Question

The position vectors of the vertices of a quadrilateral ABCD are $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ and $\overrightarrow{d}$ respectively. Area of the quadrilateral formed by joining the middle points of its sides is?

• A. $\frac{1}{4}|\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{d}+\overrightarrow{d}\times \overrightarrow{b}|$
• B. $\frac{1}{4}|\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{d}+\overrightarrow{d}\times \overrightarrow{a}|$
• C. $\frac{1}{4}|\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{d}+\overrightarrow{d}\times \overrightarrow{a}|$
• D. $\frac{1}{4}|\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{d}+\overrightarrow{a}\times \overrightarrow{d}+\overrightarrow{b}\times \overrightarrow{a}|$

Answer 1

Answer: (C)

$\frac{1}{4}|\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{d}+\overrightarrow{d}\times \overrightarrow{a}|$

It is given that $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ and $\overrightarrow{d}$ are the vertices of quadrilateral $ABCD$

Consider $E,F,G,H$ are mid points of sides $AB,BC.CD,DA$ respectively.

The position vector of these points

where $O$ i sthe mid-point of quadrilateral.

$\overrightarrow{OE}=\frac{1}{2}(a+b),\overrightarrow{OF}=\frac{1}{2}(b+c)$

$\overrightarrow{OG}=\frac{1}{2}(c+d),\overrightarrow{OH}=\frac{1}{2}(a+d)$

$\overrightarrow{EF}=\overrightarrow{OF}-\overrightarrow{OE}=\left(\frac{c-a}{2}\right)$

$\overrightarrow{FG}=\frac{1}{2}(d-b),\overrightarrow{GH}=\frac{1}{2}(a-c),\overrightarrow{GH}=\frac{1}{2}(b-d)$

it means $\overrightarrow{EF}\parallel \overrightarrow{GH}$ and $\overrightarrow{FG}\parallel \overrightarrow{HE}$

thus $EFGH$ is a parallelogram

Therefore, area of parallelogram $A=|\overline{EF}\times \overline{FG}|$

$\begin{array}{c}\overrightarrow{EF}\times \overrightarrow{FG}=\frac{1}{4}\left\{\right(c-a)\times (d-b\left)\right\}\\ =\frac{1}{4}(c\times d-c\times b-a\times d+a\times b)\\ =\frac{1}{4}(a\times b+b\times c+c\times d+d\times a)\end{array}$

Hence the option $C$ is correct answer.