# Please prove that,The vector area of a triangle whose sides are abar,bbar,cbar is 1/6(bxc+cxa+axb)

Arun
25757 Points
4 years ago
Dear student

You are wrong here as it should be ½ in stead of 1/6

Let A be the endpoint of $\underset{A}{\rightarrow}$, B be the endpoint of vector $\underset{B}{\rightarrow}$, and C be the endpoint of vector $\underset{C}{\rightarrow}$.

Then the vector from A to B is $\underset{B-A}{\rightarrow}$, and the vector from A to C is $\underset{C-A}{\rightarrow}$.

So (1/2) | $\underset{B-A}{\rightarrow}$X$\underset{C-A}{\rightarrow}$| is the area of the triangle. ( 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.)

(B-A) X(C-A) = B X C - B X A - A X C + A X A

The cross product of a vector with itself is zero, and A X B = – B X A, so(B-A) X (C-A) = B X C + A X B + C X A

which means that(1/2) | (B-A) X (C-A) | = (1/2) | B X C + A X B + C X A | = area of the triangle.