If A = <-6 ,2 ,1 >, B = <-8 ,8 ,4 > and C=A-B, what is the angle between A and C?

Jan 23, 2017

The angle is $= 127$º

Explanation:

Let's start by calculating

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

vecC=〈-6,2,1〉-〈-8,8,4〉=〈2,-6,-3〉

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=〈-6,2,1〉.〈2,-6,-3〉=-12-12-3=-27

The modulus of $\vec{A}$= ∥〈-6,2,1〉∥=sqrt(36+4+1)=sqrt41

The modulus of $\vec{C}$= ∥〈2,-6,-3〉∥=sqrt(4+36+9)=sqrt49=7

So,

costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=-27/(sqrt41*7)=-0.6

$\theta = 127$º