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  • Please solve RD Sharma class 12 chapter Scaler and Dot Product exercise 23.2 question 5 maths textbook solution

Please solve RD Sharma class 12 chapter Scaler and Dot Product exercise 23.2 question 5 maths textbook solution

Answers (1)

Answer: |P Q|=|P S|

Hint: Use vector

Given: P.T by vector the quadrilateral obtained by joining

mid point of adjacent side of rectangle is rhombus.

Solution:

ABCD is rectangle.

Let P, Q, R, S be mid-point of side AB, BC, CD, DA.

Now

\overrightarrow{P Q}=\overrightarrow{P B}+\overrightarrow{B Q}=\frac{1}{2} \overrightarrow{A B}+\frac{1}{2} \overrightarrow{B C}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{B C})=\frac{1}{2} \overrightarrow{A C}            \rightarrow (1)

\overrightarrow{S R}=\overrightarrow{S D}+\overrightarrow{D R}=\frac{1}{2} \overrightarrow{A D}+\frac{1}{2} \overrightarrow{D C}=\frac{1}{2}(\overrightarrow{A D}+\overrightarrow{D C})=\frac{1}{2} \overrightarrow{A C} \quad \rightarrow(2)

From (1) & (2)

\overrightarrow{P Q}=\overrightarrow{S R}

So side PQ and SR are equal and parallel

\begin{aligned} &(|\overrightarrow{P Q}|)^{2}=\overrightarrow{P Q} \cdot \overrightarrow{P Q} \\\\ &(|\overrightarrow{P Q}|)^{2}=(\overrightarrow{P B}+\overrightarrow{B Q})(\overrightarrow{P B}+\overrightarrow{B Q}) \\\\ &(|\overrightarrow{P Q}|)^{2}=(|\overrightarrow{P B}|)^{2}+2 \cdot \overrightarrow{P B} \overrightarrow{B Q}+(|\overrightarrow{B Q}|)^{2} \end{aligned}

\begin{aligned} &(|\overrightarrow{P Q}|)^{2}=(|\overrightarrow{P B}|)^{2}+0+(|\overrightarrow{B Q}|)^{2} \quad(\overrightarrow{P B} \perp \overrightarrow{B Q}) \\\\ &(|\overrightarrow{P Q}|)^{2}=(|\overrightarrow{P B})^{2}+(|\overrightarrow{B Q}|)^{2} \rightarrow(3) \end{aligned}

Also,

\begin{aligned} &(|\overrightarrow{P S}|)^{2}=\overrightarrow{P S} \cdot \overrightarrow{P S} \\\\ &(|\overrightarrow{P S}|)^{2}=(\overrightarrow{P A}+\overrightarrow{A S})(\overrightarrow{P A}+\overrightarrow{A S}) \\\\ &(|\overrightarrow{P S}|)^{2}=(|\overrightarrow{P A}|)^{2}+2 \cdot \overrightarrow{P A} \cdot \overrightarrow{A S}+(|\overrightarrow{A S}|)^{2} \end{aligned}

\begin{aligned} &(|\overrightarrow{P S}|)^{2}=(|\overrightarrow{P A}|)^{2}+0+(|\overrightarrow{A S}|)^{2} \quad(\overrightarrow{P A} \perp \overrightarrow{A S}) \\\\ &(|\overrightarrow{P S}|)^{2}=(|\overrightarrow{P B}|)^{2}+(|\overrightarrow{B Q}|)^{2} \rightarrow(4) \end{aligned}

From (3) & (4)

\begin{aligned} &(|\overrightarrow{P Q}|)^{2}=(|\overrightarrow{P S}|)^{2} \\\\ &|\overrightarrow{P Q}|=|\overrightarrow{P S}| \end{aligned}

 

Posted by

infoexpert26

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