Position Vectors

This lesson covers position vectors, displacement vectors, and related concepts such as midpoints and dividing a line in a given ratio — as required by the Edexcel A-Level Mathematics specification (9MA0).

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What Is a Position Vector?

The position vector of a point P is the vector from the origin O to the point P. It is written as:

OP or simply p

If P has coordinates (3, 5, -2), then the position vector of P is:

p = (3, 5, -2) = 3i + 5j - 2k

Key Point: The position vector of a point tells you where the point is relative to the origin. Every point in space has a unique position vector.

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Displacement Vectors

The displacement vector from point A to point B is the vector that takes you from A to B. It is found by:

AB = b - a

where a and b are the position vectors of A and B respectively.

Example: A has position vector (2, 1, 3) and B has position vector (5, 4, -1). Find AB.

AB = b - a = (5 - 2, 4 - 1, -1 - 3) = (3, 3, -4)

Key Properties

• AB = b - a (from A to B)
• BA = a - b = -AB (from B to A — opposite direction)
• |AB| = distance between A and B

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The Midpoint

The midpoint M of the line segment AB has position vector:

m = (a + b) / 2

This is the average of the position vectors.

Example: Find the midpoint of A(2, 6, 4) and B(8, 2, -2).

m = ((2 + 8)/2, (6 + 2)/2, (4 + (-2))/2) = (5, 4, 1)

So M is the point (5, 4, 1).

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Dividing a Line Segment in a Given Ratio

Internal Division

If a point P divides the line segment AB in the ratio m : n (internally), then the position vector of P is:

p = (na + mb) / (m + n)

Alternatively: p = a + [m/(m + n)](b - a)

Example: A has position vector (1, 2, 3) and B has position vector (7, 5, 0). Find the position vector of P, where P divides AB in the ratio 2 : 1.

p = (1 × (1, 2, 3) + 2 × (7, 5, 0)) / (2 + 1)

= ((1, 2, 3) + (14, 10, 0)) / 3

= (15, 12, 3) / 3

= (5, 4, 1)

Understanding the Ratio

If P divides AB in the ratio m : n:

• P is closer to A when m < n
• P is the midpoint when m = n
• P is closer to B when m > n

Example: P divides AB in the ratio 1 : 3. Then AP : PB = 1 : 3, so P is one-quarter of the way from A to B.

p = a + (1/4)(b - a) = (3/4)a + (1/4)b

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Using Vectors to Show Points Are Collinear

Three points A, B, C are collinear (lie on the same straight line) if AB is parallel to AC. That is:

AC = k × AB for some scalar k

Example: Show that A(1, 2, 3), B(3, 6, 7), and C(5, 10, 11) are collinear.

AB = (3 - 1, 6 - 2, 7 - 3) = (2, 4, 4) AC = (5 - 1, 10 - 2, 11 - 3) = (4, 8, 8)

AC = 2 × AB, so AB is parallel to AC and they share point A. Therefore A, B, C are collinear.

Moreover, since AC = 2 AB, the point B is the midpoint of AC.

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Vector Triangles and Paths

In many problems, you navigate between points using vector addition:

AB + BC = AC (the triangle law of addition)

This means: the displacement from A to C equals the displacement from A to B plus the displacement from B to C.

Example: In triangle OAB, M is the midpoint of AB. If OA = a and OB = b, express OM in terms of a and b.

OM = OA + AM

AM = (1/2)AB = (1/2)(b - a)

OM = a + (1/2)(b - a) = a + (1/2)b - (1/2)a = (1/2)a + (1/2)b = (a + b)/2

This confirms the midpoint formula.

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Worked Examples

Example 1: Position Vectors and Distance

Points A, B, C have position vectors a = (2, -1, 3), b = (5, 3, -1), c = (-1, 2, 7).

(a) Find the distance AB.

AB = b - a = (3, 4, -4) |AB| = √(9 + 16 + 16) = √41

(b) Find the midpoint of BC.

M = (b + c)/2 = ((5 - 1)/2, (3 + 2)/2, (-1 + 7)/2) = (2, 2.5, 3)

Example 2: Ratio Division

P divides AB in the ratio 3 : 2, where A = (1, 0, 5) and B = (6, 10, 0).

p = (2(1, 0, 5) + 3(6, 10, 0)) / 5

= ((2, 0, 10) + (18, 30, 0)) / 5

= (20, 30, 10) / 5

= (4, 6, 2)

Example 3: Vector Path

In triangle PQR, OP = p, OQ = q, OR = r. Point S lies on QR such that QS : SR = 2 : 3. Find OS in terms of q and r.

OS = OQ + (2/5)QR = q + (2/5)(r - q) = q + (2/5)r - (2/5)q = (3/5)q + (2/5)r

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Common Mistakes

1. Direction confusion — AB = b - a (finish minus start). Getting this backwards gives -AB.
2. Ratio formula errors — In ratio m : n, the formula is (na + mb)/(m + n). The coefficients are "swapped" relative to what you might expect.
3. Forgetting the origin — Position vectors are always measured from the origin O.
4. Collinearity — To show three points are collinear, you must show that two of the displacement vectors are parallel AND share a common point.

