A) \[\sqrt{180}\]
B) \[\sqrt{140}\]
C) \[\sqrt{80}\]
D) \[\sqrt{40}\]
Correct Answer: A
Solution :
Adjacent sides of parallelogram are \[\mathbf{a}=\mathbf{i}+2\mathbf{j}+3\mathbf{k}\] and \[\mathbf{b}=-\,3\,\mathbf{i}-2\mathbf{j}+\mathbf{k}\]. We know that vector area of parallelogram. \[\mathbf{a}\times \mathbf{b}=\left| \,\begin{matrix} \,\,\mathbf{i} & \,\,\mathbf{j} & \mathbf{k} \\ \,\,1 & \,\,2 & 3 \\ -3 & -2\, & 1 \\ \end{matrix}\, \right|=\mathbf{i}(2+6)-\mathbf{j}(1+9)+\mathbf{k}(-2+6)\] \[=8\mathbf{i}-10\mathbf{j}+4\mathbf{k}\]. Therefore area of parallelogram \[=|\mathbf{a}\times \mathbf{b}|=\sqrt{{{(8)}^{2}}+{{(-10)}^{2}}+{{(4)}^{2}}}=\sqrt{64+100+16}\] \[=\sqrt{180}\] sq. unit.
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