Home / IGCSE Add Maths 0606 / Vectors in two dimensions / Vector geometry: position vectors and ratios
Cambridge IGCSE · Topic 13.3
IGCSE Add Maths 0606 / Topic 13 / IGCSE

Two routes to one point, one answer.

Once every point in a diagram is written in terms of two base vectors \(\mathbf{a}\) and \(\mathbf{b}\), geometry turns into algebra. Walk along the lines of the figure, split segments in a given ratio, and when two lines cross, reach the crossing point by two different routes and compare the coefficients.

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The key idea

Write every vector as a path through known points: \(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = \mathbf{b} - \mathbf{a}\). If \(P\) divides \(AB\) in the ratio \(m : n\), then \(\overrightarrow{OP} = \mathbf{a} + \dfrac{m}{m + n}(\mathbf{b} - \mathbf{a})\). Two vectors are parallel when one is a scalar multiple of the other; three points are collinear when, in addition, the two vectors share a point. Because \(\mathbf{a}\) and \(\mathbf{b}\) are not parallel, \(p\mathbf{a} + q\mathbf{b} = r\mathbf{a} + s\mathbf{b}\) forces \(p = r\) and \(q = s\).

O A B X Y P a b OY = YA, so OY = ½a OX : XB = 2 : 1, so OX = ⅔b AX and BY meet at P OP = a + λAX = b + μBY λ = ¾, μ = ½ AP : PX = 3 : 1
Fig. 1: triangle OAB with OA = a and OB = b. Y is the midpoint of OA and X is the point on OB with OX : XB = 2 : 1. The lines AX and BY meet at P; writing OP along each line and equating coefficients gives λ = 3/4 and μ = 1/2
Section 01

Walk the path.

Pick an origin \(O\) and call \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\): these are the position vectors of \(A\) and \(B\). Any other vector in the diagram is a route made of known steps. To go from \(A\) to \(B\), go back to \(O\) and then out to \(B\): \(\overrightarrow{AB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}\). Going against an arrow changes the sign.

If \(P\) lies on \(AB\) with \(AP : PB = m : n\), then \(P\) is \(\dfrac{m}{m + n}\) of the way from \(A\) to \(B\):

\[\overrightarrow{OP} = \mathbf{a} + \frac{m}{m + n}(\mathbf{b} - \mathbf{a}) = \frac{n}{m + n}\,\mathbf{a} + \frac{m}{m + n}\,\mathbf{b}\]

For the midpoint \(M\) of \(AB\) (ratio \(1 : 1\)) this gives \(\overrightarrow{OM} = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})\). The same rule works with column vectors.

Worked example A point one quarter of the way
\(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). The point \(P\) lies on \(AB\) with \(AP : PB = 1 : 3\). Find \(\overrightarrow{AB}\) and \(\overrightarrow{OP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
  1. \(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}\).
  2. \(AP : PB = 1 : 3\), so \(AP\) is \(\dfrac{1}{1 + 3} = \dfrac{1}{4}\) of \(AB\): \(\overrightarrow{AP} = \dfrac{1}{4}(\mathbf{b} - \mathbf{a})\).
  3. \(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = \mathbf{a} + \dfrac{1}{4}\mathbf{b} - \dfrac{1}{4}\mathbf{a}\).
Answer: \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\), \(\overrightarrow{OP} = \dfrac{3}{4}\mathbf{a} + \dfrac{1}{4}\mathbf{b}\)
Worked example Column vectors
The points \(A\) and \(B\) have position vectors \(\begin{pmatrix} 2 \\ -1 \end{pmatrix}\) and \(\begin{pmatrix} 8 \\ 5 \end{pmatrix}\). Find the position vector of the point \(P\) on \(AB\) with \(AP : PB = 2 : 1\), and of the midpoint \(M\) of \(AB\).
  1. \(\overrightarrow{AB} = \begin{pmatrix} 8 \\ 5 \end{pmatrix} - \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 6 \end{pmatrix}\).
  2. \(P\) is \(\dfrac{2}{3}\) of the way along: \(\overrightarrow{OP} = \begin{pmatrix} 2 \\ -1 \end{pmatrix} + \dfrac{2}{3}\begin{pmatrix} 6 \\ 6 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \end{pmatrix} + \begin{pmatrix} 4 \\ 4 \end{pmatrix}\).
  3. \(\overrightarrow{OM} = \dfrac{1}{2}\left[\begin{pmatrix} 2 \\ -1 \end{pmatrix} + \begin{pmatrix} 8 \\ 5 \end{pmatrix}\right] = \dfrac{1}{2}\begin{pmatrix} 10 \\ 4 \end{pmatrix}\).
Answer: \(\overrightarrow{OP} = \begin{pmatrix} 6 \\ 3 \end{pmatrix}\), \(\overrightarrow{OM} = \begin{pmatrix} 5 \\ 2 \end{pmatrix}\)
Your turn
\(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\) and \(C\) lies on \(AB\) with \(AC : CB = 2 : 3\). Find \(\overrightarrow{OC}\).
Examiner trap

