The angle between the two vectors `vecA=hati+2hatj-hatk` and `vecB=-hati+hatj-2hatk`

Explanation

`vecA=(hati+2hatj-hatk)`
`|vecA|=sqrt((1)^(2)+(2)^(2)+(-1)^(2))=sqrt(6)`
Also, `vecB=(-hati+hatj-2hatk)`
`|vecB|=sqrt((-1)^(2)+(1)^(2)+(-2)^(2))=sqrt(6)`
Now, `vecA.vecB=(hati+2hatj-hatk).(-hati+hatj-2hatk)=-1+2+2=3`
Let be the angle between `vecAandvecB`
`therefore costheta=(vecA.vecB)/(|vecA|.|vecB|)=(3)/(sqrt(6)xxsqrt(6))=(3)/(6)=(1)/(2)`
`theta=60^(@)`