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I've been searching online quite a bit for a simple and elegant geometric proof that the cross product distributes over addition. That is,

$$\mathbf a \times (\mathbf b + \mathbf c) = \mathbf a \times \mathbf b + \mathbf a \times \mathbf c.$$

I've found a few (such as this one and that one), but I feel these are all poorly written / presented and do not give me a good enough intuitive feel for why the cross product is linear.

Important: I do not seek a proof that derives from the formula for computing for the cross product based on vector components. I'm looking for a coordinate-free way of gaining intuition as to why the cross product distributes over addition this way using the intuitive "area of parallelogram and right-handed orientation" conceptual definition of the cross product.

Any thoughts, ideas, or insights?

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I think homegenity should be clear. Additivity is also easy to see, at least if you assume all three vectors to be coplanar. Then it is enough to treat the cross product as a scalar. Looking at this picture, enter image description here

we have $\operatorname{area}(ABEC) = \vec{AB} \times \vec{AC}$ and $\operatorname{area}(BDFE) = \vec{BD} \times \vec{BE} = \vec{BD} \times \vec{AC}$. Now $$ \begin{align} (\vec{AB} + \vec{BD}) \times \vec{AC} &= \vec{AD} \times \vec{AC} \\ &= \operatorname{area}(ADFC) = \operatorname{area}(ABEC) + \operatorname{area}(BDFE) \\ &= \vec{AB}\times\vec{AC} + \vec{BD}\times\vec{AC} \end{align} $$ This proves the additivity of the cross product.

The equality of the areas follows from the fact that $$ \begin{align} \operatorname{area}(ADFC) &= \operatorname{area}(ADFC) - \operatorname{area}(ADB) + \operatorname{area}(CFE)\\ &= \operatorname{area}(ABEC) + \operatorname{area}(BDFE) \end{align} $$

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  • $\begingroup$ Thank you for your effort on this one. I am, though, hoping for a more general proof instead of a special case like shown here. $\endgroup$ – Trevor Kafka Oct 2 '18 at 18:17

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