# How do you find the angle between the vectors u=cos(pi/3)i+sin(pi/3)j and v=cos((3pi)/4)i+sin((3pi)/4)j?

Dec 14, 2016

$\frac{7 \pi}{12}$

#### Explanation:

The angle $\theta$ between two vectors $\vec{A}$ and $\vec{B}$ is related to the modulus (or magnitude) and scaler (or dot) product of $\vec{A}$ and $\vec{A}$ by the relationship:

$\vec{A} \cdot \vec{B} = | A | | B | \cos \theta$

By convention when we refer to the angle between vectors we choose the acute angle.

So for this problem, let the angle betwen $\vec{u}$ and $\vec{v}$ be $\theta$ then:

$\vec{u} = \cos \left(\frac{\pi}{3}\right) \underline{\hat{i}} + \sin \left(\frac{\pi}{3}\right) \underline{\hat{j}} = \frac{1}{2} \underline{\hat{i}} + \frac{\sqrt{3}}{2} \underline{\hat{j}} = \left(\begin{matrix}\frac{1}{2} \\ \frac{\sqrt{3}}{2}\end{matrix}\right)$
$\vec{v} = \cos \left(\frac{3 \pi}{4}\right) \underline{\hat{i}} + \sin \left(\frac{3 \pi}{4}\right) \underline{\hat{j}} = - \frac{\sqrt{2}}{2} \underline{\hat{i}} + \frac{\sqrt{2}}{2} \underline{\hat{j}} = \left(\begin{matrix}- \frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2}\end{matrix}\right)$

The modulus is given by;

$| \vec{u} | = | \left(\begin{matrix}\frac{1}{2} \\ \frac{\sqrt{3}}{2}\end{matrix}\right) | \setminus \setminus \setminus = \sqrt{{\left(\frac{1}{2}\right)}^{2} + {\left(\frac{\sqrt{3}}{2}\right)}^{2}} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1$
$| \vec{v} | = | \left(\begin{matrix}- \frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2}\end{matrix}\right) | = \sqrt{{\left(- \frac{\sqrt{2}}{2}\right)}^{2} + {\left(\frac{\sqrt{2}}{2}\right)}^{2}} = \sqrt{\frac{2}{4} + \frac{2}{4}} = \sqrt{1} = 1$

And the scaler product is:

$\vec{u} \cdot \vec{v} = \left(\begin{matrix}\frac{1}{2} \\ \frac{\sqrt{3}}{2}\end{matrix}\right) \cdot \left(\begin{matrix}- \frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2}\end{matrix}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\frac{1}{2}\right) \left(- \frac{\sqrt{2}}{2}\right) + \left(\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{2}}{2}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = - \frac{\sqrt{2}}{4} + \frac{\sqrt{2} \sqrt{3}}{4}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{\sqrt{2} \left(\sqrt{3} - 1\right)}{4}$

And so using $\vec{A} \cdot \vec{B} = | A | | B | \cos \theta$ we have:

$\frac{\sqrt{2} \left(\sqrt{3} - 1\right)}{4} = 1 \cdot 1 \cdot \cos \theta$
$\therefore \cos \theta = \frac{\sqrt{2} \left(\sqrt{3} - 1\right)}{4}$
$\therefore \cos \theta = 0.258819 \ldots$
$\therefore \theta = \frac{7 \pi}{12}$

So the acute angle between the vectors is $\frac{7 \pi}{12}$

We can confirm these results graphically:

Dec 14, 2016

The vector

$< a \cos \alpha , a \sin \alpha >$

is in the direction

$\theta = \alpha$.

If a = 1, it is a unit vector, in the direction.

Here the unit vectors,

$v = < \cos \left(\frac{3}{4} \pi\right) , \sin \left(\frac{3}{4} \pi\right) >$ and

$u = < \cos \frac{\pi}{3} , \sin \frac{\pi}{3} >$

are in the directions

$\theta = \frac{3}{4} \pi \mathmr{and} \frac{\pi}{3}$, respectively.

So, the angle $u \to v$, in between, is $\frac{3}{4} \pi - \frac{\pi}{3} = \frac{5}{12} \pi$.

In the sense $u \to v$, it is the supplement $\pi - \frac{5}{12} \pi = \frac{7}{12} \pi$.