Position Vectors

A position vector describes where a point lies, in relation to the origin, $boldsymbol{O}$

This allows you to write other vectors in terms of position vectors.

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Vector Basics

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Working with Position Vectors

For a point $A$ , the position vector is denoted as $overrightarrow{OA}$

Likewise the point $B$ is denoted as $overrightarrow{OB}$

Writing points as position vectors then allows you to describe vectors between points as position vectors.

If we let, $boldsymbol{a}=overrightarrow{OA}$ and $boldsymbol{b}=overrightarrow{OB}$

Then, $overrightarrow{AB}=-overrightarrow{OA}+overrightarrow{OB}=-boldsymbol{a}+boldsymbol{b}=boldsymbol{b}-boldsymbol{a}$

Describing vectors using $boldsymbol{i}$ and $boldsymbol{j}$ Units

A vector with a magnitude of $1$ unit is called a unit vector
The standard unit vectors are $boldsymbol{i}$ and $boldsymbol{j}$ , which are in the direction of the $x$ -axis and $y$ -axis respectively.
These standard unit vectors are used to express the horizontal and vertical position of the end of a vector compared to its start point

Using the diagram on the right, we can see the position vector of point $A=3boldsymbol{i}+5boldsymbol{j}$ and the position vector of point $B=5boldsymbol{i}+2boldsymbol{j}$

To write $overrightarrow{AB}$ in terms of standard unit vectors, we can just add and subtract the $boldsymbol{i}$ and $boldsymbol{j}$ components of $boldsymbol{a}$ and $boldsymbol{b}$ separately.

Thus the vector $overrightarrow{AB}=-boldsymbol{a}+boldsymbol{b}=-(3boldsymbol{i}+5boldsymbol{j})+(5boldsymbol{i}+2boldsymbol{j})=2boldsymbol{i}-3boldsymbol{j}$

This means that to go from $A$ to $B$ you go $2$ units to the right and $3$ units down.

Column Vectors

Column vectors are another way of expressing vectors.

If $A=4boldsymbol{i}-3boldsymbol{j}$ , then the column vector for $A$ is $dbinom{4}{-3}$

To add and subtract column vectors from one another you simply add or subtract the top row and then add or subtract the bottom row. For example, if $A=dbinom{5}{-2}$ and $B=dbinom{1}{3}$ , then $A+B=dbinom{5+1}{-2+3}=dbinom{6}{1}$

Multiplying a column vector by a scalar is also straight forward, you just need to multiply both the top and bottom numbers by the scalar. For instance, if $C=dbinom{-3}{4}$ , $3C=dbinom{3times-3}{3times4}=dbinom{-9}{12}$

Position Vectors Example Questions

Question 1: Point $K$ has the coordinates $(-1,3)$ .

Give the position vector of point $K$ in:

a) Standard unit vector form

b) Column vector form

[2 marks]

If $A=(x,y)$ , then $A=xboldsymbol{i}+yboldsymbol{j}=dbinom{x}{y}$

So,

a) $K=-boldsymbol{i}+3boldsymbol{j}$

b) $K=dbinom{-1}{3}$

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Question 2: $boldsymbol{a}$ has a position vector of $dbinom{7}{-2}$ and $boldsymbol{b}$ has a position vector of $dbinom{-3}{-1}$ , what is $2boldsymbol{a}-4boldsymbol{b}$ ?

Give your answer as a column vector.

[2 marks]

$2boldsymbol{a}-4boldsymbol{b}=dbinom{2times7}{2times-2}-dbinom{4times-3}{4times-1}=dbinom{14}{-4}-dbinom{-12}{-4}=dbinom{14-(-12)}{-4-(-4)}=dbinom{26}{0}$

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Question 3: Points $X$ and $Y$ have position vectors $5boldsymbol{i}-boldsymbol{j}$ and $-3boldsymbol{i}+4boldsymbol{j}$ respectively. Point $Z$ lies on the line $XY$ , such that $XZ:ZY=2:3$ .