Let $\mathbb O'$ be the split octonion algebra over $\mathbb R$. For each nonzero divisor of zero $x\in \mathbb O'$ $\mbox{($x \neq 0, N(x)=0$)}$ the kernel of the left multiplication by $x$, $Ker L_x=\{y\in \mathbb O': xy=0\}$ is maximal totally isotropic subspace in $\mathbb O'$ and is equal $\overline{x} \mathbb O'$ (since both $Ker L_x$, $\overline{x} \mathbb O'$ are totally isotropic (hence with dimension $\leq 4$) and $dim Ker L_x+Im L_x=8$). Similarly, $Ker R_x=\{y\in \mathbb O': yx=0\}=\mathbb O' \overline{x}$ is maximal totally isotropc, where $R_x(y)=yx$ for $y\in \mathbb O'$.

My question: is it maybe true that for each maximally totally isotropic $V\subset \mathbb O'$ there is a nonzero zero's divisor $x$ in $\mathbb O'$ such that $V=x\mathbb O'$ or $V=\mathbb O' x$? Is $x$ unique up to nonzero real constant?