Can a vector separating for both $\mathcal{R}$ and $\mathcal{R}'$ be annihilated by successively applying projections from each? If $\mathcal{R}$ is a von Neumann algebra acting on Hilbert space $H$, and $v \in H$ is a cyclical and separating vector for $\mathcal{R}$ (hence also for its commutant $\mathcal{R}'$), and $P \in \mathcal{R}, Q \in \mathcal{R}'$ are nonzero projections, can we have $PQv = 0$? [note i had briefly edited this to a reformulated version of the question, but am rolling it back to align with the answer below.]

Answering the original version of the question. Let $H=\mathbb C^2$, and $$ \mathcal R=\left\\{\begin{bmatrix} a&0\\\0&b\end{bmatrix}:\ a,b\in\mathbb C\right\\}\subset M_2(\mathbb C),\ \ \ \ \ \ v=\begin{bmatrix} 1\\\1\end{bmatrix}. $$ Then $v$ is cyclic and separating, and $\mathcal R'=\mathcal R$. Take $$ P=\begin{bmatrix} 1&0\\\0&0\end{bmatrix},\ \ \ Q=\begin{bmatrix} 0&0\\\0&1\end{bmatrix}. $$ Then $P\in\mathcal R$, $Q\in\mathcal R'$, and $PQ=0$, so $PQv=0$.