If `veca,vecbandvecc` determine the vertices of a triangle, show that `(1)/(2)[vecbxxvecc+veccxxveca+vecaxxvecd]` givens the vector area of the triangle. Hence, deduce the condition that the three points `veca,vecband vecc` are collinera. Also, find the unit vector normal to the plane of the triangle.

Explanation

Since, `veca,vecbandvecc` are the vertices of a `DeltaABC` as shown.

:. Area of `DeltaABC=(1)/(2)|vec(AB)xxvec(AC)|`
Now, `vec(AB)=vecb-vecaandvec(AC)=vecc-veca`
:.Area of `DeltaABC=(1)/(2)|vecb-vecaxxvecc-veca|`
`=(1)/(2)|vecbxxvecc-vecbxxveca-vecaxxvecc+vecaxxveca|`
`=(1)/(2)|vecbxxvecc+vecaxxvecb+veccxxveca+vec0|`
`=(1)/(2)|vecbxxvecc+vecaxxvecb+veccxxveca|.......(i)`
For three points to be collinear, area of the `DeltaABC` should be equal to zero.
`implies(1)/(2)[vecbxxvecc+veccxxveca+vecaxxvecb]=0`
`impliesvecbxxvecc+veccxxveca+vecaxxvecb=0.......(ii)` This is the requried conddition for collinearity of three points `veca,vecbandvecc`.
Let `hatn` be the unit normal to the plane of the `DeltaABC`.
`:.hatn=(vec(AB)xxvec(AC))/(|vec(AB)xxvec(AC)|)`
`=(vecaxxvecb+vecbxxvecc+veccxxveca)/(|vecaxxvecb+vecbxxvecc+veccxxveca|)`