A) \[10\sqrt{3}\]
B) \[5\sqrt{3}\]
C) \[8\]
D) \[4\]
Correct Answer: B
Solution :
Area of parallelogram \[=\frac{1}{2}|\vec{a}\times \vec{b}|\] \[=\frac{1}{2}\left\| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \\ \end{matrix} \right\|=\frac{1}{2}\left| \{\hat{i}(-2)-\hat{j}(14)+\hat{k}(-10)\} \right|\] \[=|-\hat{i}-7\hat{j}-5\hat{k}|=\sqrt{1+49+25}\] \[=\sqrt{75}=5\sqrt{3}\]
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