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Hello,
In a geometrical algebra exercise I have to solve the equation
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
In a geometrical algebra exercise I have to solve the equation
k F = 0,
where 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
(Theorem) - any plane F perpendicular to a vector k can be written as k wedged with another vector A
(I'm working in 4 spacetime dimensions. Probably the statement can be (dis)proven by choosing a basis - and then separating the cases where F and k are timelike, spacelike, etc - but apparently the power of geometrical algebras is that one can do without...).