# Vectors in Context

## Position Vectors

A position vector is any vector that is placed extending from the origin. Position vectors are often denoted by $\overrightarrow{OA}$, for example, to identify the vector that points from the origin to a point $A$.

Let $\overrightarrow{OA}$ and $\overrightarrow{OB}$ be the position vectors that point from the origin to the points $A$ and $B$ respectively. We can then find the vector that points from $A$ to $B$:

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

It is easy to visualise going from $A$ to $B$ by thinking of it as going backwards along $\overrightarrow{OA}$, then forwards along $\overrightarrow{OB}$.

There are a variety of problems involving position vectors. Examples 1 and 2 illustrate just two ways in which position vectors can be put into context.

## Vectors and Trigonometry

By now you will be very familiar with Pythagoras and SOHCAHTOA. This allows you to find angles and missing lengths in right-angled triangles. You may also have recently learned about non-right angled triangles. That is to say, you may have learned about the cosine rule, the sine rule and perhaps even how to find the area of a non-right angled triangle.

It is entirely possible that trigonometry may be used in the context of vectors. Examples 3 and 4 illustrate just two ways in which vectors can be using in a trigonometric setting.

## Examples

The points $A$ and $B$ have position vectors $3{\bf i}-4{\bf j}$ and $2{\bf j}-7{\bf I}$ respectively. The point $P$ splits the line vector $\overrightarrow{AB}$ in the ratio 2:3. Find the position vector $\overrightarrow{OP}$.

A lot of the information for finding the solution comes from sketching the problem. Be sure to read carefully!

The vector that joins the points $A$ and $B$ is given by

$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\left(\begin{array}{c}-7\\2\end{array}\right)-\left(\begin{array}{c}3\\-4\end{array}\right)=\left(\begin{array}{c}-10\\6\end{array}\right)$

Hence, the vector joining $A$ and $P$ is given by $\overrightarrow{OP}=\frac{2}{5}\left(\begin{array}{c}-10\\6\end{array}\right)=\left(\begin{array}{c}-4\\\frac{12}{5}\end{array}\right)$

It follows that

$\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{AP}=\left(\begin{array}{c}3\\-4\end{array}\right)+\left(\begin{array}{c}-4\\\frac{12}{5}\end{array}\right)=\left(\begin{array}{c}-1\\\frac{-8}{5}\end{array}\right)$

A postman picks up his letters from the depot (point $O$). He then heads west delivering letters for 2 km (to point $A$) before heading south down a road for 5 km (to point $B$). Calculate:

- his position vector at the end of his route relative to the depot, i.e. the vector $\overrightarrow{OB}$,
- the distance (as the crow flies) between the postman at the end of the route and the depot,
- and the bearing of the postman at the end of his route from the depot.

The route that the postman makes can be plotted using vectors:

- The position vector of the postman at the end of his route relative to the depot is $\overrightarrow{OB}\left(\begin{array}{c}-2\\-5\end{array}\right)$
- His distance from the depot can then be found using Pythagoras: $\vert \overrightarrow{OB}\vert=\sqrt{(-2)^2+(-5)^2}=\sqrt{29}$. This is approximately 5.4 km to 1 decimal place. See magnitude of vectors.
- Recall that bearings are measured clockwise from North and always have 3 digits (possible a zero 0 two at the beginning). SOHCAHTOA can be used to find angle $AOB$. This can then be subtracted from [math]270^\circ[/math] to find the bearing of the postman relative to the depot:

$\angle AOB =\tan^{-1}\left(\frac{5}{2}\right)=068.2^\circ$ to 1 decimal place. Hence, the postman’s bearing from the depot is $201.8^\circ$.

A ship leaves a port and heads to its destination for 65 km on a bearing of $071^\circ$ before receiving a distress signal. The distress signal is sent from a vessel with a position that is 54km away from and at a bearing of $122^\circ$ relative to the port. Calculate the distance between the vessel in distress and the ship.

This problem can be tackled in a number of ways but the cosine rule is the most direct. As always, sketch a diagram to help:

The cosine rule states that

$c^2=a^2+b^2-2ab\cos(C)$

where we choose $a=65$, $b=54$ and the angle between them is $C=122-71=51^\circ$. It follows that

$c^2=65^2+54^2-2\times 65\times 54\times \cos(51)=4855.51$

to two decimal places. Hence, the distance between the vessel and the ship is 69.7km to one decimal place.

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