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  • Need solution for RD sharma maths class 12 chapter 24 vector or cross product Exercise Multiple choice questions question 21 maths textbook solution

Need solution for RD sharma maths class 12 chapter 24 vector or cross product Exercise Multiple choice questions question 21 maths textbook solution

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Answer: 3\sqrt{5}

Given: \begin{aligned} &\overrightarrow{O A}=\hat{i}+2 \hat{j}+3 \hat{k} ; \overrightarrow{O B}=-3 \hat{i}-2 \hat{j}+\hat{k} \\ \end{gathered}

Hint:Area of OAB=\frac{1}{2}\left | \overrightarrow{AO}\times \overrightarrow{AB} \right |                        

Explanation:

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

              \overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A} \\

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

                =-4 \hat{i}-4 \hat{j}-2 \hat{k}

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

                         \overrightarrow{O A} \times \overrightarrow{A B}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -4 & -4 & -2 \end{array}\right| \\

                                                =\hat{i}(-4+12)-\hat{j}(-2+12)+\hat{k}(-4+8) \\

                                                =8 \hat{i}-10 \hat{j}+4 \hat{k} \\

                     |\overrightarrow{O A} \times \overrightarrow{A B}|=\sqrt{(8)^{2}+(-10)^{2}+(4)^{2}}=\sqrt{64+100+16}=\sqrt{180}                      

                    \text { Area of } \triangle \mathrm{OAB}=\frac{1}{2} \sqrt{180}=\frac{2 \sqrt{45}}{2}=\sqrt{45}=\sqrt{9 \times 5}=3 \sqrt{5} \text { units }

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