The H-O-H bond angle of H2O is 104.48 °. The Overall dipole moment of the H2O molecule is 1.855D. Calculate the dipole moment of each individual O-H bond.
$\theta = 104.48 = 52.24 °$
And I let individual O-H bond be $m_i$
The answer is: $M_{net} = 2 m_i \cos 52.24 = 1.855 $ then solve for $m_i$
I do not understand the application of dot product here. Shouldn't it be $ mi m_{nett} \cos (52.24)$
The dipole moment is a vector quantity running parallel to the OH bond. The net dipole moment is the vector sum, $\vec{p}=\vec{p_1}+\vec{p_2}. $
The magnitude squared of the dipole moments is. $p_2^2=p_1^2=m_i^2 $
$\vec{p_1}$ can be broken up into terms parralel to $\vec{p}$ and perpendicular to $\vec{p}.$ So can $\vec{p_2}$. The component parallel to the net dipole moment have the same side, the part perpendicular has opposite signs.
The component of vector $\vec{p_1}$ in the direction of $\vec{p}$ is by definition $|\vec{p_1}|\cos\theta = \frac{\vec{p}\cdot \vec{p_1}}{\sqrt{\vec{p}\cdot \vec{p}}}$
$|\vec{p_1}|=|\vec{p_2}|=m_i$. There are two fectors contributing to the total dipole moment so
$|\vec{p}|=2m_i\cos \theta=1.855D$
$\theta=104.48/2$ degrees. Solve for $m_i$.