Correct Answer - B
Clearly `veca` is perpendicular to the normals to the two planes determined by the given pairs of vectors.
We have,
`vecn_(1)=` normal vector to the plane determined by `hati` and `hati+hatj`
`impliesvecn_(1)=hatix(hati+hatj)=hatk`
`vecn_(2)=` normal vector to the plane determined by `hati-hatj` and `hati+hatk`
`impliesvecn_(2)=(hati-hatj)xx(hati+hatk)=-hati-hatj+hatk`
Since `veca` is perpendicular to `vecn_(1)` and `vecn_(2)`. Therefore,
`veca=lamda(vecn_(1)xxvecn_(2))=lamda{hatixx(-hati-hatj+hatk)}=lamda(-hatj+hati)`
Let `theta` be the angle between`veca` and `hati-2hatj+2hatk`. Then,
`cos theta=(lamda(1+2+0))/(lamdasqrt(2)sqrt(1+4+4))=1/(sqrt(2))impliestheta=pi//4`.