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Provide solution for RD Sharma maths class 12 chapter Scaler and Dot Product exercise 23.2 question 10

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Answer: A B^{2}+B C^{2}+C D^{2}+D A^{2}=A C^{2}+B D^{2}+4 P Q^{2}

Hint:  Use mid point theorem

Given: In quadrilateral ABCD

Prove A B^{2}+B C^{2}+C D^{2}+D A^{2}=A C^{2}+B D^{2}+4 P Q^{2} where P \; \& \; Q are

Middle point of diagonal AC and BD

Solution:

 

Let ABCD be quadrilateral.

Take a as origin. Let position vector B,C,D  be  \vec{b}, \vec{c}, \vec{d} respectively

Then vector P=\frac{\vec{c}}{2}  (Mid point formula)

Position vector \mathrm{Q}=\frac{\vec{b}+\vec{d}}{2}   (Mid point)

Now,

\begin{aligned} &A B^{2}+B C^{2}+C D^{2}+D A^{2} \\\\ &(|\vec{b}|)^{2}+(|\vec{c}-\vec{b}|)^{2}+(|\vec{d}-\vec{c}|)^{2}+(|\vec{d}|)^{2} \end{aligned}

\begin{aligned} &(|\vec{b}|)^{2}+(|\vec{c}|)^{2}-2 \vec{c} \cdot \vec{b}+(|\vec{b}|)^{2}+(|\vec{d}|)^{2}-2 \vec{d} \vec{c}+(|\vec{c}|)^{2}+(|\vec{d}|)^{2} \\\\ &2(|\vec{b}|)^{2}+2(|\vec{c}|)^{2}+2(|\vec{d}|)^{2}-2 \cdot \vec{b} \cdot \vec{c}-2 \cdot \vec{c} \cdot \vec{d} \quad \rightarrow(1) \end{aligned}

Also,

\begin{aligned} &A C^{2}+B D^{2}+4 P Q^{2} \\\\ &(|\overrightarrow{A C}|)^{2}+(|\overrightarrow{B D}|)^{2}+4(|\overrightarrow{P Q}|)^{2} \\\\ &(|\vec{c}|)^{2}+(|\bar{d}-\vec{b}|)^{2}+(|\vec{b}+\vec{d}|)^{2}-2(|\vec{b}+\vec{d}|)^{2} \cdot c+(|\vec{c}|)^{2} \end{aligned}

\begin{aligned} &2(|\vec{c}|)^{2}+2(|\bar{d}|)^{2}+2(|\vec{b}|)^{2}-2 \cdot \vec{b} \cdot \vec{c}-2 \vec{b} \cdot \vec{c}-2 \cdot \vec{c} \cdot \vec{d} \\\\ &A B^{2}+B C^{2}+C D^{2}+D A^{2}=A C^{2}+B D^{2}+4 P Q^{2} \end{aligned}

 

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