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1

The formula might be better written as $$\underline P(x) = (x \cdot a^{-1})a$$ So that, if $x = \alpha a + \beta a_\perp$, where $a_\perp$ is some vector perpendicular to $a$, you get $$\underline P(x) = (\alpha a \cdot a^{-1}) a + (\beta a_\perp \cdot a^{-1})a$$ But $a_\perp$ is perpendicular to $a$, and since $a^{-1}$ is just some scalar multiple of ...

0

As you said the projection is defined to be $$(x\cdot a)/a=(x \cdot a)a^{-1}$$ which is $$(x \cdot a)a^{-1}= (x \cdot a) \frac{a}{ ||a||^2 } = \frac{x \cdot a}{||a||^2}a = \frac{||a||||x||cos(\theta)}{||a||^2} a = ||x||cos(\theta)\frac{a}{||a||}$$ which, if you draw it, is the projection of x onto a.

1

There is a proof in Chapter 5 of Gracia-Bondia, Varilly, and Figueroa's book Elements of Noncommutative Geometry. They prove in Lemma 5.7, page 182 the result that bivectors in the Clifford algebra $Cl(V)$ are closed under taking commutators, and that the adjoint action of bivectors on vectors in $V$ induces an isomorphism of the Lie algebra of bivectors ...

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