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Summary

• The position vector of P is the vector from the origin to P.
• Displacement vector AB = b - a.
• Midpoint: m = (a + b)/2.
• Point dividing AB in ratio m : n: p = (na + mb)/(m + n).
• Points are collinear if the displacement vectors between them are parallel.
• The triangle law: AB + BC = AC.

A-Level Deep Dive: Position Vectors

Spec mapping

Edexcel 9MA0-01 specification section 12 — Vectors covers vectors in two dimensions and in three dimensions; calculate the magnitude and direction of a vector and convert between component form and magnitude/direction form; understand and use position vectors; calculate the distance between two points represented by position vectors (refer to the official specification document for exact wording). Position vectors sit at the centre of the Vectors section because every later result — magnitudes, midpoints, ratios, parallelism — is constructed from the basic identity $\vec{AB} = \vec{OB} - \vec{OA} = \mathbf{b} - \mathbf{a}$AB=OB−OA=b−a. They are examined principally on 9MA0-01 Paper 1 — Pure Mathematics, with synoptic appearances inside 9MA0-03 Mechanics (where particle position $\mathbf{r}(t)$r(t) is a position vector in time) and inside section 7 — Coordinate geometry (the equation of a line through two points is recast as $\mathbf{r} = \mathbf{a} + t(\mathbf{b} - \mathbf{a})$r=a+t(b−a)). The Edexcel formula booklet lists the magnitude formula but not the section-formula for division of a line in a given ratio — the candidate must derive it from position vectors on the page.

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Worked example with full mark scheme

Question (8 marks): Relative to a fixed origin $O$O, the points $A$A, $B$B and $C$C have position vectors $\mathbf{a} = 2\mathbf{i} + \mathbf{j} - \mathbf{k}$a=2i+j−k, $\mathbf{b} = 5\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}$b=5i+4j+2k and $\mathbf{c} = 11\mathbf{i} + 10\mathbf{j} + 8\mathbf{k}$c=11i+10j+8k.

(a) Show that $A$A, $B$B and $C$C are collinear. (5)

(b) Find the ratio $AB : BC$AB:BC. (3)

Solution with mark scheme:

(a) Step 1 — form $\vec{AB}$AB using position-vector subtraction.

$\vec{AB} = \mathbf{b} - \mathbf{a} = (5 - 2)\mathbf{i} + (4 - 1)\mathbf{j} + (2 - (-1))\mathbf{k} = 3\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}$AB=b−a=(5−2)i+(4−1)j+(2−(−1))k=3i+3j+3k

M1 — applying $\vec{AB} = \mathbf{b} - \mathbf{a}$AB=b−a correctly. The signed subtraction of $-\mathbf{k}$−k giving $+3\mathbf{k}$+3k is the place candidates slip; writing $2 - 1 = 1$2−1=1 for the $\mathbf{k}$k-component would lose this M1.

Step 2 — form $\vec{BC}$BC similarly.

$\vec{BC} = \mathbf{c} - \mathbf{b} = (11 - 5)\mathbf{i} + (10 - 4)\mathbf{j} + (8 - 2)\mathbf{k} = 6\mathbf{i} + 6\mathbf{j} + 6\mathbf{k}$BC=c−b=(11−5)i+(10−4)j+(8−2)k=6i+6j+6k

M1 — second correct subtraction.

Step 3 — show one is a scalar multiple of the other.

$\vec{BC} = 6\mathbf{i} + 6\mathbf{j} + 6\mathbf{k} = 2(3\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}) = 2\vec{AB}$BC=6i+6j+6k=2(3i+3j+3k)=2AB

M1 — recognising that parallel vectors must be scalar multiples; setting up $\vec{BC} = k\vec{AB}$BC=kAB for some scalar $k$k.

A1 — correct identification $k = 2$k=2 from any one component, with the other two components verified.

Step 4 — argue collinearity, not merely parallelism.

Because $\vec{AB}$AB and $\vec{BC}$BC are parallel and share the common point $B$B, the three points $A$A, $B$B, $C$C lie on the same straight line. A1 — the explicit "share point $B$B" line is the reasoning mark; without it the candidate has shown parallel direction only, which is not collinearity.

(b) Step 1 — read magnitudes off the scalar relationship.

From part (a), $\vec{BC} = 2\vec{AB}$BC=2AB, so $|\vec{BC}| = 2|\vec{AB}|$∣BC∣=2∣AB∣.

M1 — extracting magnitude ratio from the scalar relationship without recomputing.

Step 2 — state the ratio.

$AB : BC = 1 : 2$AB:BC=1:2

A1 — correct ratio in simplest form. A1 — answered in the order requested by the question ($AB : BC$AB:BC, not $BC : AB$BC:AB).

Total: 8 marks (M4 A4, split as shown).

