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In a geometrical algebra exercise I have to solve the equation

wherekF= 0, kis a vector andFis a bivector (plane).

From this it follows that also [itex]\mathbf{k} \wedge F = 0[/itex], which basically says that the vectorkis perpendicular to the planeF.

We are supposed to show (among other things) that [itex]F = \mathbf{k} \wedge \mathbf{A}[/itex] for some vectorA.

My question is: can you prove that this is always possible (or give a counterexample)? So

((I'm working in 4 spacetime dimensions. Probably the statement can be (dis)proven by choosing a basis - and then separating the cases whereTheorem) - any planeFperpendicular to a vectorkcan be written askwedged with another vectorAFandkare timelike, spacelike, etc - but apparently the power of geometrical algebras is that one can do without...).

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# Geometrical algebra's: simple equation

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