The area of the parallelogram having diagonal | vedclass

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(B) The area of a parallelogram with diagonals $\vec{d_1}$ and $\vec{d_2}$ is given by the formula: $	ext{Area} = \frac{1}{2} |\vec{d_1} 	imes \vec{

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$5\sqrt{3}$

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(B) The area of a parallelogram with diagonals $\vec{d_1}$ and $\vec{d_2}$ is given by the formula: $	ext{Area} = \frac{1}{2} |\vec{d_1} 	imes \vec{d_2}|$.Given $\vec{d_1} = 3\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{d_2} = \hat{i} - 3\hat{j} + 4\hat{k}$.First,calculate the cross product $\vec{d_1} 	imes \vec{d_2}$:$\vec{d_1} 	imes \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 1 & -2 \ 1 & -3 & 4 \end{vmatrix} = \hat{i}(4 - 6) - \hat{j}(12 - (-2)) + \hat{k}(-9 - 1) = -2\hat{i} - 14\hat{j} - 10\hat{k}$.Now,find the magnitude of the cross product:$|\vec{d_1} 	imes \vec{d_2}| = \sqrt{(-2)^2 + (-14)^2 + (-10)^2} = \sqrt{4 + 196 + 100} = \sqrt{300} = 10\sqrt{3}$.Finally,the area is $\frac{1}{2} 	imes 10\sqrt{3} = 5\sqrt{3}$ square units.

The area of the parallelogram having diagonals $\vec{d_1} = 3\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{d_2} = \hat{i} - 3\hat{j} + 4\hat{k}$ is:

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