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I was trying to find out the area of a triangle given three points in the coordinates (I didn’t search on the internet, I later found out that this was already done by someone). I got the formula as below: $$ A=\frac{1}{2}(x_2y_3-x_3y_2+x_3y_1-x_1y_3+x_1y_2-x_2y_1) $$ Where $(x_1,y_1)$ is the coordinates of the first point, and so on.

And I have rewritten it in an absurd looking way: $$ A=\frac{1}{2}\vec{x}\times\vec{y}\cdot\vec{1}$$ Where $\vec{x}= \langle x_1,x_2,x_3\rangle$, $\vec{y}=\langle y_1,y_2,y_3\rangle $, $\vec{1}=\langle 1,1,1\rangle$
This formula looks astonishingly similar to $A=\frac{1}{2}bh$.

I knew how to algebraically prove this, but when I saw the cross product suddenly appeared, I thought there must be a way to somehow prove it using vectors.

So, my question is: How to prove the formula above using vectors? (Edit) Or to be more precise, what exactly does $\mathbf{\vec{x}\boldsymbol{\times}\vec{y}}$ represent in this case?

CiaPan
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Mr. W
  • 785
  • $$ \frac{1}{2}(\vec x\times\vec y)\cdot\vec1= \frac{1}{2} \begin{vmatrix} \hat i&\hat j& \hat z\ x_1 &x_2&x_3\y_1&y_2&y_3 \end{vmatrix}\cdot \begin{pmatrix} 1\1\1 \end{pmatrix}= \frac{1}{2} \begin{vmatrix} 1&1&1\ x_1 &x_2&x_3\y_1&y_2&y_3 \end{vmatrix} $$ Now, absolute value of the last expression is the area of the triangle, proved here – Soham Saha Nov 24 '24 at 12:41

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