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

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


1 Answers

25763 Points
one year 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.

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