# How do I find the angle between vectors <3, 0> and <5, 5>?

Sep 27, 2014

Let's begin by naming our vectors.

v$= < 3 , 0 >$
w$= < 5 , 5 >$

Dot product $\to v \cdot w$

Angle between formula $\to \cos \left(\theta\right) = \frac{v \cdot w}{| | v | | \cdot | | w | |}$

Solve for $\theta$

${\cos}^{-} 1 \left(\cos \left(\theta\right)\right) = {\cos}^{-} 1 \left(\frac{v \cdot w}{| | v | | \cdot | | w | |}\right)$

$\theta = {\cos}^{-} 1 \left(\frac{v \cdot w}{| | v | | \cdot | | w | |}\right)$

Begin by finding the dot product of vectors $v \mathmr{and} w$ by adding the products of the horizontal and vertical components.

$\theta = {\cos}^{-} 1 \left(\frac{\left(3\right) \left(5\right) + \left(0\right) \left(5\right)}{| | v | | \cdot | | w | |}\right)$

$\theta = {\cos}^{-} 1 \left(\frac{15 + 0}{| | v | | \cdot | | w | |}\right)$

Now find the magnitudes of both vectors

$\theta = {\cos}^{-} 1 \left(\frac{15}{\sqrt{9 + 0} \cdot \sqrt{25 + 25}}\right)$

$\theta = {\cos}^{-} 1 \left(\frac{15}{\sqrt{9} \cdot \sqrt{50}}\right)$

$\theta = {\cos}^{-} 1 \left(\frac{15}{\sqrt{3 \cdot 3 \cdot 5 \cdot 5 \cdot 2}}\right)$

$\theta = {\cos}^{-} 1 \left(\frac{15}{15 \sqrt{2}}\right)$

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

$\theta = \frac{\pi}{4}$