Let $\vec{a}$ and $\vec{b}$ be two vectors represented by $\overrightarrow{O A}$ and $\overrightarrow{O B}$ respectively and let $\Theta$ be the angle between $\vec{a}$ and $\vec{b}$. Then,
$
\vec{b}=\overrightarrow{O M}+\overrightarrow{M B}
$

$
\vec{b}=\overrightarrow{O M}+\overrightarrow{M B}
$

Also,
$
\begin{aligned}
\overrightarrow{O M} & =(O M) \hat{a} \\
& =(O B \cos \theta) \hat{a}==(|\vec{b}| \cos \theta) \hat{a} \\
& =\left(\frac{(\vec{a} \cdot \vec{b})}{|\vec{a}|}\right) \hat{a}=\left(\frac{(\vec{a} \cdot \vec{b})}{|\vec{a}||\vec{a}|}\right) \vec{a} \quad\left[\because \hat{a}=\frac{\vec{a}}{|\vec{a}|}\right] \\
& =\left(\frac{(\vec{a} \cdot \vec{b})}{|\vec{a}|^2}\right) \vec{a}
\end{aligned}
$

As,
$
\begin{aligned}
\vec{b} & =\overrightarrow{O M}+\overrightarrow{M B} \\
\overrightarrow{M B} & =\vec{b}-\overrightarrow{O M}=\vec{b}-\left(\frac{(\vec{a} \cdot \vec{b})}{|\vec{a}|^2}\right) \vec{a}
\end{aligned}
$

Thus, the components of $\vec{b}$ along and perpendicular to $\vec{a}$ are $\left(\frac{(\vec{a} \cdot \vec{b})}{|\vec{a}|^2}\right) \vec{a}$ and $\vec{b}-\left(\frac{(\vec{a} \cdot \vec{b})}{|\vec{a}|^2}\right) \vec{a}$, respectively.