Vectors in Context

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

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

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

1. The position vector of the postman at the end of his route relative to the depot is $\overrightarrow{OB}\left(\begin{array}{c}-2\\-5\end{array}\right)$
2. 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.
3. 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 $201.8^\circ$.

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.