Let $V$ be a real $n$-dimensional vector space with inner product $(.,.)$. Let $a,b \in V$ be two vectors with corresponding orthogonal hyperplanes $H_a$ and $H_b$. Assume that $H_{a,b} = H_a \cap H_b$ is $(n-2)$-dimensional. For subspaces $X \leq Y \leq V$ with $\dim X = \dim Y-1$ and $y \in Y$, let $\pi^Y_X(y)$ be the projection of $y$ onto $X$. Clearly, $$\pi^V_{H_a}(b) = b - \frac{(a,b)}{(a,a)}a.$$ My question is, what would be the formula for $\pi^{H_a}_{H_{a,b}} \left( \pi^V_{H_a}(b)\right)$? I am trying to find an $x \in H_a$, so that $H_{a,b} = H_x$ is the corresponding orthogonal hyperplane (relative to $H_a$), but I failed to express it in terms of $a, b$ and their scalar products.
In terms of wedge products, I would like to find a vector $c \in V$ such that $$ c \wedge \ast( a\wedge b) = \ast a.$$
Edit: argh.. of course the projection of $b$ onto $H_a$ is in $\langle a,b \rangle$ and hence orthogonal to $H_{a,b}$