Area of a parallelogram, whose diagonals are `3hati+hatj-2hatk` and `hati-3hatj+4hatk` will be :

Answer

`7/2(hat(i)+hat(j)), -1/2 (hat(i)-hat(j))`

Explanation

Component along the vector `hat(i)+hat(j)`
`=(A cos theta) hat(B)=((vec(A).vec(B)))/B^(2) vec(B)=((3hat(i)+4hat(j)).(hati+hatj))/((sqrt(2))^(2))(hat(i)+hat(j))`
`=(3+4)/2(hat(i)+hat(j))=7/2(hat(i)+hat(j))`
Component along the velocity `hat(i)-hat(j)`
`=(A cos theta) hat(B)=((vec(A).vec(B)))/B^(2)vec(B)=((3hati+4hatj).(hati-hatj)(hati-hatj))/((sqrt(2))^(2))`
`=((3-4))/2(hat(i)-hat(j))=-1/2(hat(i)-hat(j))`