If #A = <5 ,6 ,-3 >#, #B = <-9 ,6 ,-2 ># and #C=A-B#, what is the angle between A and C?

1 Answer
Aug 1, 2017

The angle is #=51.6^@#

Explanation:

Let's start by calculating

#vecC=vecA-vecB#

#vecC=〈5,6,-3〉-〈-9,6,-2〉=〈14,0,-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=〈5,6,-3〉.〈14,0,-1〉=70+0+3=73#

The modulus of #vecA#= #∥〈5,6,-3〉∥=sqrt(25+36+9)=sqrt70#

The modulus of #vecC#= #∥〈14,0,-1〉∥=sqrt(196+0+1)=sqrt197#

So,

#costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=73/(sqrt70*sqrt197)=0.62#

#theta=51.6^@#