If #A = <6 ,8 ,-3 >#, #B = <7 ,1 ,-4 ># and #C=A-B#, what is the angle between A and C?

1 Answer
Nov 8, 2016

The angle is #50.9º#

Explanation:

Let's start by calculating #vecC=vecA-vecB#
#vecC=〈6,8,-3〉-〈7,1,-4〉=〈-1,7,1〉#
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# an #vecC#

#:. cos theta=vecA.vecC/(∥vecA∥*∥vecC∥)#
The dot product is #vecA.vecC=〈6,8,-3〉.〈-1,7,1〉=-6+56-3=47#
The modulus of #vecA=∥vecA∥=sqrt(36+64+9)=sqrt109#
The modulus of #vecC=∥vecC∥=sqrt(1+49+1)=sqrt51#
So #costheta=47/(sqrt109*sqrt51)=0.63#
#theta=50.9º#