Find the angle between two vector `A=2hati+hatj-hatk andB=hati-hatk.`

Explanation

`a=|A|=sqrt((2)^(2)+(1)^(2)+(-1)^(2))=sqrt(6)`
`B=|b|=sqrt((1)^(2)+(-1)^(2))=sqrt(2)`
`A.B(2hati+hatj-hatk).(hati-hatk)`
`Now , cos theta=(A.B)/(AB)=(3)/(sqrt(6).sqrt(2))=(3)/(sqrt(12))=(sqrt(3))/(2)`
`thereforetheta=30^(@)`