# What is the angle between <5,8,2 > and < 4,6,-1 >?

Jun 26, 2016

Angle between vectors $< 5 , 8 , 2 >$ and $< 4 , 6 , - 1 >$ is ${28.682}^{o}$

#### Explanation:

The angle between vectors $A = < {a}_{1} , {a}_{2} , {a}_{3} >$ and $B = < {b}_{1} , {b}_{2} , {b}_{3} >$ is given by

$A \cdot B = | A | | B | \cos \theta$, where $A \cdot B$ is the dot product and $| A |$ and $| B |$ are absolute value of the two vectors.

As $A = < 5 , 8 , 2 >$ and $B = < 4 , 6 , - 1 >$, hence

$A \cdot B = 5 \cdot 4 + 8 \cdot 6 + 2 \cdot \left(- 1\right) = 20 + 48 - 2 = 66$

$| A | = \sqrt{{5}^{2} + {8}^{2} + {2}^{2}} = \sqrt{25 + 64 + 4} = \sqrt{93}$

$| B | = \sqrt{{4}^{2} + {6}^{2} + {\left(- 1\right)}^{2}} = \sqrt{16 + 36 + 1} = \sqrt{53}$

Hence $\cos \theta = \frac{66}{\sqrt{93} \times \sqrt{53}} = \frac{66}{9.64365 \times 7.28011}$

= $\frac{66}{75.231} = 0.8773$

and $\theta = {28.682}^{o}$