# How do you find the angle between the vectors u=<3, 2> and v=<4, 0>?

Dec 13, 2016

The angle is  33.7 °(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 $\vec{B}$ 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:

$\vec{u} = \left\langle3 , 2\right\rangle$ and $\vec{v} = \left\langle4 , 0\right\rangle$

The modulus is given by;

$| \vec{u} | = | \left\langle3 , 2\right\rangle | = \sqrt{{3}^{2} + {2}^{2}} = \sqrt{9 + 4} = \sqrt{13}$
$| \vec{v} | = | \left\langle4 , 0\right\rangle | = \sqrt{{4}^{2} + {0}^{2}} = \sqrt{16} = 4$

And the scaler product is:

$\vec{u} \cdot \vec{v} = \left\langle3 , 2\right\rangle \cdot \left\langle4 , 0\right\rangle$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(3\right) \left(4\right) + \left(2\right) \left(0\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(12\right)$

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

$12 = \sqrt{13} \cdot 4 \cdot \cos \theta$
$\therefore \cos \theta = \frac{12}{4 \sqrt{13}}$
$\therefore \cos \theta = 0.83205 \ldots$
 :. theta = 33.69006 °
 :. theta = 33.7 °(3sf)