What you'll learn
Position vectors and geometric proofs form a critical component of the CIE IGCSE Mathematics Extended syllabus, testing your ability to apply vector methods to solve geometric problems algebraically. This topic requires understanding how vectors represent points in space and using vector algebra to prove geometric relationships such as parallelism, collinearity, and ratios of division. Exam questions frequently combine position vectors with geometric reasoning, demanding both computational accuracy and logical proof structure.
Key terms and definitions
Position vector — a vector that describes the location of a point relative to a fixed origin O, typically written as a or $\overrightarrow{OA}$ for point A
Displacement vector — a vector describing the movement from one point to another, independent of origin; for example, $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ where a and b are position vectors of A and B
Collinear points — three or more points that lie on the same straight line, proven when one displacement vector is a scalar multiple of another
Parallel vectors — vectors that are scalar multiples of each other, written as p = kq where k is a scalar constant
Ratio of division — when point P divides line segment AB in ratio m:n, the position vector of P can be expressed as $\mathbf{p} = \frac{n\mathbf{a} + m\mathbf{b}}{m+n}$
Linear combination — expressing a vector as the sum of scalar multiples of other vectors, such as r = λa + μb
Midpoint theorem — the position vector of midpoint M of AB is $\mathbf{m} = \frac{\mathbf{a} + \mathbf{b}}{2}$
Core concepts
Understanding position vectors
Position vectors establish a coordinate system using an origin O as the reference point. Every point in space corresponds to exactly one position vector from the origin.
Key principles:
- The position vector a of point A means $\overrightarrow{OA} = \mathbf{a}$
- Position vectors are written in bold lowercase or with an arrow notation
- Column vector notation represents position vectors: $\mathbf{a} = \begin{pmatrix} x \ y \end{pmatrix}$ in 2D or $\begin{pmatrix} x \ y \ z \end{pmatrix}$ in 3D
- Any point in a plane can be located using position vectors relative to a chosen origin
The fundamental relationship connecting position vectors is:
$$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a}$$
This means the displacement from A to B equals the position vector of B minus the position vector of A.
Converting between position and displacement vectors
CIE IGCSE Mathematics exam questions regularly test the conversion between these vector types.
From position to displacement: Given points A and B with position vectors a and b, the displacement vector is: $$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$$
Finding position vectors along a path: To find the position vector of point P on line segment AB: $$\mathbf{p} = \mathbf{a} + \overrightarrow{AP}$$
where $\overrightarrow{AP}$ is calculated as a fraction of $\overrightarrow{AB}$ based on where P lies along AB.
Multiple displacement journey: For a path O → A → B → C: $$\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AB} + \overrightarrow{BC}$$
This triangle rule application appears frequently in exam contexts involving geometric figures.
Proving collinearity using vectors
Three points A, B, and C are collinear if and only if $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel, meaning one is a scalar multiple of the other.
Method for proving collinearity:
- Calculate $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$
- Calculate $\overrightarrow{AC} = \mathbf{c} - \mathbf{a}$
- Show that $\overrightarrow{AC} = k\overrightarrow{AB}$ for some scalar k
- State: "Since $\overrightarrow{AC} = k\overrightarrow{AB}$ and the vectors share point A, points A, B, and C are collinear"
Critical exam point: Simply showing vectors are parallel is insufficient. You must explicitly state the vectors share a common point to prove collinearity.
Alternative formulation: Show that $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are parallel (both point in the same or opposite directions), proving A, B, C lie on one line.
Division of line segments in given ratios
When point P divides segment AB in ratio m:n (meaning AP:PB = m:n), the position vector formula is:
$$\mathbf{p} = \frac{n\mathbf{a} + m\mathbf{b}}{m+n}$$
Special case — midpoint: When m = n = 1, point M is the midpoint: $$\mathbf{m} = \frac{\mathbf{a} + \mathbf{b}}{2}$$
Extended application: For point P dividing AB such that AP:PB = 2:1, meaning P is $\frac{2}{3}$ of the way from A to B: $$\mathbf{p} = \mathbf{a} + \frac{2}{3}\overrightarrow{AB} = \mathbf{a} + \frac{2}{3}(\mathbf{b} - \mathbf{a}) = \frac{1}{3}\mathbf{a} + \frac{2}{3}\mathbf{b}$$
Alternatively using the ratio formula: $$\mathbf{p} = \frac{1 \times \mathbf{a} + 2 \times \mathbf{b}}{2+1} = \frac{\mathbf{a} + 2\mathbf{b}}{3}$$
Both methods are acceptable in CIE IGCSE Mathematics examinations.
Proving parallelism and identifying parallelograms
Two line segments AB and CD are parallel if and only if $\overrightarrow{AB} = k\overrightarrow{CD}$ for some scalar k.
