What is the angle between <-8,2,8> and < 2,-3,5>?

Oct 19, 2016

Angle between $< - 8 , 2 , 8 >$ and $< 2 , - 3 , 5 >$ is ${75.28}^{o}$

Explanation:

Angle between two vectors $\vec{u} = {a}_{1} \hat{i} + {b}_{1} \hat{j} + {c}_{1} \hat{k}$ or $< {a}_{1} , {b}_{1} , {c}_{1} >$

and $\vec{v} = {a}_{2} \hat{i} + {b}_{2} \hat{j} + {c}_{2} \hat{k}$ or $< {a}_{2} , {b}_{2} , {c}_{2} >$ is given by

$\cos \theta = \frac{\left(\vec{u} \cdot \vec{v}\right)}{\left(| \vec{u} | \cdot | \vec{v} |\right)}$,

where $\vec{u} \cdot \vec{v} = {a}_{1} {a}_{2} + {b}_{1} {b}_{2} + {c}_{1} {c}_{2}$

and $| \vec{u} |$ or $| \vec{v} |$ are magnitudes of vectors $\vec{u}$ or $\vec{v}$ and here they are

$\sqrt{{a}_{1}^{2} + {b}_{1}^{2} + {c}_{1}^{2}}$ and $\sqrt{{a}_{2}^{2} + {b}_{2}^{2} + {c}_{2}^{2}}$

Hence angle between $< - 8 , 2 , 8 >$ and $< 2 , - 3 , 5 >$ is given by

$\cos \theta = \frac{\left(- 8\right) \times 2 + 2 \times \left(- 3\right) + 8 \times 5}{\sqrt{{\left(- 8\right)}^{2} + {2}^{2} + {8}^{2}} \times \sqrt{{\left(2\right)}^{2} + {\left(- 3\right)}^{2} + {5}^{2}}}$

= $\frac{- 16 - 6 + 40}{\sqrt{64 + 4 + 64} \times \sqrt{4 + 9 + 25}}$

= $\frac{18}{\sqrt{132} \times \sqrt{38}}$

= $\frac{18}{11.4891 \times 6.1644}$

= $0.25415$

and $\theta = {75.28}^{o}$