Two observers moving with different velocities measures the magnetic field produced by a moving charge, at a common point A. The position vector of point A with respect to point charge is $\overrightarrow{r}$ then mark the wrong statement (s)

The magnetic field due to a moving charge at any point is given by  $\overrightarrow{\mathrm{B}}=\frac{{\mu }_0\mathrm{q}(\overrightarrow{{\mathrm{v}}_{\mathrm{r}}}\times \overrightarrow{\mathrm{r}})}{4\pi {\mathrm{r}}^3}$ 

where ${\overrightarrow{v}}_r$ . is the velocity of charge with respect to observer.

let $\overrightarrow{v_1}\mathrm{\&}\overrightarrow{v_2}$ be the velocities of two observers with respect to ground then 

$\overrightarrow{B_1}=\frac{{\mu }_0\mathrm{q}\left[\left(\overrightarrow{v}-\overrightarrow{v_1}\right)\times \overrightarrow{r}\right]}{4\pi r^3}\mathrm{\&}\overrightarrow{B_2}=\frac{{\mu }_{0}q\left[\left(\overrightarrow{v}-\overrightarrow{v_2}\right)\times \overrightarrow{r}\right]}{4\pi r^3}$

Where $\overrightarrow{v}$ is the velocity of charge with respect to ground

$\overrightarrow{B_1}=\overrightarrow{B_2}\text{ when }\left(\overrightarrow{v}-\overrightarrow{v_1}\right)\times \overrightarrow{r}=\left(\overrightarrow{v}-\overrightarrow{v_2}\right)\times \overrightarrow{r}$

$\begin{array}{l}\Rightarrow \left(\overrightarrow{v_2}-\overrightarrow{v_1}\right)\times \overrightarrow{r}=0\Rightarrow \left(\overrightarrow{v_{21}}\right)\times \overrightarrow{r}=0\\ \Rightarrow \overrightarrow{v_{21}}\left|\right|\overrightarrow{r}\end{array}$