The diagonals of a parallelogram are vec{d_1} | vedclass

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

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$\frac{\sqrt{3}}{2}$

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(D) The area of a parallelogram with diagonals $\vec{d_1}$ and $\vec{d_2}$ is given by the formula: Area $= \frac{1}{2} |\vec{d_1} 	imes \vec{d_2}|$. Given $\vec{d_1} = 0\hat{i} + 1\hat{j} + 1\hat{k}$ and $\vec{d_2} = 1\hat{i} + 1\hat{j} + 0\hat{k}$. First,calculate the cross product $\vec{d_1} 	imes \vec{d_2}$: $\vec{d_1} 	imes \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 1 & 1 \ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - 1) - \hat{j}(0 - 1) + \hat{k}(0 - 1) = -\hat{i} + \hat{j} - \hat{k}$. Now,find the magnitude of the cross product: $|\vec{d_1} 	imes \vec{d_2}| = \sqrt{(-1)^2 + (1)^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$. Finally,the area is $\frac{1}{2} 	imes \sqrt{3} = \frac{\sqrt{3}}{2}$ sq. units. Thus,the correct option is $D$.

The diagonals of a parallelogram are $\vec{d_1} = \hat{j} + \hat{k}$ and $\vec{d_2} = \hat{i} + \hat{j}$. The area of the parallelogram is . . . . . . sq. units.

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