If bar{u}=hat{imath}-2 hat{jmath}+hat{k}, bar | vedclass

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(C) First,calculate the cross product $\bar{u} 	imes \bar{v}$:$\bar{u} 	imes \bar{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 1 \

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$24$ cubic units

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(C) First,calculate the cross product $\bar{u} 	imes \bar{v}$:$\bar{u} 	imes \bar{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 1 \ 3 & 0 & 1 \end{vmatrix} = \hat{i}(-2-0) - \hat{j}(1-3) + \hat{k}(0 - (-6)) = -2\hat{i} + 2\hat{j} + 6\hat{k}$.Next,define the vectors for the edges:$\bar{a} = \bar{u} 	imes \bar{v} = -2\hat{i} + 2\hat{j} + 6\hat{k}$$\bar{b} = \bar{u} + \bar{w} = (\hat{i} - 2\hat{j} + \hat{k}) + (\hat{j} - \hat{k}) = \hat{i} - \hat{j}$$\bar{c} = \bar{v} + \bar{w} = (3\hat{i} + \hat{k}) + (\hat{j} - \hat{k}) = 3\hat{i} + \hat{j}$The volume of the parallelepiped is given by the scalar triple product $|\bar{a} \cdot (\bar{b} 	imes \bar{c})|$,which is equivalent to the absolute value of the determinant of the matrix formed by these vectors:Volume $= \left| \begin{vmatrix} -2 & 2 & 6 \ 1 & -1 & 0 \ 3 & 1 & 0 \end{vmatrix} ight|$Expanding along the third column:Volume $= |6(1 - (-3))| = |6(4)| = 24$ cubic units.

If $\bar{u}=\hat{\imath}-2 \hat{\jmath}+\hat{k}, \bar{v}=3 \hat{\imath}+\hat{k}$ and $\bar{w}=\hat{\jmath}-\hat{k}$,then the volume of the parallelepiped with $\bar{u} \times \bar{v}, \bar{u}+\bar{w}$ and $\bar{v}+\bar{w}$ as coterminus edges is

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