Proving ABCD is a parallelogram:
Method 1 — Show opposite sides are parallel and equal:
- Calculate $\overrightarrow{AB}$ and $\overrightarrow{DC}$
- Show $\overrightarrow{AB} = \overrightarrow{DC}$
- Calculate $\overrightarrow{AD}$ and $\overrightarrow{BC}$
- Show $\overrightarrow{AD} = \overrightarrow{BC}$
- Conclude: "Since opposite sides are equal and parallel, ABCD is a parallelogram"
Method 2 — Show diagonals bisect each other:
- Find midpoint M of AC: $\mathbf{m}_1 = \frac{\mathbf{a} + \mathbf{c}}{2}$
- Find midpoint N of BD: $\mathbf{m}_2 = \frac{\mathbf{b} + \mathbf{d}}{2}$
- Show $\mathbf{m}_1 = \mathbf{m}_2$
- Conclude: "Since the diagonals bisect each other, ABCD is a parallelogram"
Geometric proofs involving ratios and intersections
CIE IGCSE extended papers frequently ask you to prove that certain points divide lines in specific ratios or that lines intersect at particular points.
Finding intersection points: When two lines intersect, their position vectors at the intersection point must be equal. If P lies on both AB and CD:
- Express P on line AB: $\mathbf{p} = \mathbf{a} + s(\mathbf{b} - \mathbf{a})$ for some parameter s
- Express P on line CD: $\mathbf{p} = \mathbf{c} + t(\mathbf{d} - \mathbf{c})$ for some parameter t
- Equate and solve for s and t
- The values of s and t reveal the ratios AP:PB and CP:PD
Proving three lines meet at a point (concurrent): Calculate the intersection point of two lines using the method above, then verify this point also lies on the third line by substituting into its equation.
Worked examples
Example 1: Finding position vectors and proving collinearity
Points A, B, and C have position vectors $\mathbf{a} = \begin{pmatrix} 2 \ 1 \end{pmatrix}$, $\mathbf{b} = \begin{pmatrix} 5 \ 4 \end{pmatrix}$, and $\mathbf{c} = \begin{pmatrix} 8 \ 7 \end{pmatrix}$ respectively.
(a) Find $\overrightarrow{AB}$ and $\overrightarrow{BC}$.
(b) Prove that A, B, and C are collinear.
Solution:
(a) $$\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 5 \ 4 \end{pmatrix} - \begin{pmatrix} 2 \ 1 \end{pmatrix} = \begin{pmatrix} 3 \ 3 \end{pmatrix}$$
$$\overrightarrow{BC} = \mathbf{c} - \mathbf{b} = \begin{pmatrix} 8 \ 7 \end{pmatrix} - \begin{pmatrix} 5 \ 4 \end{pmatrix} = \begin{pmatrix} 3 \ 3 \end{pmatrix}$$
(b) $\overrightarrow{AB} = \overrightarrow{BC} = \begin{pmatrix} 3 \ 3 \end{pmatrix}$
Since $\overrightarrow{AB} = \overrightarrow{BC}$ and both vectors share point B, the points A, B, and C are collinear.
[Examiner note: Award 1 mark for each vector calculation in (a), 1 mark for showing vectors are equal or scalar multiples, 1 mark for stating collinearity with correct reasoning]
Example 2: Division of line segment in a ratio
The position vectors of points P and Q are $\mathbf{p} = 3\mathbf{i} + 2\mathbf{j}$ and $\mathbf{q} = 7\mathbf{i} + 6\mathbf{j}$ respectively. Point R divides PQ in the ratio 3:1.
(a) Find the position vector of R.
(b) Point M is the midpoint of PR. Find the position vector of M.
Solution:
(a) R divides PQ in ratio 3:1, meaning PR:RQ = 3:1.
Using the ratio formula: $$\mathbf{r} = \frac{1 \times \mathbf{p} + 3 \times \mathbf{q}}{3+1} = \frac{\mathbf{p} + 3\mathbf{q}}{4}$$
$$\mathbf{r} = \frac{(3\mathbf{i} + 2\mathbf{j}) + 3(7\mathbf{i} + 6\mathbf{j})}{4} = \frac{3\mathbf{i} + 2\mathbf{j} + 21\mathbf{i} + 18\mathbf{j}}{4}$$
$$\mathbf{r} = \frac{24\mathbf{i} + 20\mathbf{j}}{4} = 6\mathbf{i} + 5\mathbf{j}$$
(b) M is the midpoint of PR: $$\mathbf{m} = \frac{\mathbf{p} + \mathbf{r}}{2} = \frac{(3\mathbf{i} + 2\mathbf{j}) + (6\mathbf{i} + 5\mathbf{j})}{2}$$
$$\mathbf{m} = \frac{9\mathbf{i} + 7\mathbf{j}}{2} = 4.5\mathbf{i} + 3.5\mathbf{j}$$
[Examiner note: Award 2 marks for correct application of ratio formula in (a), 1 mark for correct substitution, 1 mark for final answer; 2 marks for part (b)]
Example 3: Proving a quadrilateral is a parallelogram
OABC is a quadrilateral where O is the origin. Points A, B, and C have position vectors a, b, and c respectively. It is given that $\mathbf{c} = \mathbf{a} + \mathbf{b}$.
