How to calculate $\alpha=(x \wedge b)(a \wedge b)^{-1}$ in the projection $P[x]=P[\alpha a + \beta b] = \alpha a$ using the reciprocal frame?

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In Dorst et al pg. 104, there is a projection function defined such that $P[a]=a$ and $P[b]=0$ for some specific vectors $a$ and $b$. There is a plane associated with the 2-blade $a \wedge b$. A general vector in this plane can be written $x = \alpha a + \beta b$. The action of the projection function serves to collapse these vectors as follows: $$ P[x] = P[\alpha a + \beta b] = \alpha a $$ According to the book, $\alpha$ can be computed as $(x\wedge b)/(a\wedge b)$ using the reciprocal frame. How?

Attempt: In general, reciprocal frame basic vectors $\{b^i\}$ can be calculated in terms of frame vectors $\{b_i\}$ by $$ b^i = (-1)^{i-1}(b_1 \wedge \cdots \wedge \check{b}_i \wedge \cdots b_n) \rfloor I_n^{-1} $$ We can calculate reciprocal frame vectors $\{\bar{a}, \bar{b}\}$ to the frame vectors $\{a,b\}$ (respectively) as

$$ \bar{a} = b \rfloor I_n^{-1} = b\rfloor (a\wedge b)^{-1}=b\rfloor \frac{ b\wedge a }{ (a \wedge b) * (b \wedge a) } = \frac{ (b\cdot b)a - b(b\cdot a) }{ (a \wedge b) * (b \wedge a) }$$ $$ \bar{b} = - a \rfloor I_n^{-1} $$

Take the inner product of $x=\alpha a + \beta b$ with $\bar{a}$: \begin{align*} x\cdot \bar{a} &= \alpha a\cdot \bar{a} + \beta b \cdot \bar{a}\\ &= \alpha \end{align*}

Thus $$ \alpha = \frac{ b^2x\cdot a - x\cdot b (b\cdot a) }{ (a \wedge b) * (b \wedge a) } $$ or this could be written $$ \alpha = x\cdot \bar{a} = x\cdot (b\rfloor (a\wedge b)^{-1}) $$

This is different from the answer $\alpha=(x \wedge b)(a \wedge b)^{-1}$.

Note that the book says that a SEPARATE solution method is to simply wedge $x$ with $b$, which immediately gives $x\wedge b = \alpha a\wedge b + 0$ but there is supposed to be a method that uses the reciprocal basis.

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\begin{align*} \alpha &= x\cdot (b\rfloor (a \wedge b)^{-1}) \\ &= x\rfloor (b\rfloor (a \wedge b)^{-1}) \\ &= (x\wedge b)\rfloor (a \wedge b)^{-1} \\ \end{align*}

$x\wedge b$ is proportional to $a\wedge b$, this 2-blade is the pseudoscalar of the plane, and the contraction can be replaced with the geometric product because there are no non-scalar terms in this geometric product.