If #A = <-5 ,2 ,1 >#, #B = <-8 ,4 ,2 ># and #C=A-B#, what is the angle between A and C?

1 Answer
Jun 6, 2017

The angle is #=167.4º#

Explanation:

Let's start by calculating

#vecC=vecA-vecB#

#vecC=〈-5,2,1〉-〈-8,4,2〉=〈3,-2,-1〉#

The angle between #vecA# and #vecC# is given by the dot product definition.

#vecA.vecC=∥vecA∥*∥vecC∥costheta#

Where #theta# is the angle between #vecA# and #vecC#

The dot product is

#vecA.vecC=〈-5,2,1〉.〈3,-2,-1〉=-15-4-1=-20#

The modulus of #vecA#= #∥〈-5,2,1〉∥=sqrt(25+4+1)=sqrt30#

The modulus of #vecC#= #∥〈3,-2,-1〉∥=sqrt(9+4+1)=sqrt14#

So,

#costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=-20/(sqrt30*sqrt14)=-0.98#

#theta=167.4#º