maths...vector

Referred to a fixed origin O, the position vectors of three non-collinear A, B and C are a, b and c respectively. By considering AB x AC, prove that the area of triangle ABC can be expressed in the form 1/2 |axb+bxc+cxa|.
SOLUTION
First, remember that |a×b| = |a| |b| (sin θ) and the triangle form by the vector a,b and (b-a) is 1/2|a×b| (That's because the magnitude of the cross-product is equal to the area of the parallelogram determined by the two vectors, and the area of the triangle is one-half the area of the parallelogram.)
Now A is the endpoint of vector a, B is the endpoint of vector b, and C is the endpoint of vector c.
Then the vector from A to B is b-a, and the vector from A to C is c-a.
So (1/2) | (b-a) × (c-a) | is the area of the triangle ABC.
Since (b-a) × (c-a) = b×c - b×a - a×c + a×a
The cross product of a vector with itself is zero, and a×b = -b×a, so

(b-a) × (c-a) = b×c + a×b + c×a

which means that

(1/2) | (b-a) × (c-a) | = (1/2) | b×c + a×b + c×a | = area of the triangle ABC.