If A = <4 ,-5 ,-4 >, B = <5 ,1 ,-3 > and C=A-B, what is the angle between A and C?

Jan 9, 2017

The angle is $= 49.9$º

Explanation:

Let's start by calculating

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

vecC=〈4,-5,-4〉-〈5,1,-3〉=〈-1,-6,-1〉

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=〈4,-5,-4〉.〈-1,-6,-1〉=-4+30+4=30

The modulus of $\vec{A}$= ∥〈4,-5,-4〉∥=sqrt(16+25+16)=sqrt57

The modulus of $\vec{C}$= ∥〈-1,-6,-1〉∥=sqrt(1+36+1)=sqrt38

So,

costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=30/(sqrt57*sqrt38)=0.644

$\theta = 49.9$º