IGCSE · Practice questions · Vector geometry: position vectors and ratios

Equate the coefficients.

Six original Cambridge-style questions on vector geometry: paths through a diagram, dividing a line in a ratio, collinear points, a parallelogram proof and two intersecting-lines problems solved with \(\lambda\) and \(\mu\).

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

Name the base vectors, walk the paths, then equate.

Paths: \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}\); going against an arrow changes the sign.
Ratios: \(AP : PB = m : n\) means \(\overrightarrow{AP} = \dfrac{m}{m + n}\overrightarrow{AB}\).
Collinear: one vector is a multiple of the other AND they share a point; say both.
Two lines meet: write \(\overrightarrow{OP}\) along each line with \(\lambda\) and \(\mu\), starting from a point on that line.
Equate the coefficients of \(\mathbf{a}\) and of \(\mathbf{b}\) (they are not parallel), solve, then turn \(\lambda\) into a ratio \(\lambda : (1 - \lambda)\).

Recall

[2 marks]

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

(a) \(\overrightarrow{AB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}\). ✓
(b) \(\overrightarrow{OM} = \mathbf{a} + \dfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})\). ✓

Calculation

[3 marks]

The points \(A\) and \(B\) have position vectors \(\begin{pmatrix} 3 \\ -2 \end{pmatrix}\) and \(\begin{pmatrix} 9 \\ 7 \end{pmatrix}\). The point \(P\) lies on \(AB\) with \(AP : PB = 1 : 2\). Find the position vector of \(P\) and the distance \(OP\).

\(\overrightarrow{AB} = \begin{pmatrix} 9 \\ 7 \end{pmatrix} - \begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 6 \\ 9 \end{pmatrix}\). ✓
\(\overrightarrow{OP} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} + \dfrac{1}{3}\begin{pmatrix} 6 \\ 9 \end{pmatrix} = \begin{pmatrix} 5 \\ 1 \end{pmatrix}\). ✓
\(OP = \sqrt{5^{2} + 1^{2}} = \sqrt{26}\) (about 5.10). ✓

Proof

[3 marks]

\(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\) and \(\overrightarrow{OC} = 4\mathbf{b} - 3\mathbf{a}\). Show that \(A\), \(B\) and \(C\) are collinear and find the ratio \(AB : BC\).

\(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\). ✓
\(\overrightarrow{BC} = (4\mathbf{b} - 3\mathbf{a}) - \mathbf{b} = 3\mathbf{b} - 3\mathbf{a} = 3\,\overrightarrow{AB}\). ✓
\(AB\) and \(BC\) are parallel and share the point \(B\), so \(A\), \(B\), \(C\) are collinear, with \(AB : BC = 1 : 3\). ✓

Proof

[4 marks]

\(OABC\) is any quadrilateral with \(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\) and \(\overrightarrow{OC} = \mathbf{c}\). \(P\), \(Q\), \(R\) and \(S\) are the midpoints of \(OA\), \(AB\), \(BC\) and \(CO\). Show that \(PQRS\) is a parallelogram.

\(\overrightarrow{OP} = \dfrac{1}{2}\mathbf{a}\), \(\overrightarrow{OQ} = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})\), \(\overrightarrow{OR} = \dfrac{1}{2}(\mathbf{b} + \mathbf{c})\), \(\overrightarrow{OS} = \dfrac{1}{2}\mathbf{c}\). ✓
\(\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \dfrac{1}{2}\mathbf{b}\). ✓
\(\overrightarrow{SR} = \overrightarrow{OR} - \overrightarrow{OS} = \dfrac{1}{2}\mathbf{b}\). ✓
\(\overrightarrow{PQ} = \overrightarrow{SR}\): one pair of opposite sides is parallel and equal in length, so \(PQRS\) is a parallelogram. ✓

Analysis

[5 marks]

\(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(C\) lies on \(OA\) with \(OC : CA = 1 : 2\) and \(D\) is the midpoint of \(AB\). The lines \(OD\) and \(BC\) meet at \(P\).

