# How do you find the angle between u=<-1,9> and v=<3,12>?

Nov 18, 2016

The angle between, $\theta = {\cos}^{-} 1 \left(\frac{\overline{u} \cdot \overline{v}}{| \overline{u} | | \overline{v} |}\right) \approx 1.5 \text{ radians}$

#### Explanation:

Given: $\overline{u} = < - 1 , 9 > \mathmr{and} \overline{v} = < 3 , 12 >$

Compute the dot-product:

$\overline{u} \cdot \overline{v} = \left(- 1\right) \left(3\right) + \left(9\right) \left(12\right) = 105$

Compute the magnitude of $\overline{u}$:

$| \overline{u} | = \sqrt{{\left(- 1\right)}^{2} + {9}^{2}} = \sqrt{82}$

Compute the magnitude of $\overline{v}$:

$| \overline{v} | = \sqrt{{3}^{2} + {12}^{2}} = \sqrt{153}$

Another equation for the dot-product is:

$\overline{u} \cdot \overline{v} = | \overline{u} | | \overline{v} | \cos \left(\theta\right)$

Solve for $\theta$:

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

Substitute in the computed values:

$\theta = {\cos}^{-} 1 \left(\frac{105}{\sqrt{82} \sqrt{153}}\right) \approx 1.5 \text{ radians}$