If #A = <2 ,3 ,-1 >#, #B = <6 ,1 ,2 ># and #C=A-B#, what is the angle between A and C?
1 Answer
Apr 8, 2016
- The angle between
#vecA# and#vecC# =90
Explanation:
#C=A-B=<-4,2,-3># - If you consider that the vectors are in rectangular form (the vectors' tail is on the origin), the vector difference between two vectors
#A-B# goes from the head of#vecA# to the to the head of#vecB# forming a triangle#ABC# . - By finding the magnitude of the vectors, you can find the sine of the angle between to
#vecA# and#vecC# #(angleAC)# . -
#|A|=sqrt(2^2+3^2+(-1)^2)=3.74#
#|B|=sqrt(6^2+1^2+2^2)=6.40#
#|C|=sqrt((-4)^2+2^2+(-3)^2)=5.39# -
Using the sine rule:
#|A|/(sinangleBC)=|C|/(sinangleAB)=|B|/(sinangleAC)#
#3.74/(sinangleBC)=5.39/(sinangleAB)=6.40/(sinangleAC)# -
Doing the algebra:
#angleAB=180-(angleAC+angleBC)#
# sinangleAB=sin(180(angleAC+angleBC))=sin(angleAC+angleBC)#
- The angle between
#vecA# and#vecC# =90