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Position vectors and displacement vectors

Position vectors and displacement vectors

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​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, OO. 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 PP has coordinates (a,b)(a,b). The position vector of PP from the fixed origin OO to PP, is OP→\overrightarrow{OP}.​

​OP→=ai+bj=(ab)\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 AA is the initial position, and point BB is the final position, the displacement vector AB→\overrightarrow{AB} is the difference between the two points.


The position vector of AA is OA→=a\overrightarrow{OA}=\textbf{a}.

The position vector of BB is OB→=b\overrightarrow{OB}=\textbf{b}.


AB→=OB→−OA→\boxed{\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}}​​



Example 1

The diagram shows points AA and BB. Find, in column vector form, the position vector of AA, the position vector of BB, and the vector AB→\overrightarrow{AB}.

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


The position vector of AA is its position relative to the origin, OO:

​OA→=(−12)‾\underline{\overrightarrow{OA}=\begin{pmatrix}-1\\2\end{pmatrix}}​​


The position vector of BB is its positio​n relative to the origin, OO:

​OB→=(52)‾\underline{\overrightarrow{OB}=\begin{pmatrix}5\\2\end{pmatrix}}​​


AB→\overrightarrow{AB} is the displacement vector between t​he two points:

​AB→=OB→−OA→=(52)−(−12)=(60)‾\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 AA and BB have position vectors 4i+3j4\textbf{i}+3\textbf{j} and −3i+j-3\textbf{i}+\textbf{j} respectively. Find AB→\overrightarrow{AB}.


AB→=OB→−OA→=(−3i+j)−(4i+3j)=−7i−2j‾\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 AA has coordinates (3,2)(3,2) and AB→=−2i−8j\overrightarrow{AB}=-2\textbf{i}-8\textbf{j}. Find the position vector of BB and its exact magnitude.


The position vector of AA is:

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


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

AB→=OB→−OA→OB→=AB→+OA→=(−2i−8j)+(3i+2j)=i−6j‾\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 BB:

∣OB→∣=12+(−6)2=1+36=37‾\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.


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