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  • Need solution for RD sharma maths class 12 chapter 24 vector or cross product Exercise Multiple choice questions question 19 maths textbook solution

Need solution for RD sharma maths class 12 chapter 24 vector or cross product Exercise Multiple choice questions question 19 maths textbook solution

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Answer: \frac{1}{2}\sqrt{229}

Given: The vectors from origin O to the points A and B are \overrightarrow{a}=2\hat{i}-3\hat{j}+2\hat{k};\overrightarrow{b}=2\hat{i}+3\hat{j}+\hat{k}respectively.

Hint: Area of                                                                    

Explanation: Here \overrightarrow{a}=2\hat{i}-3\hat{j}+2\hat{k};\overrightarrow{b}=2\hat{i}+3\hat{j}+\hat{k}

              \overrightarrow{O A}=( P. V of A- P.V of O)

             =(2 \hat{i}-3 \hat{j}+2 \hat{k})-(0 \hat{i}+0 \hat{j}+0 \hat{k})

              =2 \hat{i}-3 \hat{j}+2 \hat{k}

              \overrightarrow{A B}=(P . V \text { of } B-P . V \text { of } A) \\

              =(2 \hat{i}+3 \hat{j}+\hat{k})-(2 \hat{i}-3 \hat{j}+2 \hat{k}) \\

              =0 \hat{i}+6 \hat{j}-\hat{k}

              Area of \triangle O A B=\frac{1}{2}|\overrightarrow{O A} \times \overrightarrow{A B}|

              \begin{aligned} &\text { Now, }(\overrightarrow{O A} \times \overrightarrow{A B})=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -2 \\ 0 & 6 & -1 \end{array}\right| \\ \end{aligned}

                                                                         = \hat{i}(3-12)-\hat{j}(-2-0)+\hat{k}(12+0) \\

                                                                         =-9 \hat{i}+2 \hat{j}+12 \hat{k} \\

             \Rightarrow|\overrightarrow{O A} \times \overrightarrow{A B}|=\sqrt{(-9)^{2}+(2)^{2}+(12)^{2}}=\sqrt{81+4+144}=\sqrt{229}

              Area of \triangle O A B=\frac{1}{2}\sqrt{229}

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