A) \[10\sqrt{3}\]
B) \[\frac{10}{\sqrt{3}}\]
C) \[5\sqrt{3}\]
D) \[\frac{5}{\sqrt{3}}\]
E) \[\sqrt{3}\]
Correct Answer: C
Solution :
\[\because \]\[\overrightarrow{a}\times \overrightarrow{b}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \\ \end{matrix} \right|\] \[=\hat{i}(4-6)-\hat{j}(12+2)+\hat{k}(-9-1)\] \[=-2\hat{i}-14\hat{j}-10\hat{k}\] \[\therefore \]Area of required parallelogram\[=\frac{1}{2}|\overrightarrow{a}\times \overrightarrow{b}|\] \[=\frac{1}{2}\sqrt{4+196+100}=\frac{10\sqrt{3}}{2}=5\sqrt{3}\]
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