# What is the angle between <-4,3,-8 > and < 6,-3,8 >?

Oct 19, 2016

Angle between $< - 4 , 3 , - 8 >$ and $< 6 , - 3 , 8 >$ is ${170.01}^{o}$

#### 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 $< - 4 , 3 , - 8 >$ and $< 6 , - 3 , 8 >$ is given by

$\cos \theta = \frac{\left(- 4\right) \times 6 + 3 \times \left(- 3\right) + \left(- 8\right) \times 8}{\sqrt{{\left(- 4\right)}^{2} + {3}^{2} + {\left(- 8\right)}^{2}} \times \sqrt{{6}^{2} + {\left(- 3\right)}^{2} + {8}^{2}}}$

= $\frac{- 24 - 9 - 64}{\sqrt{16 + 9 + 64} \times \sqrt{36 + 9 + 64}}$

= $- \frac{97}{\sqrt{89} \times \sqrt{109}}$

= $- \frac{97}{9.43398 \times 10.44031}$

= $- 0.9848$

and $\theta = {170.01}^{o}$