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.

position vectors
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:


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.


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!

position vectors

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


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


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:

  1. his position vector at the end of his route relative to the depot, i.e. the vector \overrightarrow{OB},
  2. the distance (as the crow flies) between the postman at the end of the route and the depot,
  3. 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:

position 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 270^\circ to find the bearing of the postman relative to the depot:
    \text{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:

vectors in context

The cosine rule states that


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.

AS Maths Vectors

A2 Maths Vectors