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Reading further into Newell's Method for computing the normal of an arbitrary polygon, I came across this SO post which states:

$$(P_{1} - P_{0}) \times (P_{2} - P_{0}) \equiv (P_{0} \times P_{1}) + (P_{1} \times P_{2}) + (P_{2} \times P_{0})$$

Where:

$$P_{n>0} = (x_{n} - x_{0}, y_{n} - y_{0})$$

Meaning that each $P_{n}$ represents a vector from two points along the edge of triangle.

Algebraically how is this factor equivalent?

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The cross product is distributive over addition. So, just like when you first learned about expanding double brackets:

$$\begin{align} (P_{1} - P_{0}) \times (P_{2} - P_{0}) &\equiv P_1 \times(P_2 - P_{0}) -P_0 \times (P_{2} - P_{0}) \\ &\equiv P_1 \times P_2 - P_1\times P_0 - P_0\times P_2 + P_0 \times P_0 \tag{1} \end{align}$$

We now need some other basic properties of the cross product. For any vectors $u$ and $v$, it is true that:

$$u \times u = 0$$

$$ u \times v = - (v \times u)$$

And so, in $(1)$, the last term equals zero and we can swap vectors around to get rid of the negatives, so that:

$$\begin{align} (P_{1} - P_{0}) \times (P_{2} - P_{0}) &\equiv P_1 \times P_2 + P_0\times P_1 + P_2\times P_0 \\ &\equiv P_0\times P_1 + P_1 \times P_2 + P_2\times P_0 \tag{2} \end{align}$$

as required.

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  • $\begingroup$ By "over addition" I assume you mean "over subtraction" as well? Otherwise, how did you distribute? There is no addition in the original cross-product. $\endgroup$ – pstatix Aug 16 '18 at 18:24
  • $\begingroup$ Or is that like saying it is distributive over X + (-Y)? $\endgroup$ – pstatix Aug 16 '18 at 18:51
  • $\begingroup$ Yes to your second comment. (Maybe have a go at proving rigorously that distributivity over addition implies distributivity over subtraction.) $\endgroup$ – Malkin Aug 16 '18 at 22:03

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