# How do you find the angle between the vectors u=2i-j and v=6i+4j?

Oct 20, 2016

Angle between given vectors is ${55.55}^{o}$

#### Explanation:

If we have two vectors $\vec{u} = 2 \hat{i} - \hat{j}$ and $\vec{v} = 6 \hat{i} + 4 \hat{j}$

and angle between them is $\theta$, then

$\cos \theta = \frac{\vec{u} \cdot \vec{v}}{| u | \cdot | v |}$

= $\frac{2 \times 6 + \left(- 1\right) \times 4}{\sqrt{{2}^{2} + {\left(- 1\right)}^{2}} \times \sqrt{{6}^{2} + {4}^{2}}}$

= $\frac{12 - 4}{\sqrt{5} \times \sqrt{40}}$

= $\frac{8}{\sqrt{200}} = \frac{8}{\sqrt{5 \times 5 \times 8}}$

= $\frac{8}{5 \sqrt{8}}$

= $\frac{\sqrt{8}}{5} = 0.5657$

Hence $\theta = {55.55}^{o}$