.4
IGCSE Maths 0580 / Topic 7 / IGCSE

Prove it with vectors.

Extended. Write every line in a shape as a route made of two basic vectors, and geometry turns into algebra: parallel lines, midpoints, ratios and points in a straight line all drop out.

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The key idea

To find any vector, follow a route of known vectors: \(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b}\). Two vectors are parallel when one is a number times the other; three points are collinear when two such vectors are parallel AND share a point.

O A B M a b b − a AB = b − a AM = ½(b − a) OM = a + ½(b − a) = ½a + ½b
Fig. 1: with OA = a and OB = b, the route from A to B is b - a, and the midpoint M of AB has position vector (a + b)/2
Section 01

Position vectors.

In the lab, drag the points \(A\) and \(B\). The position vector of each point is the vector from the origin \(O\) to it, so it has the same numbers as the coordinates. Check that \(\overrightarrow{AB}\), shown in the lab, always equals \(\mathbf{b} - \mathbf{a}\).

Worked example From coordinates
\(A\) is \((2, 5)\) and \(B\) is \((7, -1)\). Find \(\overrightarrow{AB}\) and the position vector of the midpoint \(M\) of \(AB\).
  1. \(\mathbf{a} = \begin{pmatrix}2\\5\end{pmatrix}\), \(\mathbf{b} = \begin{pmatrix}7\\-1\end{pmatrix}\).
  2. \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix}5\\-6\end{pmatrix}\).
  3. \(\overrightarrow{OM} = \dfrac{1}{2}(\mathbf{a} + \mathbf{b}) = \dfrac{1}{2}\begin{pmatrix}9\\4\end{pmatrix} = \begin{pmatrix}4.5\\2\end{pmatrix}\).
Answer: \(\overrightarrow{AB} = \begin{pmatrix}5\\-6\end{pmatrix}\); \(M\) has position vector \(\begin{pmatrix}4.5\\2\end{pmatrix}\), so \(M\) is \((4.5, 2)\).
Examiner trap

Writing \(\overrightarrow{AB} = \mathbf{a} - \mathbf{b}\). The route from \(A\) to \(B\) goes BACK along \(\mathbf{a}\) (that is \(-\mathbf{a}\)) and then out along \(\mathbf{b}\): \(-\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}\).

Section 02

Routes through a shape.

Most questions give two vectors, \(\mathbf{a}\) and \(\mathbf{b}\), on a diagram and ask for other vectors in terms of them. Pick a route along lines you know, going backwards along a vector means subtracting it, and a ratio along a line means a fraction of that line's vector.

Worked example Midpoint of a side
In triangle \(OAB\), \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(M\) is the midpoint of \(AB\). Find \(\overrightarrow{AM}\) and \(\overrightarrow{OM}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
  1. \(\overrightarrow{AB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}\).
  2. \(\overrightarrow{AM} = \dfrac{1}{2}\overrightarrow{AB} = \dfrac{1}{2}(\mathbf{b} - \mathbf{a})\).
  3. \(\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} = \mathbf{a} + \dfrac{1}{2}\mathbf{b} - \dfrac{1}{2}\mathbf{a}\).
Answer: \(\overrightarrow{AM} = \dfrac{1}{2}(\mathbf{b} - \mathbf{a})\), \(\overrightarrow{OM} = \dfrac{1}{2}\mathbf{a} + \dfrac{1}{2}\mathbf{b}\) (Fig. 1)
Worked example A point dividing a line in a ratio
With \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\), \(P\) lies on \(AB\) with \(AP : PB = 1 : 3\). Find \(\overrightarrow{OP}\).
  1. \(AB\) is split into \(1 + 3 = 4\) equal parts, and \(AP\) is 1 of them: \(\overrightarrow{AP} = \dfrac{1}{4}(\mathbf{b} - \mathbf{a})\).
  2. \(\overrightarrow{OP} = \mathbf{a} + \dfrac{1}{4}(\mathbf{b} - \mathbf{a})\).
  3. \(= \mathbf{a} - \dfrac{1}{4}\mathbf{a} + \dfrac{1}{4}\mathbf{b}\).
Answer: \(\overrightarrow{OP} = \dfrac{3}{4}\mathbf{a} + \dfrac{1}{4}\mathbf{b}\)
Worked example Inside a parallelogram
\(OACB\) is a parallelogram with \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(X\) is on the diagonal \(OC\) with \(OX : XC = 1 : 2\). Find \(\overrightarrow{AX}\).
  1. Opposite sides of a parallelogram are equal vectors, so \(\overrightarrow{AC} = \mathbf{b}\) and \(\overrightarrow{OC} = \mathbf{a} + \mathbf{b}\).
  2. \(\overrightarrow{OX} = \dfrac{1}{3}(\mathbf{a} + \mathbf{b})\).
  3. \(\overrightarrow{AX} = \overrightarrow{AO} + \overrightarrow{OX} = -\mathbf{a} + \dfrac{1}{3}\mathbf{a} + \dfrac{1}{3}\mathbf{b}\).
Answer: \(\overrightarrow{AX} = -\dfrac{2}{3}\mathbf{a} + \dfrac{1}{3}\mathbf{b}\)
Your turn
With \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\), \(Q\) lies on \(AB\) with \(AQ : QB = 2 : 1\). Find \(\overrightarrow{OQ}\).
Examiner trap

