Position Vectors
A point on the plane is just a pair of coordinates — but a position vector turns that point into a direction and magnitude from the origin. Once you can read off $\vec{OP}$ from coordinates and compute $\vec{AB} = \vec{OB} - \vec{OA}$, every geometric problem in vectors becomes algebraic. That one formula unlocks the whole module.
Point $A$ is at $(3, -2)$ and point $B$ is at $(7, 4)$. Without using a formula — describe the journey from $A$ to $B$ as a vector. How far do you move horizontally? How far vertically?
Every point in the plane can be described by a position vector — an arrow from the origin $O$ to the point. Converting coordinates to position vectors lets us use algebra to solve geometry problems.
If $P$ has coordinates $(x, y)$, its position vector is $\vec{OP} = \begin{pmatrix} x \\ y \end{pmatrix}$. To get from $A$ to $B$ we write:
$\vec{AB} = \vec{OB} - \vec{OA}$ (terminal minus initial)
Key facts
- Position vector of $P(x,y)$ is $\vec{OP} = \begin{pmatrix} x \\ y \end{pmatrix}$
- $\vec{AB} = \vec{OB} - \vec{OA}$ (terminal minus initial)
- Coordinates and position vectors are interchangeable representations
Concepts
- Why $\vec{AB} = \vec{OB} - \vec{OA}$ follows from the triangle law
- The geometric meaning of a position vector as a displacement from the origin
- How to use position vectors to find midpoints and divide line segments
Skills
- Convert freely between coordinate pairs and column vector notation
- Calculate $\vec{AB}$ given the coordinates of $A$ and $B$
- Apply position vectors to find midpoints and equal vectors
A position vector describes the location of a point relative to a fixed origin $O$. If point $P$ has coordinates $(x, y)$, its position vector is:
For two points $A(x_1, y_1)$ and $B(x_2, y_2)$, we can find the vector from $A$ to $B$ using:
Why does this work? By the triangle law, travelling from $A$ to $O$ (which is $-\vec{OA}$) then from $O$ to $B$ (which is $\vec{OB}$) gives the same result as going directly from $A$ to $B$.
A position vector describes the location of a point relative to a fixed origin $O$. If point $P$ has coordinates $(x, y)$, its position vector is:
Pause — copy the position vector definition: $\overrightarrow{OP}=\begin{pmatrix}a\\b\end{pmatrix}$ for $P(a,b)$, and note that all position vectors start from the origin into your book.
Quick check: Points $P(2, 5)$ and $Q(6, 1)$. Which column vector equals $\vec{PQ}$?
We just saw that the position vector of point $P(a,b)$ from origin $O$ is $\overrightarrow{OP}=\begin{pmatrix}a\\b\end{pmatrix}$. That raises a question: if $M$ is the midpoint of segment $AB$, can you find $\overrightarrow{OM}$ without drawing a diagram, using position vectors directly? This card answers it → $\overrightarrow{OM}=\frac{1}{2}(\overrightarrow{OA}+\overrightarrow{OB})=\frac{1}{2}\begin{pmatrix}a_1+a_2\\b_1+b_2\end{pmatrix}$.
If $M$ is the midpoint of segment $AB$, then its position vector is the average of $\vec{OA}$ and $\vec{OB}$:
This follows from $\vec{OM} = \vec{OA} + \dfrac{1}{2}\vec{AB}$. More generally, the point dividing $AB$ in the ratio $m:n$ has position vector:
$\vec{OP} = \dfrac{n\,\vec{OA} + m\,\vec{OB}}{m+n}$
If $M$ is the midpoint of segment $AB$, then its position vector is the average of $\vec{OA}$ and $\vec{OB}$:
Pause — copy the midpoint formula: $\overrightarrow{OM}=\frac{1}{2}(\overrightarrow{OA}+\overrightarrow{OB})$, and the section formula $\overrightarrow{OP}=\frac{m\,\overrightarrow{OB}+n\,\overrightarrow{OA}}{m+n}$ for dividing $AB$ in ratio $m:n$ into your book.
Did you get this? True or false: the midpoint of $A(1, 3)$ and $B(5, 7)$ has position vector $\begin{pmatrix} 3 \\ 5 \end{pmatrix}$.
Worked examples · 3 in a row, reveal as you go
Points $A$ and $B$ have coordinates $A(3, -2)$ and $B(7, 4)$. Find $\vec{AB}$.
