References 1 and 2 define a pure spinor $\psi$ to be a solution of the Cartan-Penrose equation $$ \newcommand{\opsi}{{\overline\psi}} v^\mu\gamma_\mu\psi=0 \hspace{1cm} \text{with} \hspace{1cm} v^\mu\equiv \opsi\gamma^\mu\psi, \tag{1} $$ where $\gamma^0,\gamma^1,...,\gamma^{D-1}$ are Dirac matrices for a $D$-dimensional spacetime with lorentzian signature, and presumably $\opsi\equiv\psi^\dagger\gamma^0$. References 1 and 2 consider real-valued solutions of this equation, so that $v^\mu$ is an ordinary spacetime vector. Equation (1) clearly implies that $v^\mu$ is a null vector (proof: multiply equation (1) by $\opsi$), but reference 1 makes this stronger statement on page 4:
The null direction, and the orientation of ... a space-like $D - 2$ plane orthogonal to the null direction, are both encoded in a pure spinor...
In this context, "plane" means the linear span of a set of $D-2$ linearly independent vectors in spacetime. The null vector $v^\mu$ itself does not uniquely determine the orientation of such a plane. To see why it's not unique, start with any set of $D-2$ linearly independent vectors orthogonal to $v^\mu$. This defines one plane through the origin. Now add some nonzero multiple of $v^\mu$ to one of those vectors. This gives another set of $D-2$ linearly independent vectors that define a different plane through the origin, still orthogonal to $v^\mu$.
In what way does a real solution $\psi$ of equation (1) select a unique orientation for a spacelike $(D-2)$-dimensional plane orthogonal to the null vector $v^\mu\equiv \opsi\gamma^\mu\psi$?
References and notes:
Banks, Fiol, and Morisse (2006), Towards a quantum theory of de Sitter space (https://arxiv.org/abs/hep-th/0609062)
Banks, Fischler, and Mannelli (2004), Microscopic Quantum Mechanics of the $p=\rho$ Universe" (https://arxiv.org/abs/hep-th/0408076)
I posted this question on Physics SE instead of Math SE partly because reference 2 cites Misner, Thorne, and Wheeler's giant book Gravitation (1973). Maybe that book answers my question, but I don't have access to it right now.
Related: Conflicting definitions of a spinor