# How do you find the angle between the vectors u=2i-3j and v=8i+3j?

##### 1 Answer
Oct 18, 2016

${76.9}^{o}$

#### Explanation:

Starting from the saclar (dot) product

the definition:

$\vec{a} . \vec{b} = | a | | b | \cos \theta$-------(1)
where $\theta$ is the angle between $\vec{a}$ &$\vec{b}$

Also to evaluate the scalar product using components we have the result:

$\vec{a} = {x}_{\text{1"veci+y_"1}} \vec{j}$

$\vec{b} = {x}_{\text{2"veci+y_"2}} \vec{j}$

$\vec{a} . \vec{b} = {x}_{\text{1"x_"2"+y_1}}$${y}_{\text{2}}$------(2)

so for

$\vec{u} = 2 \vec{i} - 3 \vec{j}$ &$\vec{v} = 8 \vec{i} + 3 \vec{j}$

using result (2)

$\vec{u} . \vec{v} = \left(2 \times 8\right) + \left(\left(- 3\right) \times 3\right)$

$\vec{u} . \vec{v} = 16 - 9 = 7$

Now we use the definition (1)

$\vec{u} . \vec{v} = | u | | v | \cos \theta$

so, $7 = \sqrt{{2}^{2} + {3}^{2}} \times \sqrt{{8}^{2} + {3}^{2}} \times \cos \theta$

#

$7 = \sqrt{13} \times \sqrt{73} \times \cos \theta$

$\theta = {\cos}^{-} 1 \left(\frac{7}{\sqrt{13} \sqrt{73}}\right)$

$\theta = {76.9}^{o}$