Find the position vector of the midpoint $M$ of $AB$ where $A(-1, 4)$ and $B(5, 2)$.
The position vector of $A$ is $\begin{pmatrix} 2 \\ -3 \end{pmatrix}$ and $\vec{AB} = \begin{pmatrix} 5 \\ 7 \end{pmatrix}$. Find the coordinates of $B$.
Fill the gap: If $A(1, 5)$ and $B(9, 3)$, then $\vec{AB} = \begin{pmatrix}$$\\$$\end{pmatrix}$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: $\vec{BA} = \vec{OA} - \vec{OB}$ (note: the vector from $B$ to $A$, not from $A$ to $B$).
Activities · practice with the ideas
Write down the position vectors $\vec{OA}$ and $\vec{OB}$ for $A(4, -1)$ and $B(-2, 6)$, then calculate $\vec{AB}$.
Find the coordinates of the midpoint $M$ of $PQ$ where $P(-3, 5)$ and $Q(7, -1)$.
The position vector of $C$ is $\begin{pmatrix} -1 \\ 4 \end{pmatrix}$ and $\vec{CD} = \begin{pmatrix} 3 \\ -7 \end{pmatrix}$. Find the coordinates of $D$.
Given $A(2, 1)$, $B(6, 5)$, $C(4, 3)$: show that $C$ is the midpoint of $AB$ using position vectors.
Points $P$, $Q$, $R$ have position vectors $\mathbf{p}$, $\mathbf{q}$, $\mathbf{r}$. Show that $\vec{PQ} + \vec{QR} + \vec{RP} = \mathbf{0}$.
Odd one out: Three of these statements about position vectors are true. Which one is FALSE?
Earlier you estimated the vector from $A(3,-2)$ to $B(7,4)$.
The exact answer is $\vec{AB} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}$. The key is terminal minus initial: subtract $A$'s coordinates from $B$'s. Did your spatial reasoning give the right direction? The formula $\vec{AB} = \vec{OB} - \vec{OA}$ converts any geometry question about two points into a subtraction.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Points $A(-2, 3)$ and $B(4, -1)$ are given. Find $\vec{AB}$ as a column vector. (2 marks)
Q2. The position vector of $P$ is $\begin{pmatrix} 1 \\ -4 \end{pmatrix}$ and $\vec{PQ} = \begin{pmatrix} -3 \\ 6 \end{pmatrix}$. Find the coordinates of $Q$. (2 marks)
Q3. $ABCD$ is a parallelogram with $A(1, 2)$, $B(5, 4)$, $C(7, 8)$. Find the coordinates of $D$ using position vectors. (3 marks)
Comprehensive answers (click to reveal)
Activity answers: 1. $\vec{AB} = \begin{pmatrix} -6 \\ 7 \end{pmatrix}$ · 2. $M = (2, 2)$ · 3. $D = (2, -3)$ · 4. $\frac{1}{2}\begin{pmatrix} 8 \\ 6 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix} = \vec{OC}$ · 5. $(\mathbf{q}-\mathbf{p})+(\mathbf{r}-\mathbf{q})+(\mathbf{p}-\mathbf{r}) = \mathbf{0}$ ✓
Q1 (2 marks): $\vec{OA} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}$, $\vec{OB} = \begin{pmatrix} 4 \\ -1 \end{pmatrix}$ [1]. $\vec{AB} = \begin{pmatrix} 6 \\ -4 \end{pmatrix}$ [1].
Q2 (2 marks): $\vec{OQ} = \vec{OP} + \vec{PQ} = \begin{pmatrix} 1 \\ -4 \end{pmatrix} + \begin{pmatrix} -3 \\ 6 \end{pmatrix} = \begin{pmatrix} -2 \\ 2 \end{pmatrix}$ [1]. $Q(-2, 2)$ [1].
Q3 (3 marks): In a parallelogram $\vec{AB} = \vec{DC}$ [1]. $\vec{AB} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}$ [1]. $\vec{OD} = \vec{OC} - \vec{AB} = \begin{pmatrix} 7 \\ 8 \end{pmatrix} - \begin{pmatrix} 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 6 \end{pmatrix}$, so $D(3,6)$ [1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering position vector questions. Lighter alternative to the boss.
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