Suppose the scalar product of two vectors $\vec{A}$ and $\vec{B}$ is given as in figure (a)

$\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=|\overrightarrow{\mathrm{A}}||\overrightarrow{\mathrm{B}}| \cos \theta$

$\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=\mathrm{AB} \cos \theta$

$\ldots$ $(1)$ which is scalar.

where $\theta$ is the angle between $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$

This product is shown in two ways.

$\star$ Method $1:$

According to figure (b), draw a perpendicular from the head of $\vec{B}$ on $\vec{A}$ resulting OM which is shown as in figure OM is the projection of $\vec{B}$ on to $\vec{A}$ or it is called the component of $\vec{B}$ in the direction of $\overrightarrow{\mathrm{A}}$.

$\therefore\;OM=$ componant of $\vec{B}$ along $\vec{A}$

$=B \cos \theta$

$\therefore \overrightarrow{ A } \cdot \overrightarrow{ B }$$=A B \cos \theta$

$=A(B \cos \theta)$

$=A(O M)$

$=\text { Magnitude of } \vec{A} \times \operatorname{component} \text { of } \vec{B} \text { along } \vec{A}$