If #A = <5 ,-1 ,3 >#, #B = <2 ,-2 ,5 ># and #C=A-B#, what is the angle between A and C?

1 Answer
Feb 14, 2017

The angle is #=68.8#º

Explanation:

Let's start by calculating

#vecC=vecA-vecB#

#vecC=〈5,-1,3〉-〈2,-2,5〉=〈3,1,-2〉#

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,-1,3〉.〈3,1,-2〉=15-1-6=8#

The modulus of #vecA#= #∥〈5,-1,3〉∥=sqrt(25+1+9)=sqrt35#

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

So,

#costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=8/(sqrt35*sqrt14)=0.36#

#theta=68.8#º