# What is the angle between <2,9,-2>  and <0,3?0 >?

${17.446}^{\setminus} \circ$

#### Explanation:

The angle $\setminus \theta$ between the vectors $< 2 , 9 , - 2 >$ & $< 0 , 3 , 0 >$ is given as

$\setminus \cos \setminus \theta = \setminus \frac{\left(2 i + 9 j - 2 k\right) \setminus \cdot \left(0 i + 3 j + 0 k\right)}{\setminus \sqrt{{2}^{2} + {9}^{2} + {\left(- 2\right)}^{2}} \setminus \sqrt{{0}^{2} + {3}^{2} + {0}^{2}}}$

$\setminus \cos \setminus \theta = \setminus \frac{0 + 27 + 0}{\setminus \sqrt{89} \setminus \cdot 3}$

$\setminus \cos \setminus \theta = \setminus \frac{9}{\setminus \sqrt{89}}$

$\setminus \theta = \setminus {\cos}^{- 1} \left(\frac{9}{\setminus} \sqrt{89}\right)$

$\setminus \theta = {17.446}^{\setminus} \circ$

Jul 26, 2018

The angle is $= {17.45}^{\circ}$

#### Explanation:

The angle between $\vec{A}$ and $\vec{B}$ is given by the dot product definition.

vecA.vecB=∥vecA∥*∥vecB∥costheta

Where $\theta$ is the angle between $\vec{A}$ and $\vec{B}$

The dot product is

vecA.vecB=〈2,9,-2〉.〈0,3,0〉=0+27-0=27

The modulus of $\vec{A}$= ∥〈2,9,-2〉∥=sqrt(4+81+4)=sqrt89

The modulus of $\vec{B}$= ∥〈0,3,0〉∥=sqrt(9)=3

So,

costheta=(vecA.vecB)/(∥vecA∥*∥vecB∥)=27/(sqrt89*3)=0.954

$\theta = \arccos \left(0.954\right) = {17.45}^{\circ}$