Prove that OABC is a parallelogram.
Solution:
To prove OABC is a parallelogram, we need to show that opposite sides are parallel and equal.
Find $\overrightarrow{OA}$: $$\overrightarrow{OA} = \mathbf{a}$$
Find $\overrightarrow{BC}$: $$\overrightarrow{BC} = \mathbf{c} - \mathbf{b} = (\mathbf{a} + \mathbf{b}) - \mathbf{b} = \mathbf{a}$$
Therefore $\overrightarrow{OA} = \overrightarrow{BC}$, so OA is parallel to BC and OA = BC.
Find $\overrightarrow{OB}$: $$\overrightarrow{OB} = \mathbf{b}$$
Find $\overrightarrow{AC}$: $$\overrightarrow{AC} = \mathbf{c} - \mathbf{a} = (\mathbf{a} + \mathbf{b}) - \mathbf{a} = \mathbf{b}$$
Therefore $\overrightarrow{OB} = \overrightarrow{AC}$, so OB is parallel to AC and OB = AC.
Since both pairs of opposite sides are equal and parallel, OABC is a parallelogram.
[Examiner note: Award 1 mark for each correct displacement vector, 1 mark for showing each pair of opposite sides equal, 1 mark for correct conclusion with proper reasoning — total 6 marks]
Common mistakes and how to avoid them
Mistake: Forgetting to subtract position vectors when finding displacement vectors — Students write $\overrightarrow{AB} = \mathbf{a} + \mathbf{b}$ instead of $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$. Remember: displacement FROM A TO B means "end minus start" = b − a.
Mistake: Not stating common point when proving collinearity — Showing $\overrightarrow{AB} = 2\overrightarrow{CD}$ only proves the lines are parallel, not that A, B, C are collinear. Correction: Use vectors that share a point, such as $\overrightarrow{AB}$ and $\overrightarrow{AC}$, then explicitly state "these vectors share point A".
Mistake: Incorrect ratio formula application — For point P dividing AB in ratio m:n, students mix up which position vector gets which multiplier. Correction: Use $\mathbf{p} = \frac{n\mathbf{a} + m\mathbf{b}}{m+n}$ where m corresponds to the ratio from A, so the far point b is multiplied by m.
Mistake: Incomplete parallelogram proof — Only showing one pair of opposite sides are equal. Correction: You must prove either both pairs of opposite sides are equal and parallel, OR that diagonals bisect each other. One pair is insufficient.
Mistake: Sign errors in column vector arithmetic — When subtracting column vectors, students often make sign errors. Correction: Write out each component subtraction explicitly: $\begin{pmatrix} 5 \ 4 \end{pmatrix} - \begin{pmatrix} 2 \ 7 \end{pmatrix} = \begin{pmatrix} 5-2 \ 4-7 \end{pmatrix} = \begin{pmatrix} 3 \ -3 \end{pmatrix}$.
Mistake: Concluding vectors are equal when only parallel — Writing $\overrightarrow{AB} = 3\overrightarrow{CD}$ then concluding the lines have equal length. Correction: Equal vectors must have the same magnitude and direction; scalar multiples other than ±1 indicate different magnitudes.
Exam technique for "Vectors: Position vectors and geometric proofs using vectors"
Command word "Prove" or "Show that" — You must show complete algebraic working leading to the required result. Stating the answer without derivation scores zero marks. Set out your proof logically with clear steps, and always make a concluding statement that explicitly answers what was asked.
Working with given information — CIE questions often state relationships like "P is the midpoint of AB" or "X divides YZ in ratio 2:3". Translate these immediately into vector equations: midpoint means $\mathbf{p} = \frac{\mathbf{a}+\mathbf{b}}{2}$; ratio 2:3 means $\mathbf{x} = \frac{3\mathbf{y}+2\mathbf{z}}{5}$. Mark schemes reward accurate formula application.
Geometric proof structure — Extended response questions worth 4-6 marks require: (1) calculate relevant displacement vectors, (2) show algebraic relationship between them, (3) interpret geometrically, (4) state conclusion with reasoning. Missing the final explanatory sentence typically costs the last mark.
Accuracy in notation — Distinguish between vectors a (bold or underlined in handwriting) and scalars a. Use correct arrow notation $\overrightarrow{AB}$ for displacement vectors. Mark schemes deduct for inconsistent or incorrect notation, particularly in proof questions where precision matters.
Quick revision summary
Position vectors locate points relative to origin O. Displacement vector $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ connects two points. Points are collinear when displacement vectors between them are scalar multiples sharing a common point. Point P dividing AB in ratio m:n has position vector $\mathbf{p} = \frac{n\mathbf{a}+m\mathbf{b}}{m+n}$; midpoint is $\frac{\mathbf{a}+\mathbf{b}}{2}$. Parallelograms proven by showing opposite sides equal ($\overrightarrow{AB} = \overrightarrow{DC}$) or diagonals bisect. Geometric proofs require clear working, algebraic justification, and explicit concluding statements. Practice converting between position and displacement vectors fluently for exam success.