# 9.1 Distances

Consider the vector $\vec{a}$. Let’s say that $a=(1,1)$. How far is $a$ from the origin?

11 1 The distance between a vector and the origin is the same thing as the magnitude of the vector.

To find the distance between the origin (the vector $(0,0)$) and $\vec{a}$ we can use Pythagoras’ theorem.

$\left\lVert a\right\rVert=\sqrt{(a_{x})^{2}+(a_{y})^{2}}$ (9.1)

How about if we want to find the distance between the vector $\vec{a}=(1,1)$ and another vector $\vec{b}=(2,0)$? What we can do is find the vector between the two points (this is written as $\vec{ab}$), and then use Pythagoras’ theorem in the same way we did above?

First, we can draw a diagram:

We have no clue how to find the vector between $\vec{a}$ and $\vec{b}$ ($\vec{ab}$). To find the vector between $\vec{a}$ and $\vec{b}$ can draw a diagram and think about what we do know. Remember that we can read $\vec{a}$ as "move from the origin to $(1,1)$" and $\vec{b}$ as "move from the origin to $(2,0)$". Then, to move from $\vec{a}$ to $\vec{b}$ we want to move "from $\vec{a}$ to the origin" and "from the origin to $\vec{b}$". This is $-\vec{a}+\vec{b}$. To find the distance (aka magnitude of this vector), we just use Pythagoras’s theorem.

Therefore the distance between $\vec{a}$ and $\vec{b}$ is

 $\displaystyle\left\lVert-\vec{a}+\vec{b}\right\rVert$ $\displaystyle=\left\lVert\begin{bmatrix}-1\\ -1\end{bmatrix}+\begin{bmatrix}2\\ 0\end{bmatrix}\right\rVert$ $\displaystyle=\sqrt{(-1+2)^{2}+(-1)^{2}}$ $\displaystyle=\sqrt{2}$