The fraction comes from the ratio's TOTAL: \(AP : PB = 1 : 3\) means \(AP = \dfrac{1}{4}AB\), not \(\dfrac{1}{3}AB\). And \(\overrightarrow{AB}\) is \(\mathbf{b} - \mathbf{a}\) (end minus start), not \(\mathbf{a} - \mathbf{b}\).

Section 02

Parallel, collinear, shaped.

Vectors that are scalar multiples of each other, such as \(2\mathbf{a} - \mathbf{b}\) and \(-6\mathbf{a} + 3\mathbf{b}\), point along parallel lines; the scalar also compares their lengths. Parallel alone is not enough for points to lie on one line: the lines could be side by side. Three points \(A\), \(B\), \(C\) are collinear when \(\overrightarrow{AB} = k\,\overrightarrow{BC}\) AND the two vectors share the point \(B\).

Worked example Show three points are collinear
\(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\) and \(\overrightarrow{OC} = 3\mathbf{b} - 2\mathbf{a}\). Show that \(A\), \(B\) and \(C\) are collinear and find the ratio \(AB : BC\).
  1. \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\).
  2. \(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = 3\mathbf{b} - 2\mathbf{a} - \mathbf{b} = 2\mathbf{b} - 2\mathbf{a} = 2(\mathbf{b} - \mathbf{a})\).
  3. So \(\overrightarrow{BC} = 2\,\overrightarrow{AB}\): the lines \(AB\) and \(BC\) are parallel, and they share the point \(B\), so \(A\), \(B\), \(C\) lie on one straight line.
  4. \(BC\) is twice as long as \(AB\).
Answer: Collinear, with \(AB : BC = 1 : 2\)
Worked example Name the quadrilateral
\(\overrightarrow{OA} = 4\mathbf{a}\), \(\overrightarrow{OB} = 4\mathbf{a} + 2\mathbf{b}\) and \(\overrightarrow{OC} = 2\mathbf{a} + 2\mathbf{b}\). Show that \(OABC\) is a trapezium.
  1. Sides in order: \(\overrightarrow{OA} = 4\mathbf{a}\), \(\overrightarrow{AB} = (4\mathbf{a} + 2\mathbf{b}) - 4\mathbf{a} = 2\mathbf{b}\).
  2. \(\overrightarrow{CB} = (4\mathbf{a} + 2\mathbf{b}) - (2\mathbf{a} + 2\mathbf{b}) = 2\mathbf{a}\), and \(\overrightarrow{OC} = 2\mathbf{a} + 2\mathbf{b}\).
  3. \(\overrightarrow{OA} = 2\,\overrightarrow{CB}\), so \(OA\) is parallel to \(CB\) and twice as long.
  4. \(\overrightarrow{OC} = 2\mathbf{a} + 2\mathbf{b}\) is not a multiple of \(\overrightarrow{AB} = 2\mathbf{b}\), so the other pair of sides is not parallel.
Answer: Exactly one pair of parallel sides (\(OA\) and \(CB\), with \(OA = 2CB\)), so \(OABC\) is a trapezium
Your turn
\(\mathbf{a}\) and \(\mathbf{b}\) are not parallel. Find \(k\) if \((k - 1)\mathbf{a} + 4\mathbf{b}\) is parallel to \(\mathbf{a} + 2\mathbf{b}\).
Examiner trap

Showing \(\overrightarrow{BC} = 2\,\overrightarrow{AB}\) proves only that the lines are parallel. For collinear you must also state the common point \(B\); examiners withhold the mark without it.

