$\text{ Given a parallelogram }\mathrm{ABCD}\text{ . If }|\overrightarrow{AB\mid }=a,|\overrightarrow{AD}\mid =b\text{ and }\left|\overrightarrow{AC}\right|=c,\text{ then }\overrightarrow{DB}\cdot \overrightarrow{AB}$ $\text{ has the value : }$

$\text{ To find }(\overrightarrow{a}-\overrightarrow{b})\cdot \overrightarrow{a},\text{ i.e. }|\overrightarrow{a}{|}^2-\overrightarrow{a}\cdot \overrightarrow{b}$     ………..(1)

$\begin{array}{r}\text{ Now }\\ \overrightarrow{a}+\overrightarrow{b}=\overrightarrow{c}\end{array}$

$\Rightarrow |\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+2\overrightarrow{a}\cdot \overrightarrow{b}=|\overrightarrow{c}{|}^2$

$$\begin{gathered} \Rightarrow 2\overrightarrow{a}\cdot \overrightarrow{b}=|\overrightarrow{c}{|}^2-|\overrightarrow{a}{|}^2-|\overrightarrow{b}{|}^2 \\ \overrightarrow{a}\cdot \overrightarrow{b}=\frac{|\overrightarrow{c}{|}^2-|\overrightarrow{a}{|}^2-|\overrightarrow{b}{|}^2}{2} \end{gathered}$$

$\text{ Substitute the value of }\overrightarrow{a}\cdot \overrightarrow{b}\text{ in eqn. (1)}$

$\left|\overrightarrow{a^2}\right|-$$\frac{|\overrightarrow{c}{|}^2-|\overrightarrow{a}{|}^2-|\overrightarrow{b}{|}^2}{2}$

$\frac{3{\overrightarrow{a}}^2+{\overrightarrow{b}}^2-{\overrightarrow{c}}^2}{2}$