Show that area of the parallelogram whose diagonals are given by `veca and vecb` is `|vecaxxvecb|/2` Also, find the area of the parallelogram whose diagonals are `2i -j+k and i +3j - k.`

Explanation

Let ABCD be a parallelogram such that

`vec(AB)=vecp,vec(AD)=vecqimpliesvec(BC)=vecq`
By triangle law of addition, we get
`vec(AC)=vecp+vecp=veca" "[say......(i)]`
Similarly. `vec(BD)=-vecp+vecp=vecb" "[say......(ii)]`
By adding Eqs. (i) and (ii), we get
`veca+vecb=2vecqimpliesvecq=(1)/(2)(veca+vecb)`
On subtracing Eq.(ii) from (i) we get
`veca-vecb=2vecbimpliesvecp=(1)/(2)(veca+vecb)`
Now , `vecpxxvecq=(1)/(4)(veca-vecb)xx(veca-vecb)`
`=(1)/(4)(vecaxxveca+vecaxxvecb-vecbxxveca-vecbxxvecb)`
`=(1)/(4)[vecaxxvecb+vecaxxvecb]`
`=(1)/(2)(vecaxxvecb)`
So, aera fo a parallelogram `ABCD=|vecpxxvecq|=(1)/(2)|vecaxxvecb|`
Now, area of a parallelogram, whhose diagonals are `2hati-hatj+hatkandhati+3hatj-hatk`.
`=(1)/(2)|(2hati-hatj+hatk)xx(hati+3hatj-hatk)|`
`=(1)/(2)|[hati,hatj,hatk],[2,-1,1],[1,3,-1]|`
`=(1)/(2)|[hati(1-3)-hatj(-2-1)+hatk(6+1)]|`
`=(1)/(2)|-2hati+3hatj+7hatk|`
`=(1)/(2)sqrt(4+9+49)`
`(1)/(2)sqrt(62)` sq units