## Two-Dimensional Vectors

Vectors are arrays of vertically stored values and they have a magnitude and a direction. A two-dimensional vector has two entries – one for displacement in the x direction and one for the y direction. A vector identifies an amount of displacement and this displacement can refer to anywhere in the plane. A position vector, however, points from the origin.

## Addition & Scalar Multiplication of Vectors

Example 1 –  $\left(\begin{array}{c}2\\5\end{array}\right)+\left(\begin{array}{c}4\\3\end{array}\right)=\left(\begin{array}{c}6\\8\end{array}\right)$

Example 2 –  $\left(\begin{array}{c}3\\-5\end{array}\right)+\left(\begin{array}{c}-4\\-2\end{array}\right)=\left(\begin{array}{c}-1\\-7\end{array}\right)$

Example 3 –  $2\left(\begin{array}{c}-1\\6\end{array}\right)=\left(\begin{array}{c}-2\\12\end{array}\right)$

Example 4 –  $-3\left(\begin{array}{c}3\\-2\end{array}\right)=\left(\begin{array}{c}-9\\6\end{array}\right)$

## Magnitude and Direction of Vectors

Vectors have both magnitude and direction. The overall displacement, with all directions considered, identifies the direction of the vector. The magnitude of the vector is given by the length of the overall displacement – this is found using Pythagoras.

## Position Vectors & Distances

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:

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

The length of a vector can be found using pythagoras.

Example – If

$\overrightarrow{OA}=\left(\begin{array}{c}2\\-4\end{array}\right)$, $\overrightarrow{OB}=\left(\begin{array}{c}3\\1\end{array}\right)$

then

$\overrightarrow{AB}=\left(\begin{array}{c}3\\1\end{array}\right)-\left(\begin{array}{c}2\\-4\end{array}\right)=\left(\begin{array}{c}1\\5\end{array}\right)$.

The length of this vector can then be found using pythagoras:

$\vert\overrightarrow{AB}\vert=\sqrt{1^2+5^2}=\sqrt{26}$