If #A = <-5 ,2 ,1 >#, #B = <-8 ,4 ,9 ># and #C=A-B#, what is the angle between A and C?

1 Answer
Nov 17, 2016

The angle is #=124.2#º

Explanation:

Let's calculate #vecC#

#vecC=vecA-vecB#

Therefore, #vecC=〈-5,2,1〉-〈-8,4,9〉=〈3,-2,-8〉#

The angle #theta# between #vecA# and #vecC# is calculated from the dot product definition.

#vecA.vecC=∥vecA∥.∥vecC∥costheta#

The dot product is
#vecA.vecC=〈-5,2,1〉.〈3,-2,-8〉=-15-4-8=-27#

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

The modulus of #vecC=∥〈3,-2,-8〉∥=sqrt(9+4+64)=sqrt77#

Therefore, #costheta=vecA.vecC/(∥vecA∥.∥vecC∥)#

#=-27/(sqrt30*sqrt77)=-0.56#

#theta=124.2#º