If `veca+vecb+vecc=0`, then show that `vecaxxvecbxxvecc=veccxxveca`. Interpret the result geometrically.

Explanation

Since, `veca+vecb+vecc=0`
`impliesvecb=-vecc-veca`
Now, `vecaxxvecbxx=vecaxx(-vecc-veca)`
`=vecaxx(-vecc)+vecaxx(-veca)=-vecaxxvecc`
`impliesvecaxxvecb=veccxxveca.....(i)`
Also, `vecbxxvecc=(-vecc-veca)xxvecc`
`=(-veccxxvecc)+(-vecaxxvecc)=-vecaxxvecc`
`impliesvecbxxvecc=veccxxveca......(ii)`
From Eqs. (i) and (ii), `vecaxxvecb=vecbxxvecc=veccxxveca`
Geometrical interpretation of the result

If ABCD is a parallelogram such that `vec(AB)=vecaandvec(AD)=vecb` and these adjacent sides are making angle `theta` between each other then we say that
Area of parallelogram `ABCD=|veca||vecb||sintheta|=|vecaxxvecb|` Since, parallelogram on the same base and between the same parallels are equal in area.
We can say that, `|vecaxxvecb|=|vecaxxvecc|=|vecbxxvecc|`
This also implies that, `vecaxxvecb=vecaxxvecc=vecbxxvecc`
So, area of the parallelograms formed by taking any two sides represented by `veca,vecbandvecc` as adjacent are equal.