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

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


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

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

Algebraically how is this factor equivalent?


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.

  • $\begingroup$ 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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