Stage 1 · Learn

Check your understanding

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

Quick check+10 XP

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

Quick check+10 XP

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

Quick check+10 XP

\(P\) lies on \(AB\) with \(AP : PB = 2 : 3\). Find \(\overrightarrow{OP}\) in terms of \(\mathbf{a} = \overrightarrow{OA}\) and \(\mathbf{b} = \overrightarrow{OB}\).

Quick check+10 XP

\(\overrightarrow{AB} = 2\mathbf{a} - \mathbf{b}\). Which \(\overrightarrow{BC}\) makes \(A\), \(B\), \(C\) collinear?

Section 03

Where two lines meet.

The classic question: two lines cross at \(P\) and you must locate \(P\). Write \(\overrightarrow{OP}\) twice, once along each line, using an unknown scalar for each (\(\lambda\) and \(\mu\)). Both expressions describe the same vector, and \(\mathbf{a}\) and \(\mathbf{b}\) are not parallel, so the coefficients of \(\mathbf{a}\) must match and so must the coefficients of \(\mathbf{b}\). That gives two simultaneous equations.

For \(P\) on the line from \(A\) to \(X\), \(\overrightarrow{OP} = \overrightarrow{OA} + \lambda\,\overrightarrow{AX}\), where \(\lambda\) is the fraction of the way from \(A\) to \(X\). So \(AP : PX = \lambda : (1 - \lambda)\).

