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# SirPls prove that the vector area of a triangle whose vertices are a vector bvector and cvector is0.5(bvec cross cvector +cvector cross a vec+ a vector cross b vector)Thanks and Rgds,Jai

bharat bajaj IIT Delhi
7 years ago

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.

Thanks & Regards

Bharat Bajaj

IIT Delhi

Sumit Majumdar IIT Delhi
7 years ago
Dear student,
Let the vertices be given by the vectors:
$\overrightarrow{A}=\left ( a_{1}, a_{2}, a_{3} \right ), \overrightarrow{B}=\left ( b_{1}, b_{2}, b_{3} \right ), \overrightarrow{C}=\left ( c_{1}, c_{2}, c_{3} \right )$
So, the area would be given by:
$\Delta =\frac{1}{2}\sqrt{\begin{vmatrix} a_{2} & a_{3} & 1\\ b_{2} & b_{3} & 1\\ c_{2} & c_{3}& 1 \end{vmatrix}^{2}+\begin{vmatrix} a_{3} & a_{1} & 1\\ b_{3} & b_{1} & 1\\ c_{3} & c_{1}& 1 \end{vmatrix}^{2}+\begin{vmatrix} a_{1} & a_{2} & 1\\ b_{1} & b_{2} & 1\\ c_{1} & c_{2}& 1 \end{vmatrix}^{2}}$$\Delta =\frac{1}{2}\sqrt{\begin{vmatrix} a_{2} & a_{3} & 1\\ b_{2} & b_{3} & 1\\ c_{2} & c_{3}& 1 \end{vmatrix}^{2}+\begin{vmatrix} a_{3} & a_{1} & 1\\ b_{3} & b_{1} & 1\\ c_{3} & c_{1}& 1 \end{vmatrix}^{2}+\begin{vmatrix} a_{1} & a_{2} & 1\\ b_{1} & b_{2} & 1\\ c_{1} & c_{2}& 1 \end{vmatrix}^{2}}=\frac{1}{2}\left | \left ( A\times B \right )\cdot C \right |$
Regards
Sumit