# If A = <8 ,0 ,4 >, B = <6 ,7 ,4 > and C=A-B, what is the angle between A and C?

Jun 12, 2017

The angle is =75.8º

#### Explanation:

Let's start by calculating

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

vecC=〈8,0,4〉-〈6,7,4〉=〈2,-7,0〉

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=〈8,0,4〉.〈2,-7,0〉=16+0+0=16

The modulus of $\vec{A}$= ∥〈8,0,4〉∥=sqrt(64+0+16)=sqrt80

The modulus of $\vec{C}$= ∥〈2,-7,0〉∥=sqrt(4+49+0)=sqrt53

So,

costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=16/(sqrt80*sqrt53)=0.25

$\theta = 75.8$º