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(B) Given the adjacent sides of the parallelogram are $\vec{a} = \hat{i} + \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} + \hat{k}$.The area of a paralle

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$\sqrt{3}$

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(B) Given the adjacent sides of the parallelogram are $\vec{a} = \hat{i} + \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} + \hat{k}$.The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product: $|\vec{a} 	imes \vec{b}|$.First,calculate the cross product $\vec{a} 	imes \vec{b}$:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 0 & 1 \ 2 & 1 & 1 \end{vmatrix}$$= \hat{i}(0(1) - 1(1)) - \hat{j}(1(1) - 1(2)) + \hat{k}(1(1) - 0(2))$$= \hat{i}(-1) - \hat{j}(-1) + \hat{k}(1)$$= -\hat{i} + \hat{j} + \hat{k}$.Now,calculate the magnitude of the resulting vector:$|\vec{a} 	imes \vec{b}| = \sqrt{(-1)^2 + (1)^2 + (1)^2}$$= \sqrt{1 + 1 + 1} = \sqrt{3}$.Thus,the area of the parallelogram is $\sqrt{3}$ square units.

The area of the parallelogram whose adjacent sides are $\hat{i}+\hat{k}$ and $2\hat{i}+\hat{j}+\hat{k}$ is

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