Let's consider a vector: $$ \vec V = (x\, y\, z)^T $$
$\vec W$ is the vector obtained by rotating $\vec V$ about $z-$axis by $90^\circ$.
$$ \vec W = R_z\left(90^{\circ}\right)\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{ccc} \cos 90^{\circ} & -\sin 90^{\circ} & 0 \\ \sin 90^{\circ} & \cos 90^{\circ} & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{ccc} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} -y \\ x \\ z \end{array}\right] $$
But the dot product of $\vec V$ and $\vec W$ is clearly non zero for arbitrary $z$:
$$ \vec V \cdot \vec W = -xy + xy + z^2 \neq 0 $$
What's wrong with my reasoning?
If you take an arbitrary vector (in a space with 3 dimensions), and rotate it $90^\circ$ about an arbitrary axis, the result is usually not orthogonal to what you started with. In this particular case, consider the vector $[0,0,1]^T$, and think about what should happen as you rotate it.
You can even see this in your calculations: $\vec V \cdot \vec W = -xy + xy + z^2 = z^2$. That tells you that the clue is the $z$-coordinate of $\vec V$. It doesn't tell you yet whether it is your calculations that are wrong, or your conceptions, but it tells you exactly where the two are different, and thus why I and Hyperplane above tell you to use the exact same example vector.