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

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

Answers (1)

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}

 

Posted by

infoexpert26

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