9.1 Distances

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

11 1 The distance between a vector and the origin is the same thing as the magnitude of the vector.
ax→\vec{a_{x}}ay→\vec{a_{y}}

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

∥a∥=(ax)2+(ay)2\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 a→=(1,1)\vec{a}=(1,1) and another vector b→=(2,0)\vec{b}=(2,0)? What we can do is find the vector between the two points (this is written as a⁢b→\vec{ab}), and then use Pythagoras’ theorem in the same way we did above?

First, we can draw a diagram:

a→\vec{a}b→\vec{b}

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

a→\vec{a}b→\vec{b}−a→+b→-\vec{a}+\vec{b}

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

∥−a→+b→∥\displaystyle\left\lVert-\vec{a}+\vec{b}\right\rVert =∥[−1−1]+[20]∥\displaystyle=\left\lVert\begin{bmatrix}-1\\ -1\end{bmatrix}+\begin{bmatrix}2\\ 0\end{bmatrix}\right\rVert
=(−1+2)2+(−1)2\displaystyle=\sqrt{(-1+2)^{2}+(-1)^{2}}
=2\displaystyle=\sqrt{2}