(a) Given \(\overrightarrow{OP} = \lambda\,\overrightarrow{OD}\), write \(\overrightarrow{OP}\) in terms of \(\lambda\), \(\mathbf{a}\) and \(\mathbf{b}\).

(b) Given \(\overrightarrow{BP} = \mu\,\overrightarrow{BC}\), write \(\overrightarrow{OP}\) in terms of \(\mu\), \(\mathbf{a}\) and \(\mathbf{b}\).

(a) \(\overrightarrow{OD} = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})\), so \(\overrightarrow{OP} = \dfrac{1}{2}\lambda\,\mathbf{a} + \dfrac{1}{2}\lambda\,\mathbf{b}\). ✓
(b) \(\overrightarrow{BC} = \dfrac{1}{3}\mathbf{a} - \mathbf{b}\), so \(\overrightarrow{OP} = \mathbf{b} + \mu\left(\dfrac{1}{3}\mathbf{a} - \mathbf{b}\right) = \dfrac{1}{3}\mu\,\mathbf{a} + (1 - \mu)\,\mathbf{b}\). ✓
(c) \(\mathbf{a}\): \(\dfrac{1}{2}\lambda = \dfrac{1}{3}\mu\), so \(\mu = \dfrac{3}{2}\lambda\). \(\mathbf{b}\): \(\dfrac{1}{2}\lambda = 1 - \dfrac{3}{2}\lambda\), so \(2\lambda = 1\). ✓
\(\lambda = \dfrac{1}{2}\), \(\mu = \dfrac{3}{4}\), and \(\overrightarrow{OP} = \dfrac{1}{4}\mathbf{a} + \dfrac{1}{4}\mathbf{b}\). ✓
\(BP : PC = \dfrac{3}{4} : \dfrac{1}{4} = 3 : 1\). ✓

Analysis

[6 marks]

In triangle \(OAB\), \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(Y\) lies on \(AB\) with \(AY : YB = 1 : 3\) and \(X\) is the midpoint of \(OB\). The lines \(OY\) and \(AX\) meet at \(P\).

(a) Find \(\overrightarrow{OY}\) and \(\overrightarrow{AX}\).

(b) By writing \(\overrightarrow{OP} = \lambda\,\overrightarrow{OY}\) and \(\overrightarrow{AP} = \mu\,\overrightarrow{AX}\), find \(\lambda\) and \(\mu\).

(a) \(\overrightarrow{OY} = \mathbf{a} + \dfrac{1}{4}(\mathbf{b} - \mathbf{a}) = \dfrac{3}{4}\mathbf{a} + \dfrac{1}{4}\mathbf{b}\); \(\overrightarrow{AX} = \dfrac{1}{2}\mathbf{b} - \mathbf{a}\). ✓
(b) Route 1: \(\overrightarrow{OP} = \dfrac{3}{4}\lambda\,\mathbf{a} + \dfrac{1}{4}\lambda\,\mathbf{b}\). Route 2: \(\overrightarrow{OP} = \mathbf{a} + \mu\left(\dfrac{1}{2}\mathbf{b} - \mathbf{a}\right) = (1 - \mu)\,\mathbf{a} + \dfrac{1}{2}\mu\,\mathbf{b}\). ✓
\(\mathbf{b}\): \(\dfrac{1}{4}\lambda = \dfrac{1}{2}\mu\), so \(\lambda = 2\mu\). \(\mathbf{a}\): \(\dfrac{3}{4}(2\mu) = 1 - \mu\), so \(\dfrac{5}{2}\mu = 1\). ✓
\(\mu = \dfrac{2}{5}\) and \(\lambda = \dfrac{4}{5}\). ✓
(c) \(\overrightarrow{OP} = \dfrac{4}{5}\left(\dfrac{3}{4}\mathbf{a} + \dfrac{1}{4}\mathbf{b}\right) = \dfrac{3}{5}\mathbf{a} + \dfrac{1}{5}\mathbf{b}\). ✓
\(OP : PY = \dfrac{4}{5} : \dfrac{1}{5} = 4 : 1\) and \(AP : PX = \dfrac{2}{5} : \dfrac{3}{5} = 2 : 3\). ✓

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.