Q.Show that the area of triangle ABC, the position vectors of whose vertices are $\bar a,\bar b$ and $\bar c$, is $\frac{1}{2}\left|\bar a\times\bar b+\bar b\times\bar c+\bar c\times\bar a\right|$.

If a and b are two vectors placed with the same initial point, they span a parallelogram of base |a| and height |b| sin(theta); since |a x b| = |a||b| sin(theta), the area of that parallelogram is simply |a x b|, and the area of the triangle with sides a and b (from the same vertex) is half of that, (1/2)|a x b|. For a triangle with vertices given by position vectors a, b, c, taking AB=b-a and AC=c-a as the two sides gives area = (1/2)|(b-a) x (c-a)|, which expands to the symmetric form (1/2)|a x b + b x c + c x a|. The same cross-p …