I'm just learning this Latex(sic) formatting, so it's not ideal.(adsbygoogle = window.adsbygoogle || []).push({});

I was trying to explore the geometrical significance of the cross product when I happened upon an interesting observation. I've seen things like this before, but never had time to really examine them.

I define two vectors:

[itex]\pmb{A}=A^x\overset{\pmb{{}^{\wedge}}}{\pmb{i}}+A^y\overset{\pmb{{}^{\wedge}}}{\pmb{j}}=A\left(\cos (\alpha )\overset{\pmb{{}^{\wedge}}}{\pmb{i}}+\sin (\alpha )\overset{\pmb{{}^{\wedge}}}{\pmb{j}}\right)[/itex]

[itex]\pmb{B}=B^x\overset{\pmb{{}^{\wedge}}}{\pmb{i}}+B^y\overset{\pmb{{}^{\wedge}}}{\pmb{j}}=B\left(\cos (\beta )\overset{\pmb{{}^{\wedge}}}{\pmb{i}}+\sin (\beta )\overset{\pmb{{}^{\wedge}}}{\pmb{j}}\right)[/itex]

Express a new basis with the x axis aligned with the first vector:

[itex]\pmb{\hat{i}}=\cos (\alpha )\pmb{\hat{i}'}\pmb{-}\sin (\alpha )\overset{\pmb{{}^{\wedge}}}{\pmb{j}}'[/itex]

[itex]\pmb{\hat{j}}=\sin (\alpha )\pmb{\hat{i}'}+\cos (\alpha )\overset{\pmb{{}^{\wedge}}}{\pmb{j}}'[/itex]

Write the second vector in terms of the new basis and fiddle with it some:

[itex]\pmb{B}=B^x\left(\cos (\alpha )\pmb{\hat{i}'}\pmb{-}\sin (\alpha )\overset{\pmb{{}^{\wedge}}}{\pmb{j}}'\right)[/itex]

[itex]+B^y\left(\sin (\alpha )\pmb{\hat{i}'}+\cos (\alpha )\overset{\pmb{{}^{\wedge}}}{\pmb{j}}'\right)[/itex]

[itex]=B(\cos (\beta )(\cos (\alpha )\pmb{\hat{i}'}\pmb{-}\sin (\alpha )\overset{\pmb{{}^{\wedge}}}{\pmb{j}}')+\sin (\beta )(\sin (\alpha )\pmb{\hat{i}'}+\cos (\alpha )\overset{\pmb{{}^{\wedge}}}{\pmb{j}}'))[/itex]

[itex]=B\left(\cos (\beta )\cos (\alpha )\pmb{\hat{i}'}\pmb{-}\cos (\beta )\sin (\alpha )\overset{\pmb{{}^{\wedge}}}{\pmb{j}}'+\sin (\beta )\sin (\alpha )\pmb{\hat{i}'}+\sin (\beta )\cos (\alpha )\overset{\pmb{{}^{\wedge}}}{\pmb{j}}'\right)[/itex]

[itex]=B\left[(\cos (\beta )\cos (\alpha )+\sin (\beta )\sin (\alpha ))\pmb{\hat{i}}'+(\pmb{-}\cos (\beta )\sin (\alpha )+\sin (\beta )\cos (\alpha ))\overset{\pmb{{}^{\wedge}}}{\pmb{j}}'\right][/itex]

[itex]=B\left[\cos (\beta -\alpha )\pmb{\hat{i}}'+\sin (\beta -\alpha )\overset{\pmb{{}^{\wedge}}}{\pmb{j}}'\right][/itex]

[itex]=B^{x'}\overset{\pmb{{}^{\wedge}}}{\pmb{i}}'+B^{y'}\overset{\pmb{{}^{\wedge}}}{\pmb{j}}'[/itex]

Notice that the [itex]B^{x'}[/itex] component is just the magnitude of the cross product divided by the magnitude of [itex]\pmb{A}[/itex].

Use my own definition of a "complete product" and see that the first term is the dot product, and the second term is the cross product:

[itex]\pmb{AB}=AB\left[(\cos (\beta )\cos (\alpha )+\sin (\beta )\sin (\alpha ))\pmb{\hat{i}}'+(\pmb{-}\cos (\beta )\sin (\alpha )+\sin (\beta )\cos (\alpha ))\overset{\pmb{{}^{\wedge}}}{\pmb{j}}'\right][/itex]

[itex]\pmb{AB}=\left.(A^xB^x+A^yB^y\right)\overset{\pmb{{}^{\wedge}}}{\pmb{i}}'+\left.(A^xB^y-A^yB^x\right)\overset{\pmb{{}^{\wedge}}}{\pmb{j}}'[/itex]

I believe this works in 3 dimensions. Has anybody seen a development of this line of reasoning regarding vector products?

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# Dot and Cross Product from Rotation Matrix

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