If #A = <4 ,1 ,8 >#, #B = <5 ,7 ,3 ># and #C=A-B#, what is the angle between A and C?

1 Answer
Mar 31, 2018

The angle is #=65.0^@#

Explanation:

Start by calculating

#vecC=vecA-vecB#

#vecC=〈4,1,8〉-〈5,7,3〉=〈-1,-6,5〉#

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=〈4,1,8〉.〈-1,-6,5〉=-4-6+40=30#

The modulus of #vecA#= #∥〈4,1,8〉∥=sqrt(16+1+64)=sqrt81=9#

The modulus of #vecC#= #∥〈-1,-6,5〉∥=sqrt(1+36+25)=sqrt62#

So,

#costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=30/(9*sqrt62)=0.42#

#theta=arccos(0.42)=65.0^@#