# What is the angle between <0,9,-6 >  and <1,3,5> ?

Dec 2, 2017

The angle is $= {92.7}^{\circ}$

#### Explanation:

The vectors are $\vec{A} = < 0 , 9 , - 6 >$ and $\vec{B} = < 1 , 3 , 5 >$

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=〈0,9,-6〉.〈1,3,5〉=(0xx1)+(9xx3)+(-6xx5)=0+27-30=-3

The modulus of $\vec{A}$= ∥〈0,9,-6〉∥=sqrt(0+9^2+(-6)^2)=sqrt(81+36)=sqrt(117)

The modulus of $\vec{B}$= ∥〈1,3,5〉∥=sqrt(1^2+3^2+5^2)=sqrt(1+9+25)=sqrt35

So,

costheta=(vecA.vecB)/(∥vecA∥*∥vecB∥)=-3/(sqrt117*sqrt35)=-0.0469

$\theta = \arccos \left(- 0.0469\right) = {92.7}^{\circ}$