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)

6.40sinangleAB=5.39sinangleAC
6.40sin(angleAC+angleBC)=5.39sinangle AC

sin(angleAC+angleBC)=8.41sinangleAC

cancel(sinangleAC)*cosangleBC+cancel(cosangleAC*sinangleBC)=8.41cancel(sinangleAC)

cosangleBC=8.41=>angleBC=32.75

cosangleAC*sinangleBC=0 ; sinangleBC!=0

:.cosangleAC=0=>angleAC=90

  • The angle between vecA and vecC =90