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Need solution for RD Sharma maths class 12 chapter 22 Algebra of Vectors exercise Multiple choice question 11

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Answer: Parallelogram

Hint: Find ABCD

Given: If  \vec{a}, \vec{b}, \vec{c} \; \&\; \vec{d}  are position vectors of points A,B,C and D  such that no three of them are collinear  \vec{a}+\vec{c}=\vec{b}+\vec{d}  then ABCD is

Solution: Given \vec{a}+\vec{c}=\vec{b}+\vec{d}

                        \begin{aligned} &\vec{c}-\vec{d}=\vec{b}-\vec{a} \\ \\ &\overrightarrow{A B}=\overrightarrow{D C} \text { and } \vec{a}+\vec{c}=\vec{b}+\vec{d} \\ \\ &\vec{c}-\vec{d}=\vec{b}-\vec{a} \\ \\ &\overrightarrow{A D}=\overrightarrow{B C} \end{aligned}

               Since \vec{a}+\vec{c}=\vec{b}+\vec{d}

            \frac{1}{2}(\vec{a}+\vec{c})=\frac{1}{2}(\vec{b}+\vec{d})

            So position vector midpoint of BD =  position vector of midpoint of AC

            Hence bisect each other

            The given point ABCD is parallelogram.

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