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

1 Answer
Apr 6, 2017

The angle is #=66.6#º

Explanation:

Let's start by calculating

#vecC=vecA-vecB#

#vecC=〈4,2,5〉-〈-8,2,8〉=〈12,0,-3〉#

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=〈4,2,5〉.〈12,0,-3〉=48+0-15=33#

The modulus of #vecA#= #∥〈4,2,5〉∥=sqrt(16+4+25)=sqrt45#

The modulus of #vecC#= #∥〈12,0,-3〉∥=sqrt(144+0+9)=sqrt153#

So,

#costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=33/(sqrt45*sqrt153)=0.4#

#theta=66.6#º