Question

If `A = <2 ,3 ,-1 >`, `B = <6 ,1 ,2 >` and `C=A-B`, what is the angle between A and C?

Answer 1

• 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