A) \[4\sqrt{6}\]
B) \[\frac{1}{2}\sqrt{21}\]
C) \[\frac{\sqrt{6}}{2}\]
D) \[\sqrt{6}\]
Correct Answer: A
Solution :
\[\mathbf{a}+\mathbf{b}=2\mathbf{i}+4\mathbf{j}+6\mathbf{k},\] \[\mathbf{b}+\mathbf{c}=8\mathbf{i}+12\mathbf{j}+16\,\mathbf{k}\] Area of parallelogram \[=\frac{1}{2}|\overrightarrow{A}\times \overrightarrow{B}|\], where \[\overrightarrow{A}\] and \[\overrightarrow{B}\] are diagonals \[=\frac{1}{2}\left| \,\left| \,\begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 4 & 6 \\ 8 & 12 & 16 \\ \end{matrix}\, \right|\, \right|\] \[=\frac{1}{2}|\mathbf{i}(64-72)-\mathbf{j}(32-48)+\mathbf{k}(24-32)|\] \[=\frac{1}{2}|-8\mathbf{i}+16\mathbf{j}-8\mathbf{k}|\,=\,|-4\mathbf{i}+8\mathbf{j}-4\mathbf{k}|\] \[=\sqrt{16+64+16}=\sqrt{96}=4\sqrt{6}\].
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