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Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): Relative to a fixed origin $O$O, points $P$P and $Q$Q have position vectors $\mathbf{p}$p and $\mathbf{q}$q respectively. The point $R$R lies on the line segment $PQ$PQ such that $PR : RQ = 1 : 3$PR:RQ=1:3.

(a) Show that the position vector of $R$R is $\dfrac{3\mathbf{p} + \mathbf{q}}{4}$43p+q​. (3)

(b) Given that $\mathbf{p} = 4\mathbf{i} + 8\mathbf{j}$p=4i+8j and $\mathbf{q} = 12\mathbf{i} - 4\mathbf{j}$q=12i−4j, find $|\vec{OR}|$∣OR∣ in exact form. (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — writing $\vec{OR} = \vec{OP} + \vec{PR}$OR=OP+PR and $\vec{PR} = \tfrac{1}{4}\vec{PQ} = \tfrac{1}{4}(\mathbf{q} - \mathbf{p})$PR=41​PQ​=41​(q−p).
• M1 (AO1.1b) — substituting and combining: $\vec{OR} = \mathbf{p} + \tfrac{1}{4}(\mathbf{q} - \mathbf{p})$OR=p+41​(q−p).
• A1 (AO2.1) — algebraic simplification to the printed form $\dfrac{3\mathbf{p} + \mathbf{q}}{4}$43p+q​, with the rearrangement step shown.

(b)

• M1 (AO1.1b) — substituting components: $\vec{OR} = \dfrac{3(4\mathbf{i} + 8\mathbf{j}) + (12\mathbf{i} - 4\mathbf{j})}{4} = \dfrac{24\mathbf{i} + 20\mathbf{j}}{4} = 6\mathbf{i} + 5\mathbf{j}$OR=43(4i+8j)+(12i−4j)​=424i+20j​=6i+5j.
• M1 (AO1.1b) — applying the magnitude formula: $|\vec{OR}| = \sqrt{6^2 + 5^2} = \sqrt{61}$∣OR∣=62+52​=61​.
• A1 (AO2.5) — final answer in exact surd form $\sqrt{61}$61​, not a decimal.

Total: 6 marks split AO1 = 4, AO2 = 2. Section-12 questions reliably reward AO2 marks for the derivation of the section-formula on the page — Edexcel will not let candidates quote the formula $\dfrac{m\mathbf{q} + n\mathbf{p}}{m + n}$m+nmq+np​ without showing where it comes from.

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Synoptic links

Connects to:

1. Section 12 — 2D and 3D vectors: position vectors generalise immediately from $(x, y)$(x,y) to $(x, y, z)$(x,y,z) with no change in algebraic form. The same identity $\vec{AB} = \mathbf{b} - \mathbf{a}$AB=b−a holds in any dimension; only the magnitude formula gains a $z^2$z2 term: $|\mathbf{r}| = \sqrt{x^2 + y^2 + z^2}$∣r∣=x2+y2+z2​.

2. Vector geometry — parallelism and collinearity: two vectors are parallel iff one is a scalar multiple of the other. Three points are collinear iff the vectors joining them are parallel and share a common point. The "common point" qualifier is what distinguishes collinearity from mere directional agreement and is the source of most lost A1 marks.

3. Section 7 — Coordinate geometry: the Cartesian equation of the line through $A$A and $B$B is the parametric form $\mathbf{r} = \mathbf{a} + t(\mathbf{b} - \mathbf{a})$r=a+t(b−a) with $t \in \mathbb{R}$t∈R. At $t = 0$t=0 you are at $A$A; at $t = 1$t=1 you are at $B$B; at $t = \tfrac{1}{2}$t=21​ you are at the midpoint $\tfrac{1}{2}(\mathbf{a} + \mathbf{b})$21​(a+b). This recovers the midpoint formula directly from the position-vector identity.

4. 9MA0-03 Mechanics — position-time: a particle's position at time $t$t is the position vector $\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^2$r(t)=r0​+ut+21​at2 for constant acceleration. Velocity is $\dot{\mathbf{r}}(t)$r˙(t), displacement between times $t_1$t1​ and $t_2$t2​ is $\mathbf{r}(t_2) - \mathbf{r}(t_1)$r(t2​)−r(t1​) — the same subtraction identity reused.

5. Section 8 — Differentiation under linear-motion modelling: differentiating a position vector componentwise gives the velocity vector. The position-vector formalism makes it routine to read off motion in any direction by projecting onto unit vectors $\mathbf{i}$i, $\mathbf{j}$j, $\mathbf{k}$k.

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Mark-scheme literacy

Position-vector questions on 9MA0 split AO marks toward AO1 with a healthy AO2 share for derivations:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	55–65%	Forming $\vec{AB} = \mathbf{b} - \mathbf{a}$AB=b−a, computing magnitudes, performing component arithmetic, identifying scalar multiples
AO2 (reasoning / interpretation)	30–40%	Justifying collinearity via "shared point", deriving the section formula on the page, choosing the correct ratio order, recognising parallelism vs collinearity distinction
AO3 (problem-solving)	0–10%	Multi-stage geometric configurations (e.g. centroid, locus problems) — rare at AS, more common at A2