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

1 Answer
Oct 21, 2017

The angle is #=131^@#

Explanation:

Let's start by calculating

#vecC=vecA-vecB#

#vecC=〈-6,2,1〉-〈-8,7,2〉=〈2,-5,-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# and #vecC#

The dot product is

#vecA.vecC=〈-6,2,1〉.〈2,-5,-1〉 = (-6)* (2)+(2)* (-5)+(1)* (-1) #

#=-12-10-1=-23#

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

The modulus of #vecC#= #∥〈2,-5,-1〉∥=sqrt(4+25+1)=sqrt30#

So,

#costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=-23/(sqrt41*sqrt30)=-0.656#

#theta=arccos(-0.656)=131^@#