A) \[8\sqrt{3}\] sq units
B) 8 sq units
C) \[8\sqrt{2}\]sq units
D) None of these
Correct Answer: A
Solution :
Adjacent sides of a parallelogram are \[\overrightarrow{a}=3\hat{i}+\hat{j}+2\hat{k}\] and \[\overrightarrow{b}=2\hat{i}-2\hat{j}+4\hat{k}\] \[\overrightarrow{a}\times \overrightarrow{b}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 3 & 1 & 2 \\ 2 & -2 & 4 \\ \end{matrix} \right|\] \[=\hat{i}(4+4)-\hat{j}(12-4)+\hat{k}(-6-2)\] \[=8\hat{i}-8\hat{j}-8\hat{k}\] \[\therefore \]Area\[=|\overrightarrow{a}\times \overrightarrow{b}|=\sqrt{{{(8)}^{2}}+{{(-8)}^{2}}+{{(-8)}^{2}}}\] \[=8\sqrt{3}\]sq units
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