# If A = <8 ,9 ,4 >, B = <6 ,-9 ,-1 > and C=A-B, what is the angle between A and C?

Jun 21, 2018

The angle is $= {33.4}^{\circ}$

#### Explanation:

Start by calculating

$\vec{C} = \vec{A} - \vec{B}$

vecC=〈8,9,4〉-〈6,-9,-1〉=〈2,18,6〉

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

vecA.vecC=∥vecA∥*∥vecC∥costheta

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

The dot product is

vecA.vecC=〈8,9,4〉.〈2,18,6〉=16+162+24=202

The modulus of $\vec{A}$= ∥〈8,9,4〉∥=sqrt(64+81+16)=sqrt161

The modulus of $\vec{C}$= ∥〈2,18,6〉∥=sqrt(4+324+36)=sqrt364

So,

costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=202/(sqrt161*sqrt364)=0.83

$\theta = \arccos \left(0.83\right) = {33.4}^{\circ}$