If #A = <6 ,4 ,9 >#, #B = <3 ,-1 ,0 ># and #C=A-B#, what is the angle between A and C?

1 Answer
May 19, 2017

The angle is #=15.8#º

Explanation:

Let's start by calculating

#vecC=vecA-vecB#

#vecC=〈6,4,9〉-〈3,-1,0〉=〈3,5,9〉#

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=〈6,4,9〉.〈3,5,9〉=18+20+81=119#

The modulus of #vecA#= #∥〈6,4,9〉∥=sqrt(36+16+81)=sqrt133#

The modulus of #vecC#= #∥〈3,5,9〉∥=sqrt(9+25+81)=sqrt115#

So,

#costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=119/(sqrt133*sqrt115)=0.96#

#theta=15.8#º