Question

Two forces act at the vertex $A$ of quadrilateral $ABCD$ represented by $\overline{AB},\overline{AD}$ and two at $C$ represented by $\overline{CD}$ and $\overline{CB}$. If $E,F$ are mid points of $\overline{AC}$ and $\overline{BD}$ respectively, then their resultant is

• A. $\overline{EF}$
• B. $2\overline{EF}$
• C. $\frac{3}{2}\overline{EF}$
• D. $4\overline{EF}$

Answer 1

The correct option is D $4\overline{EF}$

Let position vector of $A,B,C,D$ are $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c},\overrightarrow{d}$
Position Vector of $E=\frac{\overrightarrow{a}+\overrightarrow{c}}{2}$
Position Vector of $F=\frac{\overrightarrow{b}+\overrightarrow{d}}{2}$

So, $\overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a}$, $\overrightarrow{AD}=\overrightarrow{d}-\overrightarrow{a}$, $\overrightarrow{CD}=\overrightarrow{d}-\overrightarrow{c}$ and $\overrightarrow{CB}=\overrightarrow{b}-\overrightarrow{c}$
Resultant=$\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{CD}+\overrightarrow{CB}$
$=2(\overrightarrow{b}-a+\overrightarrow{d}-\overrightarrow{c})$
$=2(2F-2E)$
$=4\overrightarrow{FE}$