# Vectors

Vectors are quantities that have both magnitude and direction. Velocity is a vector (it has magnitude and direction), but speed is a scalar (it only has magnitude). We need to use vectors to solve two-dimensional problems.

There are many different ways to represent the direction of a vector. In this course, we will use a system based on compass directions. The direction is placed in square brackets after the magnitude and the unit, and it contains an angle in degrees placed between two compass directions. For example, [N 12º W] means that you start pointing north and then rotate 12º towards west to point in the correct direction. This is equivalent to [W 78º N].

To add or subtract vectors, we have to follow four steps:

- Break each vector up into its
*x*and*y*components using sine and cosine. - Combine the
*x*components and*y*components (add or subtract). - Use the Pythagorean theorem to find the resultant magnitude.
- Use the inverse tangent function to find the resultant direction.

## Example

If $\displaystyle \vec{a} = 65 \, \text{m [E 36º N]}$ and $\displaystyle \vec{b} = 99 \, \text{m [N 24º W]}$, calculate $\displaystyle \vec{a} + \vec{b}$.

First we should always draw a sketch. I’ve labelled the *x* and *y* components for each vector with a sign (north and east are positive, south and west are negative) and a trig function (cosine for adjacent, sine for opposite).

Now we can make our table. We calculate the components of each vector, and then we add them up in the last row.

x | y | |
---|---|---|

$\displaystyle \vec{a}$ | $\displaystyle 65 \cos{36} º = 52.59$ | $\displaystyle 65 \sin{36} º = 38.21$ |

$\displaystyle \vec{b}$ | $\displaystyle - 99 \sin{24} º = - 40.27$ | $\displaystyle 99 \cos{24} º = 90.44$ |

$\displaystyle \vec{a} + \vec{b}$ | $\displaystyle 12.32$ | $\displaystyle 128.65$ |

Now we can use the Pythagorean theorem to calculate the magnitude of our resultant vector, rounding off to two significant digits (because 65 and 99 have two):

$\displaystyle \left \lvert \vec{a} + \vec{b} \right \rvert = \sqrt{12.32^{2} + 128.65^{2}} = 1.3 \times 10^{2} \, \text{m}$.

We can draw another sketch to help with finding the angle. If you draw the largest component first, then you will always get an angle below 45º. If you always do it this way, then you’ll know that you’ve made a mistake if your angle is above 45º.

Now we can use inverse tangent to find theta (opposite over adjacent):

$\displaystyle \theta = \tan^{- 1}{\left ( \frac{12.32}{128.65} \right )} = 5.5 º$.

Therefore, the resultant vector
$\displaystyle \vec{a} + \vec{b}$
is 1.3 × 10^{2} m [N 5.5º E].