Pure 10.4 - Position vectors and distance between points | Edexcel A-Level Mathematics

A position vector locates a point relative to a chosen origin. You will subtract positions to find a directed displacement, calculate distances in two and three dimensions, and use a distance condition to find an unknown coordinate. Keeping vector and length notation distinct also explains why changing the origin cannot change a distance.

Position vectors

If you move the origin, does the distance between two fixed points change? Take $A=(2, 3)$ and $B=(7, 1)$ . Choosing a new origin ten units to the right changes their coordinates to $(-8, 3)$ and $(-3, 1)$ , but the points themselves have not moved. We will see the cancellation that protects their separation.

You already know

Coordinates name a point relative to axes and an origin. A vector gives signed component changes; its magnitude is the square root of the sum of squared components.

An arrow starting at the chosen origin and ending at a point is that point's position vector.

Position vector

The position vector of $A$ , relative to a fixed origin $O$ , is $\overrightarrow{OA}$ . For $A=(x, y, z)$ , it is $x\mathbf i+y\mathbf j+z\mathbf k$ . For example, $A=(2, 3)$ has position vector $2\mathbf i+3\mathbf j$ ; the point and its vector are related but are different kinds of object.

The origin matters for a position vector. A general displacement arrow need not begin there.

[DIAGRAM: asset_slug: p1_10_4_understand_and_use_position_vectors_and_calculate_the_distance_between_two_points__diagram_01; asset_name: Positions and displacement between two points; caption: Both position vectors start at O but the displacement starts at A; description: Cartesian axes showing A at (2, 3), B at (7, 1), arrows OA and OB from O, and the directed arrow AB labelled b minus a]

In three dimensions, $C=(5, -2, 0)$ gives $\overrightarrow{OC}=5\mathbf i-2\mathbf j$ ; the zero third term may be omitted. Conversely, a point with position vector $-\mathbf i+3\mathbf k$ has coordinates $(-1, 0, 3)$ , including the missing $y$ -coordinate.

Key idea

A position vector answers “where is the point relative to this origin?”; a displacement answers “how do I move from one point to another?”

Check which arrow is actually attached to the chosen origin.

Head minus tail

To travel from $A$ to $B$ , you can go back from $A$ to $O$ , then out from $O$ to $B$ . With position vectors $\mathbf a$ and $\mathbf b$ , that route is $-\mathbf a+\mathbf b$ .

Displacement from positions

Relative to one fixed origin,

\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\mathbf b-\mathbf a

Read this as finish minus start, or head minus tail.

Not in the formula booklet: learn it.

Do not subtract in the order in which the question happens to print the data. The arrow's name determines the order.

Worked example 1: displacement from A to B

Question. Points $A$ and $B$ have position vectors $2\mathbf i-3\mathbf j$ and $-4\mathbf i+5\mathbf j$ . Find $\overrightarrow{AB}$ .

Plan. Subtract the start position from the finish position.

\begin{aligned} \overrightarrow{AB}&=(-4\mathbf i+5\mathbf j)-(2\mathbf i-3\mathbf j) &&\text{finish minus start}\\ &=(-4-2)\mathbf i+(5+3)\mathbf j &&\text{subtract both components}\\ &=-6\mathbf i+8\mathbf j &&\text{simplify} \end{aligned}

Why this step? Subtracting $-3\mathbf j$ adds $3\mathbf j$ , since the return from $A$ to the origin reverses that component.

Answer. $\overrightarrow{AB}=-6\mathbf i+8\mathbf j$ .

Check. Adding this displacement to $2\mathbf i-3\mathbf j$ gives $-4\mathbf i+5\mathbf j$ , the position of $B$ .

Change only the requested direction: now travel from $B$ to $A$ .

Worked example 2: reversing the displacement

Question. With the same positions $\overrightarrow{OA}=2\mathbf i-3\mathbf j$ and $\overrightarrow{OB}=-4\mathbf i+5\mathbf j$ , find $\overrightarrow{BA}$ .

Plan. Reverse the start and finish in the subtraction.

\begin{aligned} \overrightarrow{BA}&=(2\mathbf i-3\mathbf j)-(-4\mathbf i+5\mathbf j) &&\text{finish at }A\\ &=(2+4)\mathbf i+(-3-5)\mathbf j &&\text{subtract corresponding components}\\ &=6\mathbf i-8\mathbf j &&\text{simplify} \end{aligned}

Why this step? Reversing a directed journey reverses every component, rather than changing its length.

Answer. $\overrightarrow{BA}=6\mathbf i-8\mathbf j$ .

Check. $\overrightarrow{BA}=-\overrightarrow{AB}$ , as it must.

This is also a quick way to detect a direction error in a longer problem.

Watch out

Wrong: $\overrightarrow{AB}=\mathbf a-\mathbf b$ . That is $\overrightarrow{BA}$ . Right: $\overrightarrow{AB}=\mathbf b-\mathbf a$ .

Find the reversed displacement for a new pair of points.

Distance in two dimensions

Once the displacement is known, its magnitude is the direct distance between the points. A displacement of six units horizontally and four vertically has length $\sqrt{6^2+4^2}=2\sqrt{13}$ .

Two-dimensional distance

For $A=(x_1, y_1)$ and $B=(x_2, y_2)$ ,

AB=\lvert\overrightarrow{AB}\rvert=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

The non-negative square root gives the distance.

Not in the formula booklet: learn it.

Each squared difference is a perpendicular contribution to Pythagoras' theorem.

[DIAGRAM: asset_slug: p1_10_4_understand_and_use_position_vectors_and_calculate_the_distance_between_two_points__diagram_02; asset_name: Distance from perpendicular component differences; caption: Two dimensions use two squared differences and three dimensions use three; description: A two-dimensional right triangle with component differences delta x and delta y and diagonal d, alongside a projected cuboid with three component differences and body diagonal d; the projected diagonal avoids intermediate vertices]
Two dimensions use two squared differences and three dimensions use three

Use exact surds unless a decimal is requested.

Worked example 3: a distance as a simplified surd

Question. Given $\overrightarrow{OP}=-4\mathbf i+\mathbf j$ and $\overrightarrow{OQ}=2\mathbf i+5\mathbf j$ , find the exact distance $PQ$ .

Plan. Find $\overrightarrow{PQ}$ , then take its magnitude.

\begin{aligned} \overrightarrow{PQ}&=(2-(-4))\mathbf i+(5-1)\mathbf j &&\text{position differences}\\ &=6\mathbf i+4\mathbf j &&\text{simplify}\\ PQ&=\sqrt{6^2+4^2}=\sqrt{52}=2\sqrt{13} &&\text{magnitude, then simplify surd} \end{aligned}

Why this step? The question asks for a distance, so the directed vector is an intermediate result rather than the final answer.

Answer. $PQ=2\sqrt{13}$ .

Check. The distance exceeds each individual component length, $6$ and $4$ , but is less than their sum $10$ .

Change only the horizontal coordinate of $Q$ into an unknown, keeping the same distance. A sign condition identifies the appropriate position.

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Last Accessed

Pure Mathematics (Papers 1 and 2)

Statistics (Paper 3, Section A)

Mechanics (Paper 3, Section B)

Pure 10.4 - Position vectors and distance between points

Position vectors

Head minus tail

Distance in two dimensions

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