# What is the angle between <9,7,1> and <8,3,6> ?

Mar 2, 2016

$\alpha \approx {34.06}^{\circ}$

#### Explanation:

The angle $\alpha$ between the two vectors

$\vec{u} = < 9 , 7 , 1 >$ and $\vec{v} = < 8 , 3 , 6 >$

can be computed with the formula

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

where the product in the numerator is a dot product.

$\cos \alpha = \frac{9 \cdot 8 + 7 \cdot 3 + 1 \cdot 6}{\sqrt{{9}^{2} + {7}^{2} + {1}^{2}} \cdot \sqrt{{8}^{2} + {3}^{2} + {6}^{2}}} = \frac{99}{\sqrt{131} \cdot \sqrt{109}} \approx 0.8285$
$\alpha \approx {34.06}^{\circ}$