Question

In a $||gm\,ABCD;|\overrightarrow{AB}|=a,|\overrightarrow{AD}|=b $ and $|AC|=c.$ Then $\overrightarrow{DB}\,.\overrightarrow{AB}$ has the value

• A. $\frac{3a^2+b^2-c^2}{2}$
• B. $\frac{a^2+3b^2-c^2}{2}$
• C. $\frac{a^2-b^2+3c^2}{2}$
• D. $\frac{a^2+3b^2+c^2}{2}$

Answer 1

Answer: (A)

Since $\overrightarrow{DB}=\overrightarrow{DA}+\overrightarrow{AB}$
$\therefore \overrightarrow{DB}=\overrightarrow{DB}-\overrightarrow{AB}$

$[$Hence $\overrightarrow{AB}=\vec{a}, \overrightarrow{AD}=\vec{b}, \overrightarrow{AC}=\vec{c}]$
$\therefore \left(\overrightarrow{DA}\right)^{2}+\left(\overrightarrow{DB}\right)^{2}+\left(\overrightarrow{AB}\right)^{2}-2\,\overrightarrow{DB}\cdot\overrightarrow{AB}\quad\ldots\left(1\right)$
Also in the $\left|\right| gm$
$2\left(a^{2}+b^{2}\right)=c^{2}+\overrightarrow{DB^{2}}$
$\therefore \overrightarrow{DB^{2}}=2a^{2}+2b^{2}-c^{2}$
Putting in $\left(1\right)$,
$b^{2}=2a^{2}+2b^{2}-c^{2}+a^{2}-2\,\overrightarrow{AB}\cdot\overrightarrow{DB}$
$\therefore \overrightarrow{AB}\cdot\overrightarrow{DB}=\frac{3a^{2}+b^{2}-c^{2}}{2}$