Question

If `A = <2 ,-5 ,9 >`, `B = <-9 ,4 ,-5 >` and `C=A-B`, what is the angle between A and C?

Answer 1

`vecA=<2,-5,9>` and `vecB=<-9,4,-5>`
Hence `vecC=vecA-vecB=<11,-9,14>`
If `theta` be the angle between`vec A` and `vec C`
then
`cos theta=(vecA.vecC)/(|vecA||vecC|)=(11.2+(-5)(-9)+9.14)/(sqrt(2^2+(-5)^2+9^2)sqrt(11^2+(-9)^2+14^2 ))=0.9224`
`theta =22.72^0`