A) \[\frac{1}{2}\sqrt{17}\]
B) \[\frac{1}{2}\sqrt{14}\]
C) \[\sqrt{41}\]
D) \[\frac{1}{2}\sqrt{7}\]
Correct Answer: C
Solution :
The area of parallelogram is given by \[=|\overrightarrow{AB}\times \overrightarrow{AD}|\] \[=\frac{1}{2}|\overrightarrow{AC}\times \overrightarrow{BD}|\] Here we are given adjacent sides and so \[\overrightarrow{AB}\times \overrightarrow{AD}=\left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 0 & -1 \\ 1 & 2 & 0 \\ \end{matrix} \right|=2\mathbf{i}-\mathbf{j}+6\mathbf{k}\] Hence required area is \[=|2\mathbf{i}-\mathbf{j}+6\mathbf{k}|=\sqrt{41}.\]
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