Prove that the area of the parallelogram with diagonal `veca` and `vecb` is `(1)/(2)|vecaxxvecb|`

Explanation

Let ABCD is a parallelogram with A as the origin. Let the position vector of B and D be `vecp` and `vecq` respectively. Then
`vec(AB)=vecp" and " vec(AD)=vecq`
But `vec(BC)=vec(AD)` Therefore `vec(BC)=vecq`
By triangle law of addition
`vec(AB)+vec(BC)=vec(AC)`

`vecp+vecq=vec(AC)`
`vecp+vecq=veca" "`.......(i)
`vec(BD)`=postion vector of D-positon vector of B.
`rArr " " vecb=vecq-vecp" "` .......(ii)
Adding (i) and (ii), we get
`2vecq=veca+vecb rArr vecq=(1)/(2)(veca+vecb)`
Subtracting (ii) from (i), we get
`2vecp=veca-vecb rArr vecp=(1)/(2)(veca-vecb)`
Now `vecpxxvecq=(1)/(2)(veca-vecb)xx(1)/(2)(veca+vecb)`
`rArr vecpxxvecq=(1)/(4)[(veca-vecb)xx(veca+vecb)]`
`rArr vecpxxvecq=(1)/(4)[vecaxxveca+vecaxxvecb-vecbxxveca-vecbxxvecb]`
`rArr " "vecpxxvecq=(1)/(4)[vecaxxvecb+vecaxxvecb]`
`rArr " " vecpxxvecq=(1)/(2)(vecaxxvecb)`
So, the area of parallelogram ABCD `=|vecpxxvecq|=(1)/(2)|vecaxxvecb|`