Worked example Two lines in a triangle (Fig. 1)
In triangle \(OAB\), \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(Y\) is the midpoint of \(OA\) and \(X\) lies on \(OB\) with \(OX : XB = 2 : 1\). The lines \(AX\) and \(BY\) meet at \(P\). Find \(\overrightarrow{OP}\) and the ratio \(AP : PX\).
  1. \(\overrightarrow{OX} = \dfrac{2}{3}\mathbf{b}\) and \(\overrightarrow{OY} = \dfrac{1}{2}\mathbf{a}\), so \(\overrightarrow{AX} = \dfrac{2}{3}\mathbf{b} - \mathbf{a}\) and \(\overrightarrow{BY} = \dfrac{1}{2}\mathbf{a} - \mathbf{b}\).
  2. Along \(AX\): \[\overrightarrow{OP} = \mathbf{a} + \lambda\left(\tfrac{2}{3}\mathbf{b} - \mathbf{a}\right) = (1 - \lambda)\,\mathbf{a} + \tfrac{2}{3}\lambda\,\mathbf{b}\]
  3. Along \(BY\): \[\overrightarrow{OP} = \mathbf{b} + \mu\left(\tfrac{1}{2}\mathbf{a} - \mathbf{b}\right) = \tfrac{1}{2}\mu\,\mathbf{a} + (1 - \mu)\,\mathbf{b}\]
  4. Equate coefficients: \(\mathbf{a}\): \(1 - \lambda = \dfrac{1}{2}\mu\); \(\mathbf{b}\): \(\dfrac{2}{3}\lambda = 1 - \mu\), so \(\mu = 1 - \dfrac{2}{3}\lambda\).
  5. Substitute: \(1 - \lambda = \dfrac{1}{2} - \dfrac{1}{3}\lambda\), so \(\dfrac{1}{2} = \dfrac{2}{3}\lambda\), giving \(\lambda = \dfrac{3}{4}\) and \(\mu = \dfrac{1}{2}\).
  6. \(\overrightarrow{OP} = \dfrac{1}{4}\mathbf{a} + \dfrac{1}{2}\mathbf{b}\). Since \(\lambda = \dfrac{3}{4}\), \(P\) is three quarters of the way from \(A\) to \(X\).
Answer: \(\overrightarrow{OP} = \dfrac{1}{4}\mathbf{a} + \dfrac{1}{2}\mathbf{b}\), \(AP : PX = 3 : 1\) (and \(\mu = \frac{1}{2}\), so \(P\) is the midpoint of \(BY\))
Worked example A line from the origin
\(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\). \(C\) lies on \(AB\) with \(AC : CB = 1 : 2\) and \(D\) is the midpoint of \(OA\). The lines \(OC\) and \(BD\) meet at \(Q\). Find \(OQ : QC\).
  1. \(\overrightarrow{OC} = \mathbf{a} + \dfrac{1}{3}(\mathbf{b} - \mathbf{a}) = \dfrac{2}{3}\mathbf{a} + \dfrac{1}{3}\mathbf{b}\), and \(\overrightarrow{BD} = \dfrac{1}{2}\mathbf{a} - \mathbf{b}\).
  2. \(Q\) on \(OC\): \(\overrightarrow{OQ} = \lambda\,\overrightarrow{OC} = \dfrac{2}{3}\lambda\,\mathbf{a} + \dfrac{1}{3}\lambda\,\mathbf{b}\).
  3. \(Q\) on \(BD\): \(\overrightarrow{OQ} = \mathbf{b} + \mu\left(\dfrac{1}{2}\mathbf{a} - \mathbf{b}\right) = \dfrac{1}{2}\mu\,\mathbf{a} + (1 - \mu)\,\mathbf{b}\).
  4. \(\mathbf{a}\): \(\dfrac{2}{3}\lambda = \dfrac{1}{2}\mu\), so \(\mu = \dfrac{4}{3}\lambda\). \(\mathbf{b}\): \(\dfrac{1}{3}\lambda = 1 - \dfrac{4}{3}\lambda\), so \(\dfrac{5}{3}\lambda = 1\).
  5. \(\lambda = \dfrac{3}{5}\) (and \(\mu = \dfrac{4}{5}\)), so \(\overrightarrow{OQ} = \dfrac{3}{5}\,\overrightarrow{OC} = \dfrac{2}{5}\mathbf{a} + \dfrac{1}{5}\mathbf{b}\).
Answer: \(OQ : QC = 3 : 2\)
Examiner trap

\(\lambda = \dfrac{3}{4}\) gives \(AP : PX = 3 : 1\), not \(3 : 4\): the ratio compares the two PARTS, \(\lambda\) and \(1 - \lambda\). Also start each route at a point on its own line (\(A\) for line \(AX\), \(B\) for line \(BY\)), or the expression does not describe that line.

Stage 2 · Exam

Exam-style questions

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Exam style+20 XP

In triangle \(OAB\), \(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\), \(Y\) is the midpoint of \(OA\) and \(X\) is on \(OB\) with \(OX : XB = 3 : 1\). \(AX\) and \(BY\) meet at \(P\). Find \(AP : PX\).

Exam style+20 XP

\(\mathbf{a}\) and \(\mathbf{b}\) are not parallel and \((\lambda + 1)\mathbf{a} + 2\mu\,\mathbf{b} = 5\mathbf{a} + (\mu + 3)\mathbf{b}\). Find \(\lambda\) and \(\mu\).

Exam style+20 XP

\(\overrightarrow{OA} = 6\mathbf{a}\), \(\overrightarrow{OC} = 2\mathbf{b}\) and \(\overrightarrow{CB} = 3\mathbf{a}\). Which statement about the quadrilateral \(OABC\) is true?

Exam style+20 XP

\(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\) and \(\overrightarrow{OC} = 3\mathbf{a} + k\mathbf{b}\). Given that \(A\), \(B\) and \(C\) are collinear, find \(k\).

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Two routes, one point, equal coefficients.

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Six original Cambridge-style questions
Paths and midpoints, a ratio with column vectors, a collinearity proof, the midpoints of any quadrilateral, and two problems where lines meet and \(\lambda\) and \(\mu\) give the ratios.