If #A = <4 ,1 ,8 >#, #B = <6 ,7 ,-2 ># and #C=A-B#, what is the angle between A and C?

1 Answer
May 20, 2018

The angle is #=51.7^@#

Explanation:

Start by calculating

#vecC=vecA-vecB#

#vecC=〈4,1,8〉-〈6,7,-2〉=〈-2,-6,10〉#

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〉.〈-2,-6,10〉=-8-6+80=66#

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

The modulus of #vecC#= #∥〈-2,-6,10〉∥=sqrt(4+36+100)=sqrt140#

So,

#costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=66/(9*sqrt140)=0.6198#

#theta=arccos(0.6198)=51.7^@#