CompuChip

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## Main Question or Discussion Point

Hello,

In a geometrical algebra exercise I have to solve the equation

where

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

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

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

In a geometrical algebra exercise I have to solve the equation

**k**

*F*= 0,

**k**is a vector and*F*is a bivector (plane).From this it follows that also [itex]\mathbf{k} \wedge F = 0[/itex], which basically says that the vector

**k**is perpendicular to the plane*F*.We are supposed to show (among other things) that [itex]F = \mathbf{k} \wedge \mathbf{A}[/itex] for some vector

**A**.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 where *Theorem*) - any plane*F*perpendicular to a vector**k**can be written as**k**wedged with another vector**A***F*and**k**are timelike, spacelike, etc - but apparently the power of geometrical algebras is that one can do without...).