# What is the angle between <2,7,-1 > and < 3,2,-8 >?

May 28, 2016

$\alpha = {89.61}^{o}$

#### Explanation:

$\text{find :} A \cdot B$

$A = < 2 , 7 , - 1 >$

${A}_{x} = 2$
${A}_{y} = 7$
${A}_{z} = - 1$

$B = < 3 , 2 , - 8 >$

${B}_{x} = 3$
${B}_{y} = 2$
${B}_{z} = - 8$

$A \cdot B = {A}_{x} \cdot {B}_{x} + {A}_{y} \cdot {B}_{y} + {A}_{z} \cdot {B}_{z}$

$A \cdot B = 2 \cdot 3 + 7 \cdot 2 + \left(\left(- 1\right) \cdot \left(- 8\right)\right)$
$A \cdot B = 6 + 14 + 8$
$A \cdot B = 28 \text{ ;dot product for A and B}$

$\text{find :"||A||" magnitude of A}$

$| | A | | = \sqrt{{A}_{x}^{2} + {A}_{y}^{2} + {A}_{z}^{2}}$

$| | A | | = \sqrt{{2}^{2} + {7}^{2} + {\left(- 1\right)}^{2}}$

$| | A | | = \sqrt{4 + 49 + 1} \text{ ;} | | A | | = \sqrt{54}$

$\text{find :"||B||" magnitude of B}$

$| | B | | = \sqrt{{B}_{x}^{2} + {B}_{y}^{2} + {B}_{z}^{2}}$

$| | B | | = \sqrt{{3}^{2} + {2}^{2} + {\left(- 8\right)}^{2}} = \sqrt{9 + 4 + 64}$

$| | B | | = \sqrt{77}$

$\cos \alpha = \frac{A \cdot B}{| | A | | \cdot | | B | |}$

$\cos \alpha = \frac{28}{\sqrt{54} \cdot \sqrt{77}}$

$\cos \alpha = \frac{28}{\sqrt{54 \cdot 77}}$

$\cos \alpha = \frac{28}{4158}$

$\cos \alpha = 0.0067340067$

$\alpha = {89.61}^{o}$