# Define an inner product between p-blades so that 0= completely orthogonal and 1=completely overlapping for their subspaces

$$\DeclareMathOperator\span{span}$$Denote two $$p$$-blades $$\nu=v_1\wedge \dots \wedge v_p$$ and $$\omega=w_1\wedge \dots \wedge w_p$$ $$\in \bigwedge^p X$$, where $$X$$ is an inner product space. How to define the inner product of $$p$$-blades $$\langle \nu, \omega\rangle$$ from the inner product of $$X$$ so that $$(1)\qquad \frac{\langle \nu, \omega\rangle}{\|\nu\|\|\omega\|}=1 \iff \span(v_1,\dots, v_p)=\span(w_1,\dots, w_p);$$ $$(2)\qquad\langle \nu, \omega\rangle=0 \iff \span(v_1,\dots, v_p)\perp \span(w_1,\dots, w_p).$$ I know one may define $$\langle \nu, \omega\rangle=\det(\langle v_i,w_j\rangle)$$, but this defintion doesn't satisfy (2). See the arXiv paper Grassmann angle formulas and identities by Mandolesi.

It is not possible. Let (i,j,k) be the canonical basis of $$\mathrm{R}^3$$, $$\nu=i\wedge j$$ and $$\omega_t = (\cos (t) i + \sin(t) k)\wedge j$$. For $$t=0$$ your condition (1) requires $$\langle\nu,\omega_0\rangle=1$$, and as $$\omega_\pi = -\omega_0$$ we have $$\langle\nu,\omega_\pi\rangle=-1$$. By continuity $$\langle\nu,\omega_t\rangle=0$$ for some $$0, but $$\omega_t$$ is never completely orthogonal to $$\nu$$.