Algebraically we can get #C=A-B# by subtracting the components of #B# from the components of #A#
#vecC=vecA-vecB=<3-2,-1-(-2),3-5>#
# = <1,-1+2,-2> = <1,1,-2>#
Also, we can find the dot product of the vector A anc C
#vecA*vecC=3(1)+(-1)(1)+3(-2)=3+-1+-6=-4#
Another definition of the dot product is
#vecA*vecC=|vecA||vecC|cos(theta_(A,C))#
So, we need to find #|A|# and #|C|#
#|vecA|=sqrt(3^2+(-1)^2+3^2)=sqrt(9+1+9)=sqrt(10)#
#|vecC|=sqrt(1^2+1^2+(-2)^2)=sqrt(1+1+4)=sqrt(6)#
#=> vecA*vecC=|vecA||vecC|cos(theta_(A,C))=(sqrt(10))(sqrt(6))cos(theta_(A,C))#
Then we plug in
#<=> (vecA*vecC)/((sqrt(10))(sqrt(6))##= (-4)/(sqrt(60)##= (-4)/(2sqrt(15))##=(-2)/(sqrt(15)##=cos(theta_(A,C))#
#<=> cos^(-1)(-2/(sqrt(15)))=underline(theta_(A,C)approx 121.091^@)#
And we have the angle
