If #A = <6 ,1 ,5 >#, #B = <3 ,6 ,-2 ># and #C=A-B#, what is the angle between A and C?

1 Answer
Nov 9, 2016

The angle is #=48º#

Explanation:

Let's calculate #vecC=vecA-vecB=〈6,1,5〉-〈3,6,-2〉#
#=〈3,-5,7〉#
To calculate the angle #theta#, we need to calculate the dot product.
#vecA.vecC=∥vecA∥*∥vecC∥costheta#
Let's calculate the dot product #vecA.vecC#
#=〈6,1,5〉.〈3,-5,7〉=18-5+35=48#
The modulus of #vecA=∥vecA∥=∥〈6,1,5〉∥#
#=sqrt(36+1+25)=sqrt62#
The modulus of #vecC=∥vecC∥=∥〈3,-5,7〉∥#
#=sqrt(9+25+49)=sqrt83#

#:. costheta=48/(sqrt62*sqrt83)=0.67#

#theta=48º#