# If vecu=(3,-1) and vecv=(-6,4), then how do you find the angle between u and v vectors?

## If $\vec{u} = \left(3 , - 1\right)$ and $\vec{v} = \left(- 6 , 4\right)$, then how do you find the angle between $u$ and $v$ vectors?

Jun 14, 2018

See below

#### Explanation:

We know that dot product of two vectors is defined as

vecv·vecu=abs(u)·abs(v)·costheta (1)

with $\theta$ angle between both vectors

By other hand, we know also that absu=sqrt(u_1^2+u_2^2

Then $\left\mid u \right\mid = \sqrt{9 + 1} = \sqrt{10}$
$\left\mid v \right\mid = \sqrt{36 + 16} = \sqrt{52}$

And vecv·vecu=v_1u_1+v_2u_2=abs(u)·abs(v)·costheta (1)

Applying this to our case

vecu·vecv=-18-4=-22 (2)

vecu·vecv=sqrt10·sqrt52·costheta (3)

Combining (2) and (3) costheta=-22/(sqrt10·sqrt52)=-0.96476382

theta=arccos -0.96476382=164º 44´41.57´´

Jun 14, 2018

Compute the dot product using $\vec{u} \cdot \vec{v} = \left({u}_{x}\right) \left({v}_{x}\right) + \left({u}_{y}\right) \left({v}_{y}\right)$
Compute the magnitudes, $| \vec{u} | \mathmr{and} | \vec{v} |$
Use the dot product formula, $\vec{u} \cdot \vec{v} = | \vec{u} | | \vec{v} | \cos \left(\theta\right)$, to find the angle.

#### Explanation:

Compute the dot product using $\vec{u} \cdot \vec{v} = \left({u}_{x}\right) \left({v}_{x}\right) + \left({u}_{y}\right) \left({v}_{y}\right)$

$\vec{u} \cdot \vec{v} = \left(3\right) \left(- 6\right) + \left(- 1\right) \left(4\right)$

$\vec{u} \cdot \vec{v} = - 22$

Compute the magnitudes, $| \vec{u} | \mathmr{and} | \vec{v} |$

$| \vec{u} | = \sqrt{{u}_{x}^{2} + {u}_{y}^{2}}$

$| \vec{u} | = \sqrt{{3}^{2} + {\left(- 1\right)}^{2}}$

$| \vec{u} | = \sqrt{10}$

$| \vec{v} | = \sqrt{{v}_{x}^{2} + {v}_{y}^{2}}$

$| \vec{v} | = \sqrt{{\left(- 6\right)}^{2} + {4}^{2}}$

$| \vec{v} | = 2 \sqrt{13}$

Use the dot product formula, $\vec{u} \cdot \vec{v} = | \vec{u} | | \vec{v} | \cos \left(\theta\right)$, to find the angle.

$- 22 = \left(\sqrt{10}\right) \left(2 \sqrt{13}\right) \cos \left(\theta\right)$

$\theta = {\cos}^{-} 1 \left(- \frac{22}{\left(\sqrt{10}\right) \left(2 \sqrt{13}\right)}\right)$

$\theta = {\cos}^{-} 1 \left(- \frac{22}{\left(\sqrt{10}\right) \left(2 \sqrt{13}\right)}\right)$

$\theta = {164.7}^{\circ}$