Find the angle between`2(hati+hatj)` and `hati`

Explanation

Let `vecA=2hati+2hatj` and `vecB=hati" "|vecA|=sqrt(2^2+2^2)=sqrt8,|vecB|=|hati|=1`
but `vecA.vecB=AB cosetheta` so that `costheta=(vecA.vecB)/(AB)`
where `vecA.vecB=AB costheta` so that `costheta=(vecA.vecB)/(AB)`
where `vecA.vecB=(2hati+2hatj).hati=2hati.hati+2hatj.hati=2(1)+0=2`
Hence `costheta=2/(sqrt8xx1)=1/sqrt2`
This implies that `theta=45^(@)`