Position vectors and displacement vectors

In a nutshell

Vectors are used to describe a position of a point in space. A position vector gives the position of a point relative to a fixed origin, $O$ . A displacement vector describes movement from one position to another.

Position vectors

A position vector gives the position of a point relative to a fixed origin.

Maths; Vectors I; KS5 Year 12; Position vectors and displacement vectors

Point $P$ has coordinates $(a,b)$ . The position vector of $P$ from the fixed origin $O$ to $P$ , is $\overrightarrow{OP}$ .

$\boxed{\overrightarrow{OP}=a\textbf{i}+b\textbf{j}=\begin{pmatrix}a\\b\end{pmatrix}}$

Displacement vectors

A displacement vector describes movement from one position to another. In a two-coordinate system, the difference between the position vectors of the final point and the initial point, relative to the origin, gives your displacement.

Maths; Vectors I; KS5 Year 12; Position vectors and displacement vectors

If point $A$ is the initial position, and point $B$ is the final position, the displacement vector $\overrightarrow{AB}$ is the difference between the two points.

The position vector of $A$ is $\overrightarrow{OA}=\textbf{a}$ .

The position vector of $B$ is $\overrightarrow{OB}=\textbf{b}$ .

$\boxed{\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}}$

Example 1

The diagram shows points $A$ and $B$ . Find, in column vector form, the position vector of $A$ , the position vector of $B$ , and the vector $\overrightarrow{AB}$ .

Maths; Vectors I; KS5 Year 12; Position vectors and displacement vectors

The position vector of $A$ is its position relative to the origin, $O$ :

$\underline{\overrightarrow{OA}=\begin{pmatrix}-1\\2\end{pmatrix}}$

The position vector of $B$ is its position relative to the origin, $O$ :

$\underline{\overrightarrow{OB}=\begin{pmatrix}5\\2\end{pmatrix}}$

$\overrightarrow{AB}$ is the displacement vector between the two points:

$\begin{aligned}\overrightarrow{AB}&=\overrightarrow{OB}-\overrightarrow{OA}\\&=\begin{pmatrix}5\\2\end{pmatrix}-\begin{pmatrix}-1\\2\\\end{pmatrix}\\&=\underline{\begin{pmatrix}6\\0\end{pmatrix}}\end{aligned}$

Example 2

Points $A$ and $B$ have position vectors $4\textbf{i}+3\textbf{j}$ and $-3\textbf{i}+\textbf{j}$ respectively. Find $\overrightarrow{AB}$ .

$\begin{aligned}\overrightarrow{AB}&=\overrightarrow{OB}-\overrightarrow{OA}\\&=(-3\textbf{i}+\textbf{j})-(4\textbf{i}+3\textbf{j})\\&=\underline{-7\textbf{i}-2\textbf{j}}\end{aligned}$

Example 3

Point $A$ has coordinates $(3,2)$ and $\overrightarrow{AB}=-2\textbf{i}-8\textbf{j}$ . Find the position vector of $B$ and its exact magnitude.

The position vector of $A$ is:

$\overrightarrow{OA}=3\textbf{i}+2\textbf{j}$

We know that $\begin{aligned}\overrightarrow{AB}&=\overrightarrow{OB}-\overrightarrow{OA}\end{aligned}$ . So:

$\begin{aligned}\overrightarrow{AB}&=\overrightarrow{OB}-\overrightarrow{OA}\\\overrightarrow{OB}&=\overrightarrow{AB}+\overrightarrow{OA}\\&=(-2\textbf{i}-8\textbf{j})+(3\textbf{i}+2\textbf{j})\\&=\underline{\textbf{i}-6\textbf{j}}\end{aligned}$

For the magnitude of the position vector of $B$ :

$\begin{aligned}|\overrightarrow{OB}|&=\sqrt{1^2+(-6)^2}\\&=\sqrt{1+36}\\&=\underline{\sqrt{37}}\end{aligned}$ 

Note: You have been asked for the exact value, so leave your answer in surd form.