\bar{\nu}_{\ell}$ is determined by the form factor $A_{0}$ at $q^{2}=0$, just as in the case of the $D$ decays.

The ratio of $B\rightarrow  \rho  \ell   \bar{\nu}_{\ell}$ to $B\rightarrow   K^{*}  \nu  \bar{\nu}$ gives

$$\begin{aligned}
\frac{[d\Gamma    (B\rightarrow   \rho   \ell    \bar{\nu}_{\ell}
)/dq^{2}]_{q^{2} \rightarrow 0}}{[d\Gamma (B\rightarrow K^{*} \nu
\bar{\nu}    )/dq^{2}]_{q^{2}    \rightarrow    0}}    &    =   &
\frac{|V_{ub}|^{2}}{|V^{*}_{ts}V_{tb}|^{2}}
\left(\frac{m^{2}_{B}-m^{2}_{\rho}}{m^{2}_{B}-m^{2}_{K^{*}}}
\right)^{3}   \left(\frac{   2   \pi   \sin^{2}\theta_{W}}{\alpha
}\right)^{2} \nonumber \\[0.4cm]
& \cdot & \frac{|A^{B\rightarrow \rho}_{0}(0)|^{2}}{|A^{B\rightarrow
K^{*}}_{0}(0)|^{2}} \frac{1}{|X(x_{t})|^{2}}  \nonumber  \\[0.4cm] 
 & = & (1.86  \times  10^{4})  \,  \frac{|V_{ub}|^{2}}{|V^{*}_{ts}|^{2}}
\frac{|A^{B\rightarrow \rho}_{0}(0)|^{2}}{|A^{B\rightarrow 
K^{*}}_{0}(0)|^{2}} \label{eq:OT5} \end{aligned}$$

The form factors $A^{B\rightarrow    \rho}_{0}(0)$ and $A^{B\rightarrow  K^{*}}_{0}(0)$ have already been calculated [@OXT] in the light- front quark model as shown in the table. Thus, we see $A^{B\rightarrow
\rho}_{0}(0)/ A^{B\rightarrow  K^{*}}_{0}(0)=0.75$. That is, the $SU(3)_{F}$ breaking of the form factors becomes larger as the mass of the decaying meson increases. In ref. [@LW] the double ratio of form factors $(f^{(B\rightarrow\rho   )}/
f^{(B\rightarrow
K^{*})})/(f^{(D\rightarrow\rho)}/f^{(D\rightarrow  K^{*})})$, was considered. In our notation, this would correspond to the ratios of the form factors $A_{1}$. They chose to write everything in terms of $A_{1}$ and ratios of the other form factors to $A_{1}$. As mentioned above, they did not calculate at $q^2=0$, a kinematic point at which only one form factor is 