# If A = <-7 ,1 ,5 >, B = <3 ,6 ,1 > and C=A-B, what is the angle between A and C?

Apr 4, 2017

The angle is $= 34.3$º

#### Explanation:

Let's start by calculating

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

vecC=〈-7,1,5〉-〈3,6,1〉=〈-10,-5,4〉

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=〈-7,1,5〉.〈-10,-5,4〉=70-5+20=85

The modulus of $\vec{A}$= ∥〈-7,1,5〉∥=sqrt(49+1+25)=sqrt75

The modulus of $\vec{C}$= ∥〈-10,-5,4〉∥=sqrt(100+25+16)=sqrt141

So,

costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=85/(sqrt75*sqrt141)=0.83

$\theta = 34.3$º