In a parallelogram ABCD, $|\vec{AB}|=a,\,|\vec{AD}|=b$ and $|\vec{AC}|=c$, then $\vec{DB}.\vec{AB}$ has the value:
Answer & explanation
Correct answer: option 1
$\vec{DB}.\vec{AB}=(\vec{AB}-\vec{AD}).(\vec{AB})=|\vec{AB}|^2-\vec{AD}.\vec{AB}$
$=a^2-ab\cosθ=a^2+ab[\frac{b^2+a^2-c^2}{2ab}]=\frac{2a^2+b^2+a^2-c^2}{2}=\frac{3a^2+b^2-c^2}{2}$