Find the angle between the vectors `hati+hatj and hati-hatk`.

Explanation

Let us consider, `vecA=(hati+hatj) and vecB=(hati-hatk)`
`therefore vecA*vecA=(hati+hatj)*(hati+hatj)=1^2+1^2=2=A^2`
`therefore A= sqrt(2)` units
and `vecB*vecB=(hati-hatk)*(hati-hatk)=1^2+(-1)^2=2=B^2`
`therefore B =sqrt(2)` units
`vecA*vecB=(hati+hatj)*(hati-hatk)=hati*hati-hati*hatk+hatj*hati-hatj*hatk = 1-0+0-0=1`
If the anlge between the two vectors is `theta` ,
`vecA*vecB=AB cos theta`
`or, cos theta =(vecA*vecB)/(AB)=1/(sqrt(2)*sqrt(2))=1/2= cos 60^@`
`therefore theta =60^@`.