# How do you find the angle between the vectors u=6i+3j and v=-4i+4j?

Dec 13, 2016

72°(3sf)

#### 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:

$\vec{u} = 6 \underline{\hat{i}} + 3 \underline{\hat{j}} = \left(\begin{matrix}6 \\ 3\end{matrix}\right)$ and $\vec{v} = - 4 \underline{\hat{i}} + 4 \underline{\hat{j}} = \left(\begin{matrix}- 4 \\ 4\end{matrix}\right)$

The modulus is given by;

$| \vec{u} | = | \left(\begin{matrix}6 \\ 3\end{matrix}\right) | \setminus \setminus \setminus = \sqrt{{6}^{2} + {3}^{2}} = \sqrt{36 + 9} = \sqrt{45} = 3 \sqrt{5}$
$| \vec{v} | = | \left(\begin{matrix}- 4 \\ 4\end{matrix}\right) | = \sqrt{- {4}^{2} + {4}^{2}} = \sqrt{16 + 16} = \sqrt{32} = 4 \sqrt{2}$

And the scaler product is:

$\vec{u} \cdot \vec{v} = \left(\begin{matrix}6 \\ 3\end{matrix}\right) \cdot \left(\begin{matrix}- 4 \\ 4\end{matrix}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(6\right) \left(- 4\right) + \left(3\right) \left(4\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = - 24 + 12$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = - 12$

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

$- 12 = 3 \sqrt{5} \cdot 4 \sqrt{2} \cdot \cos \theta$
$\therefore \cos \theta = \frac{- 12}{12 \sqrt{5} \sqrt{2}}$
$\therefore \cos \theta = - 0.316227 \ldots$
 :. theta = 108.4349 °
 :. theta = 108 °(3sf)

So the acute angle between the vectors is 72°(3sf)