A) \[3\sqrt{2}\]
B) \[\frac{3}{\sqrt{2}}\]
C) \[\sqrt{2}\]
D) None of these
Correct Answer: B
Solution :
Let \[\mathbf{p}=2\mathbf{a}-\mathbf{b}\] and \[\mathbf{q}=4\mathbf{a}-5\mathbf{b}.\] Then \[\mathbf{p}\times \mathbf{q}=(2\mathbf{a}-\mathbf{b})\times (4\mathbf{a}-5\mathbf{b})=-6(\mathbf{a}\times \mathbf{b})\] \[=-6|\mathbf{a}||\mathbf{b}|\sin \frac{\pi }{4}\mathbf{\hat{n}}=-6\times \frac{1}{\sqrt{2}}\mathbf{\hat{n}}=-3\sqrt{2}\,\mathbf{\hat{n}}.\] Hence the area of the given parallelogram \[=\frac{1}{2}|\mathbf{p}\times \mathbf{q}|\,=\frac{3}{\sqrt{2}}.\]
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