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  • Please solve RD Sharma class 12 chapter Scaler and Dot Product exercise 23.2 question 1 maths textbook solution

Please solve RD Sharma class 12 chapter Scaler and Dot Product exercise 23.2 question 1 maths textbook solution

Answers (1)

Answer: O P^{2}+O Q^{2}=\frac{5}{9} A B^{2}

Hint: Use \Delta O A B

Given: \text { In } \Delta O A B, \angle A O B=90^{\circ} \text { and if } P \; \& \; Q are points of trisection of AB-PT

O P^{2}+O Q^{2}=\frac{5}{9} A B^{2}

Solution:

\overrightarrow{O A}, \overrightarrow{O B}

\begin{aligned} &\overrightarrow{O A}, \overrightarrow{O B}=0 \\\\ &\overrightarrow{A B} \cdot \overrightarrow{A B}=\overrightarrow{O A} \cdot \overrightarrow{O B} \\\\ &\overrightarrow{O P}=\overrightarrow{O B}-\frac{2}{3} \overrightarrow{A B} \\\\ &\overrightarrow{O Q}=\overrightarrow{O B}-\frac{1}{3} \overrightarrow{A B} \end{aligned}

\begin{aligned} &\overrightarrow{O P}=\overrightarrow{O B}-\frac{2}{3}(\overrightarrow{O B}-\overrightarrow{O A}) \\\\ &\overrightarrow{O P}=\frac{\overrightarrow{O B}}{3}+\frac{2}{3} \overrightarrow{O A} \\\\ &\overrightarrow{O Q}=\overrightarrow{O B}-\frac{1}{3}(\overrightarrow{O B}-\overrightarrow{O A}) \\ \end{aligned}

\frac{2}{3} \overrightarrow{O B}+\frac{\overrightarrow{O A}}{3}

\begin{aligned} &O P^{2}+O Q^{2}=\frac{5}{9} A B^{2} \Rightarrow O P^{2}=|\overrightarrow{O P}|^{2}=\overrightarrow{O P} \cdot \overrightarrow{O P} \\\\ &O Q^{2}=|\overrightarrow{O Q}|^{2}=\overrightarrow{O Q} \cdot \overrightarrow{O Q} \\\\ &\overrightarrow{O P} \cdot \overrightarrow{O P}=\left(\frac{\overrightarrow{O B}}{3}+\frac{2}{3} \overrightarrow{O A}\right)\left(\frac{\overrightarrow{O B}}{3}, \frac{2}{3} \overrightarrow{O A}\right) \end{aligned}

\begin{aligned} &=\frac{|\overrightarrow{O B}|^{2}}{9}+\frac{4}{9}|\overrightarrow{O A}|^{2} \\\\ &O Q \cdot O Q=\frac{4}{9}|\overrightarrow{O B}|^{2}+\frac{|\overrightarrow{O A}|^{2}}{9} \end{aligned}

\begin{aligned} &O P^{2}+O Q^{2}=\frac{5}{9}\left(|O B|^{2}+|\overrightarrow{O A}|^{2}\right)=|A B|^{2} \\\\ &O P^{2}+O Q^{2}=\frac{5}{9}|A B|^{2} \end{aligned}

 

 

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