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

Question

$\DeclareMathOperator\span{span}$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.

Answer 1

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 $0<t<\pi$, but $\omega_t$ is never completely orthogonal to $\nu$.