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(D) The area of a parallelogram formed by two vectors $\overrightarrow A$ and $\overrightarrow B$ is given by the magnitude of their cross product,i.e

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$5$

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(D) The area of a parallelogram formed by two vectors $\overrightarrow A$ and $\overrightarrow B$ is given by the magnitude of their cross product,i.e.,$Area = |\overrightarrow A 	imes \overrightarrow B|$.Given $\overrightarrow A = 2\hat i + 3\hat j$ and $\overrightarrow B = \hat i + 4\hat j$.$\overrightarrow A 	imes \overrightarrow B = (2\hat i + 3\hat j) 	imes (\hat i + 4\hat j)$.Using the cross product rules $(\hat i 	imes \hat i = 0, \hat j 	imes \hat j = 0, \hat i 	imes \hat j = \hat k, \hat j 	imes \hat i = -\hat k)$:$\overrightarrow A 	imes \overrightarrow B = 2(4)(\hat i 	imes \hat j) + 3(1)(\hat j 	imes \hat i) = 8\hat k - 3\hat k = 5\hat k$.The magnitude is $|\overrightarrow A 	imes \overrightarrow B| = |5\hat k| = 5$.Thus,the area is $5 	ext{ units}^2$.

The area of the parallelogram represented by the vectors $\overrightarrow A = 2\hat i + 3\hat j$ and $\overrightarrow B = \hat i + 4\hat j$ is ....... $units^2$.

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