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Cross Product

$\vec{p_0 p_1} \;\times\; \vec{p_0 p_2}$

Assuming p₀ = (0, 0), compute the signed area within (0,0), p₁, p₂, p₁ + p₂ = (x₁ + x₂, y₁ + y₂).

$\vec{p_1} \;\times\; \vec{p_2} = x_1 y_2 - x_2 y_1$ , the area of the parallelogram

This is the determinant of the matrix

$\begin{displaymath}\left[ \begin{array}{cc} x_1 & y_1 \\ x_2 & y_2 \end{array} \right]\end{displaymath}$

Note if p₁ = (4,0) and p₂ = (2,4), we are computing the signed area within (origin, p₁, p₂, p₁ + p₂) which is $\vec{p_1} \;\times\; \vec{p_2}$ = 16 - 0 = 16.

If we compute $\vec{p_2} \;\times\; \vec{p_1}$ the signed area is 0 - 16 = -16.

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