Find out the angle made by `(hati+hatj)` vector from X and Y axes respectively.

Explanation

`a=sqrt(a_(x)^(2)+a_(y)^(2))=sqrt(1^(2)+1^(2))=sqrt(2)`
`cos alpha=(a_(x))/(a)=(1)/sqrt(2) therefore alpha=45^(@)`
`cos beta=(a_(y))/(a)=(1)/sqrt(2) therefore beta=45^(@) hati+hatj` is at bisector of X and Y axes.