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  • Need solution for RD Sharma maths class 12 chapter 22 Algebra of Vectors exercise Multiple choice question 15

Need solution for RD Sharma maths class 12 chapter 22 Algebra of Vectors exercise Multiple choice question 15

Answers (1)

Answer: (2,-3)

Hint: (x,y) value find

Given: If three points A, B , C  have position vectors  \begin{aligned} &\hat{i}+x \hat{j}+3 \hat{k} \\\\ &3 \hat{i}+4 \hat{j}+7 \hat{k} \\ \\ &y \hat{i}-2 \hat{j}-5 \hat{k} \end{aligned}    respective are collinear , then (x,y)=

 Solution: If

        \begin{aligned} &\hat{i}+x \hat{j}+3 \hat{k} \\\\ &3 \hat{i}+4 \hat{j}+7 \hat{k} \\\\ &y \hat{i}-2 \hat{j}-5 \hat{k} \end{aligned}

        \begin{aligned} &\overrightarrow{A B}=3 \hat{i}+4 \hat{j}+7 \hat{k}-\hat{i}-x \hat{j}-3 \hat{k}=2 \hat{i}+(4-x) \hat{j}+4 \hat{k} \\\\ &\overrightarrow{B C}=y \hat{i}-2 \hat{j}-5 \hat{k}-3 \hat{i}-4 \hat{j}-7 \hat{k}=(y-3) \hat{i}-6 \hat{j}-12 \hat{k} \end{aligned}

Since vectors are collinear.

        \begin{aligned} &\overrightarrow{A B}=\lambda \overrightarrow{B C} \\\\ &2 \hat{i}+(4-x) \hat{j}+4 \hat{k}=\lambda((y-3) \hat{i}-6 \hat{j}-12 \hat{k}) \\\\ &2=\lambda(y-3) \end{aligned}                    ...................(1)

        (4-x)=-6 \lambda                                                                             ....................(2)

        4=-12 \lambda \Rightarrow \lambda=\frac{-1}{3}

Substitute \lambda in (1) and (2)

        \begin{aligned} &2=\frac{-1}{3}(y-3) \Rightarrow-6=y-3 \Rightarrow y=-3 \\\\ &4-x=-6 \times \frac{-1}{3} \Rightarrow 4-x=2 \Rightarrow x=2 \\\\ &x=2, y=-3 \end{aligned}

 

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