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

1 Answer
Jan 23, 2017

The angle is #=127#º

Explanation:

Let's start by calculating

#vecC=vecA-vecB#

#vecC=〈-6,2,1〉-〈-8,8,4〉=〈2,-6,-3〉#

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=〈-6,2,1〉.〈2,-6,-3〉=-12-12-3=-27#

The modulus of #vecA#= #∥〈-6,2,1〉∥=sqrt(36+4+1)=sqrt41#

The modulus of #vecC#= #∥〈2,-6,-3〉∥=sqrt(4+36+9)=sqrt49=7#

So,

#costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=-27/(sqrt41*7)=-0.6#

#theta=127#º