Turning a ratio straight into a fraction. \(AP : PB = 1 : 3\) means \(AP\) is \(\dfrac{1}{4}\) of \(AB\), not \(\dfrac{1}{3}\): add the parts of the ratio to find the denominator.

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Four quick checks. Each correct answer earns XP and records this skill on your profile.

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\(A\) is the point \((3, -2)\). What is its position vector \(\overrightarrow{OA}\)?

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\(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). What is \(\overrightarrow{AB}\)?

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\(M\) is the midpoint of \(AB\), where \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). What is \(\overrightarrow{OM}\)?

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How are the vectors \(2\mathbf{a} - 6\mathbf{b}\) and \(\mathbf{a} - 3\mathbf{b}\) related?

Section 03

Proving with vectors.

In the lab, drag the heads of \(\mathbf{a}\) and \(\mathbf{b}\) and watch the proof hold for every shape. Parallel lines show up as vectors that are multiples of each other; if the parallel vectors also share a point, the points lie on one straight line.

Worked example Showing lines are parallel
\(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\), \(\overrightarrow{OC} = 3\mathbf{a}\) and \(\overrightarrow{OD} = 3\mathbf{b}\). Show that \(CD\) is parallel to \(AB\) and find the ratio of their lengths.
  1. \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\).
  2. \(\overrightarrow{CD} = -3\mathbf{a} + 3\mathbf{b} = 3(\mathbf{b} - \mathbf{a})\).
  3. So \(\overrightarrow{CD} = 3\overrightarrow{AB}\).
Answer: \(CD\) is parallel to \(AB\) and three times as long, so \(CD : AB = 3 : 1\). (Triangles \(OAB\) and \(OCD\) are similar.)
Worked example Showing points are collinear
\(\overrightarrow{OP} = \mathbf{a} + 2\mathbf{b}\), \(\overrightarrow{OQ} = 3\mathbf{a} + 4\mathbf{b}\) and \(\overrightarrow{OR} = 4\mathbf{a} + 5\mathbf{b}\). Show that \(P\), \(Q\) and \(R\) are collinear.
  1. \(\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = 2\mathbf{a} + 2\mathbf{b} = 2(\mathbf{a} + \mathbf{b})\).
  2. \(\overrightarrow{QR} = \overrightarrow{OR} - \overrightarrow{OQ} = \mathbf{a} + \mathbf{b}\).
  3. \(\overrightarrow{PQ} = 2\overrightarrow{QR}\), so \(PQ\) is parallel to \(QR\), and both pass through \(Q\).
Answer: \(P\), \(Q\), \(R\) lie on one straight line, with \(PQ : QR = 2 : 1\).
Your turn
Use vectors to show that \(A(1, 2)\), \(B(3, 5)\) and \(C(7, 11)\) are collinear.
Examiner trap

Stopping at 'parallel'. Two parallel vectors could lie on two different lines. For collinear you must also say that the vectors share a common point.

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\(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(P\) lies on \(AB\) with \(AP : PB = 1 : 3\). Find \(\overrightarrow{OP}\).

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\(A\) is \((2, 5)\) and \(B\) is \((8, -1)\). What is the position vector of the midpoint of \(AB\)?

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\(\overrightarrow{OP} = \mathbf{a} + 2\mathbf{b}\), \(\overrightarrow{OQ} = 3\mathbf{a} + 4\mathbf{b}\) and \(\overrightarrow{OR} = 4\mathbf{a} + 5\mathbf{b}\). The points are collinear. Find the ratio \(PQ : QR\).

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\(OACB\) is a parallelogram with \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(X\) lies on \(OC\) with \(OX : XC = 1 : 2\). Find \(\overrightarrow{AX}\).

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Six original Cambridge-style questions
Position vectors, routes through triangles, points dividing lines in a ratio, parallel and collinear proofs, and the crossbar of a sign frame.