Consider a $3$-dimensional real inner product space, then each bivector can be written in the form $v\wedge w$ with $v,w\in V$ and is interpreted as an oriented area. The inner product of two bivectors $A=v\wedge w$ and $B=x\wedge y$ is defined by$$A\cdot B=\det\begin{pmatrix}v\cdot x&v\cdot y\\ v\cdot y&w\cdot y\end{pmatrix}$$ On the one hand, the inner product defines an angle $$\theta_1=\arccos\frac{A\cdot B}{\sqrt{A\cdot A}\sqrt{B\cdot B}}$$ between $A$ and $B$ and on the other hand, we can consider the angle $\theta_2$ between the planes spannend by $v,w$ and $x,y$.
Are they equal?
My issue is that I don't know how to express the angle $\theta_2$ in a way that is suitable for solving the problem...
I think the answer is yes:
Consider the hodge star with respect to one of the two orientations (the final result is independent of our choice). Then $*A=v\times w$ and $*B=x\times y$, so the fact that the hodge star is isometric implies the desired result: $$\theta_1=\arccos\frac{A\cdot B}{\sqrt{A\cdot A}\sqrt{B\cdot B}}=\arccos\frac{*A\cdot *B}{\sqrt{*A\cdot *A}\sqrt{*B\cdot *B}}=\angle(v\times w,x\times y)=\theta_2$$