A) \[\sqrt{171}\,\text{unit}\]
B) \[\sqrt{72}\,\text{unit}\]
C) \[171\,\,\text{unit}\]
D) \[\sqrt{191}\,\,\text{unit}\]
E) None of the above
Correct Answer: A
Solution :
Area of a parallelogram is \[=\frac{1}{2}|\overrightarrow{a}\times \overrightarrow{b}|\] where \[\overrightarrow{a}\] and \[\overrightarrow{b}\] are diagonals. Given, \[\overrightarrow{a}=\overrightarrow{p}=5\hat{i}-4\hat{j}+3\hat{k},\] \[\overrightarrow{b}=\overrightarrow{q}=3\hat{i}+2\hat{j}-\hat{k},\] \[\therefore \] \[\overrightarrow{a}\times \overrightarrow{b}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 5 & -\,4 & 3 \\ 3 & 2 & -1 \\ \end{matrix} \right|\] \[\overrightarrow{a}\times \overrightarrow{b}=\hat{i}(4-6)-\hat{j}(-5-9)+\hat{k}(10+12)\] \[\overrightarrow{a}\times \overrightarrow{b}=-\,2\hat{i}+14\hat{j}+22\hat{k}\] \[|\overrightarrow{a}\times \overrightarrow{b}|\,\,=\sqrt{{{(2)}^{2}}+{{(14)}^{2}}+{{(22)}^{2}}}=\sqrt{684}\] Area\[=\frac{|\overrightarrow{a}\times \overrightarrow{b}|}{2}=\frac{\sqrt{684}}{2}=\sqrt{\frac{684}{4}}=\sqrt{171}\]
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