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  • Provide solution for RD Sharma maths class 12 chapter 22 Algebra of Vectors exercise Multiple choice question 14

Provide solution for RD Sharma maths class 12 chapter 22 Algebra of Vectors exercise Multiple choice question 14

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Answer: Form as isosceles triangle.

Hint: ABC for which triangle

Given: The position vector of pointsA, B and Care 2 \vec{i}+\vec{j}-\vec{k}, 3 \vec{i}-2 \vec{j}+\vec{k}, i+4 j-3 k

Solution:

        \begin{aligned} &\overrightarrow{O A}=2 \hat{i}+\hat{j}-\hat{k}, \\\\ &\overrightarrow{O B}=3 \hat{i}-2 \hat{j}+\hat{k}, \\\\ &\overrightarrow{O C}=\hat{i}+4 \hat{j}-3 \hat{k} \end{aligned}

        \begin{aligned} &\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A} \\\\ &=(3-2) \hat{i}+(-2-1) \hat{j}+(1+1) \hat{k} \\\\ &=\hat{i}-3 \hat{j}+2 \hat{k} \end{aligned}

        \begin{aligned} &|\overrightarrow{A B}|=\sqrt{1^{2}+3^{2}+2^{2}}=\sqrt{1+9+4}=\sqrt{14} \\\\ &\overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B} \\\\ &=(1-3) \hat{i}+(4+2) \hat{j}+(-3-1) \hat{k} \\\\ &=-2 \hat{i}+6 \hat{j}-4 \hat{k} \end{aligned}

        \begin{aligned} &|\overrightarrow{B C}|=\sqrt{(2)^{2}+6^{2}+(-4)^{2}}=\sqrt{4+36+16}=\sqrt{56}=2 \sqrt{14} \\\\ &\overrightarrow{C A}=\overrightarrow{O A}-\overrightarrow{O C} \\\\ &=(2-1) \hat{i}+(1-4) \hat{j}+(-1+3) \hat{k} \end{aligned}

        \begin{aligned} &=\hat{i}-3 \hat{j}+2 \hat{k} \\\\ &|\overrightarrow{C A}|=\sqrt{1^{2}+3^{2}+2^{2}}=\sqrt{1+9+4}=\sqrt{14} \end{aligned}

       

        Hence \overrightarrow{A B}\; \& \; \overrightarrow{C A}=\sqrt{14}

        It forms isosceles triangle.

 

 

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