 # What Are Basis Vectors Used For?

Knowledge of basis vectors is important for your basic understanding of vectors. Theory

### ImportantConcepts

• A basis is a set of vectors you can combine in different ways to create all the other vectors in a plane.

• A basis vector is a vector from this set.

Theory

### BasisVector

If you have two vectors $\stackrel{\to }{x}$ and $\stackrel{\to }{y}$ that are not parallel, you can use them as basis vectors. These basis vectors can be multiplied by any real number. This means that all other vectors in the plane can be written as a sum of these vectors multiplied by a number.

Example 1

Given the basis vectors $\phantom{\rule{-0.17em}{0ex}}\left(1,-1\right)$ and $\phantom{\rule{-0.17em}{0ex}}\left(3,4\right)$, write $\phantom{\rule{-0.17em}{0ex}}\left(11,10\right)$ using the basis vectors.

$\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}\left(11,10\right)& =k\phantom{\rule{-0.17em}{0ex}}\left(1,-1\right)+l\phantom{\rule{-0.17em}{0ex}}\left(3,4\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(k,-k\right)+\phantom{\rule{-0.17em}{0ex}}\left(3l,4l\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $\phantom{\rule{-0.17em}{0ex}}\left(11,10\right)=k\phantom{\rule{-0.17em}{0ex}}\left(1,-1\right)+l\phantom{\rule{-0.17em}{0ex}}\left(3,4\right)=\phantom{\rule{-0.17em}{0ex}}\left(k,-k\right)+\phantom{\rule{-0.17em}{0ex}}\left(3l,4l\right)$

$\begin{array}{llll}\hfill k+3l& =11\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill k& =11-3l\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill \\ \hfill -k+4l& =10\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ⇓\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -\left(11-3l\right)+4l& =10\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -11+3l+4l& =10\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 7l& =21\phantom{\rule{1em}{0ex}}|:7\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill l& =3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill \\ \hfill k& =11-3l\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ⇓\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill k& =11-3\cdot 3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill k& =2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llllllll}\hfill k+3l& =11\phantom{\rule{2em}{0ex}}& \hfill -k+4l& =10\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill k& =11-3l\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill \\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill -\phantom{\rule{-0.17em}{0ex}}\left(11-3l\right)+4l& =10\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill -11+3l+41& =10\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill 7l& =21\phantom{\rule{1em}{0ex}}|:7\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill l& =3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill k& =11-3\cdot 3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill k& =2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

This allows you to write
$\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}\left(11,10\right)& =2\phantom{\rule{-0.17em}{0ex}}\left(1,-1\right)+3\phantom{\rule{-0.17em}{0ex}}\left(3,4\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(2,-2\right)+\phantom{\rule{-0.17em}{0ex}}\left(9,12\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $\phantom{\rule{-0.17em}{0ex}}\left(11,10\right)=2\phantom{\rule{-0.17em}{0ex}}\left(1,-1\right)+3\phantom{\rule{-0.17em}{0ex}}\left(3,4\right)=\phantom{\rule{-0.17em}{0ex}}\left(2,-2\right)+\phantom{\rule{-0.17em}{0ex}}\left(9,12\right)$

You can see that the vector $\phantom{\rule{-0.17em}{0ex}}\left(11,10\right)$ can be written as a sum of the two basis vectors multiplied by the constants 2 and 3.

Example 2

Given the vectors $\begin{array}{llll}\hfill \stackrel{\to }{u}& =2\stackrel{\to }{a}-\stackrel{\to }{b},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \stackrel{\to }{v}& =-6\stackrel{\to }{a}-2\stackrel{\to }{b},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

find the angle between the vectors $\stackrel{\to }{u}$ and $\stackrel{\to }{v}$ when $\phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{a}|=2$, $\phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{b}|=3$ and $\stackrel{\to }{a}\cdot \stackrel{\to }{b}=3$

The angle between $\stackrel{\to }{u}$ and $\stackrel{\to }{v}$ is

 $\mathrm{cos}\alpha =\frac{\stackrel{\to }{u}\cdot \stackrel{\to }{v}}{\phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{u}|\cdot \phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{v}|}$

First, you need to find $\stackrel{\to }{u}\cdot \stackrel{\to }{v}$: $\begin{array}{llll}\hfill \stackrel{\to }{u}\cdot \stackrel{\to }{v}& =\phantom{\rule{-0.17em}{0ex}}\left(2\stackrel{\to }{a}-\stackrel{\to }{b}\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(-6\stackrel{\to }{a}-2\stackrel{\to }{b}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-12{\stackrel{\to }{a}}^{2}-4\stackrel{\to }{a}\cdot \stackrel{\to }{b}+6\stackrel{\to }{a}\cdot \stackrel{\to }{b}+2{\stackrel{\to }{b}}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-12{\stackrel{\to }{a}}^{2}+2\stackrel{\to }{a}\cdot \stackrel{\to }{b}+2{\stackrel{\to }{b}}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-12\phantom{\rule{-0.17em}{0ex}}{|\stackrel{\to }{a}|}^{2}+2\stackrel{\to }{a}\cdot \stackrel{\to }{b}+2\phantom{\rule{-0.17em}{0ex}}{|\stackrel{\to }{b}|}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-12\cdot {2}^{2}+2\cdot 3+2\cdot {3}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-48+6+18\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-24\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

You also need to calculate the length of $\stackrel{\to }{u}$ and $\stackrel{\to }{v}$. Remember that ${\stackrel{\to }{a}}^{2}=\phantom{\rule{-0.17em}{0ex}}{|\stackrel{\to }{a}|}^{2}$. $\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}{|\stackrel{\to }{u}|}^{2}& =\left(2\stackrel{\to }{a}-\stackrel{\to }{b}\right)\cdot \left(2\stackrel{\to }{a}-\stackrel{\to }{b}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =4\phantom{\rule{-0.17em}{0ex}}{|\stackrel{\to }{a}|}^{2}-4\stackrel{\to }{a}\cdot \stackrel{\to }{b}+\phantom{\rule{-0.17em}{0ex}}{|\stackrel{\to }{b}|}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =4\cdot {2}^{2}-4\cdot 3+{3}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =16-12+9\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =13\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{u}|& =\sqrt{\phantom{\rule{-0.17em}{0ex}}{|\stackrel{\to }{u}|}^{2}}=\sqrt{13},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \phantom{\rule{-0.17em}{0ex}}{|\stackrel{\to }{v}|}^{2}& =\left(-6\stackrel{\to }{a}-2\stackrel{\to }{b}\right)\cdot \left(-6\stackrel{\to }{a}-2\stackrel{\to }{b}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =36\phantom{\rule{-0.17em}{0ex}}{|\stackrel{\to }{a}|}^{2}-24\stackrel{\to }{a}\cdot \stackrel{\to }{b}+4\phantom{\rule{-0.17em}{0ex}}{|\stackrel{\to }{b}|}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =36\cdot {2}^{2}-24\cdot 3+4\cdot {3}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =144-72+36\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =108\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{v}|& =\sqrt{\phantom{\rule{-0.17em}{0ex}}{|\stackrel{\to }{v}|}^{2}}=\sqrt{108}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Now, you can insert the expressions back into the formula and find the angle. The angle between $\stackrel{\to }{u}$ and $\stackrel{\to }{v}$ becomes $\begin{array}{llll}\hfill \mathrm{cos}\alpha & =\frac{-24}{\sqrt{13}\cdot \sqrt{108}}\approx -0.64,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \alpha & \approx {\mathrm{cos}}^{-1}\phantom{\rule{-0.17em}{0ex}}\left(-0.64\right)\approx 130.1\text{°}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$