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Explanation

`Let veca= xhati+yhatj+zhatk`
Given `|veca|=1` therefore,
`x^(2)+y^(2)+z^(2)=1`
Angle between `veca` and z-axis is `pi//4` , therefore,
`cos (pi/4)= (veca.hatk)/(|veca||hatk|)`
`z=1/sqrt2`
`Now veca+hati+ahtj= (x+1)hati+(y+1)hatj+zhatk`
Given that `veca+hati+hatj` is a unit vector. therefore,
`|veca+hati+hatj|=sqrt([(x+1)^(2)+(y+1)^(2)z^(2)])=1`
`x^(2)+y^(2)=z^(2)+2x+2y+1=0`
1+2x+2y+1=0
y=-(x+1)
from (i), we have
`x^(2)+(x+1)^(2)+(1//2)=1`
`Rightarrow 4x^(2)+4x+1=0or(2x + 1)^(2)=0`
`x=1/2 Rightarrowy=1/2`
Hence,` veca=-1/2hati-1/2hatj+1/sqrt2hatk`