Question

In a parallelogram $ABCD ,|\overrightarrow{A B}|=a,|\overrightarrow{AD}|=b$ and $|\overrightarrow{AC}|=c$, the value of $\overrightarrow{ DB } \cdot \overrightarrow{ AB }$ is

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

Answer 1

Answer: (A)

Let $\overrightarrow{AB}=\vec{a}, \overrightarrow{A D}=\vec{b}$ and $\overrightarrow{AC}=\vec{c}$ when $\overrightarrow{ a }, \overrightarrow{ b }$
and $\vec{c}$ are non-collinear coplanar vectors.
$\overrightarrow{ DB }=\overrightarrow{ AB }-\overrightarrow{ AC }=\overrightarrow{ a }-\overrightarrow{ b }$
Now, $\vec{D} B \cdot \overrightarrow{AB}=(\vec{a}-\vec{b}) \cdot(\vec{a})=\vec{a} \cdot \vec{a}-\vec{b} \cdot \vec{a}$
$a^{2}-a b \cos \theta =a^{2}-\frac{c^{2}-a^{2}-b^{2}}{2}=\frac{3 a^{2}+b^{2}-c^{2}}{2} $
$[\because$ In $\Delta ABC, \cos (\pi-\theta)=\frac{a^{2}+b^{2}-c^{2}}{2 a b}]$