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  • Please solve RD Sharma class 12 chapter 22 Algebra of Vectors exercise Multiple choice question 1 maths textbook solution

Please solve RD Sharma class 12 chapter 22 Algebra of Vectors exercise Multiple choice question 1 maths textbook solution

Answers (1)

Answer: (1-\sqrt{3}) \hat{i}+(1+\sqrt{3}) \hat{j}

Hint: You have to find circumcenter of \Delta ABC

Given: In \Delta A B C, A=(0,0), B=(3,3 \sqrt{3}), C=(-3 \sqrt{3}, 3), magnitude 2\sqrt{2}

Solution:

        \begin{aligned} &|\overrightarrow{A O}|=2 \sqrt{2} \\ &|\overrightarrow{A O}|=|\overrightarrow{B O}|=|\overrightarrow{C O}|=2 \sqrt{2}=R \end{aligned}

Let Positive vector of be  x \hat{i}+y\hat{j}

        \begin{aligned} &|\overrightarrow{A O}|=\sqrt{x^{2}+y^{2}} \\\\ &x^{2}+y^{2}=8 \end{aligned}        ...................(1)

        \begin{aligned} &\text { Also }|\overrightarrow{B O}|=|\overrightarrow{C O}|\\\\ &\sqrt{(x-3)^{2}+(y-3 \sqrt{3})^{2}}=\sqrt{(x+3 \sqrt{3})^{2}+(y-3)^{2}}\\\\ &x^{2}-6 x+9+y^{2}-6 \sqrt{3} y+27=x^{2}+6 \sqrt{3} x+27+y^{2}-6 y+9\\\\ &y(6-6 \sqrt{3})=x(6 \sqrt{3}+6) \end{aligned}

        y=\frac{x(\sqrt{3}+1)}{1-\sqrt{3}}                ......................(2)

Substituting from (2) and (1) we get

        \begin{aligned} &(1-\sqrt{3})^{2} x^{2}+(1+\sqrt{3})^{2} x^{2}=8(1-\sqrt{3})^{2} \\\\ &x^{2} \times 8=8(1-\sqrt{3})^{2} \\\\ &x=1-\sqrt{3} \\\\ &y=1+\sqrt{3} \end{aligned}

The positive vector of O is (1-\sqrt{3}) \hat{i}+(1+\sqrt{3}) \hat{j}

        \overrightarrow{A O}=(1-\sqrt{3}) \hat{i}+(1+\sqrt{3}) \hat{j}

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