# How do you find the angle between the vectors u=2i-3j and v=i-2j?

##### 1 Answer
Apr 14, 2017

So the acute angle between the vectors is 7.122°(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:

$\vec{u} = 2 \hat{i} - 3 \hat{j}$ and $\vec{v} = \hat{i} - 2 \hat{j}$

The modulus is given by;

$| | \vec{u} | | = \sqrt{{\left(2\right)}^{2} + {\left(- 3\right)}^{2}} = \sqrt{4 + 9} = \sqrt{13}$
$| | \vec{v} | | = \sqrt{{\left(1\right)}^{2} + {\left(- 2\right)}^{2}} = \sqrt{1 + 4} = \sqrt{5}$

And the scaler product is:

$\vec{u} \cdot \vec{v} = \left(2\right) \left(1\right) + \left(- 3\right) \left(- 2\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 2 + 6$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 8$

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

$8 = s q t \left(13\right) \sqrt{5} \cos \theta$
$\therefore \cos \theta = \frac{8}{\sqrt{13} \sqrt{5}}$
$\therefore \cos \theta = 0.992277 \ldots$
 :. theta = 7.1250 ... °
 :. theta = 7.13 °(3sf)

So the acute angle between the vectors is 7.13°(3sf)