Let vec{OD} = hat{i} + 2hat{j} + 6hat{k} and | vedclass

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(C) In parallelogram $OBDC$, diagonals are $\vec{d_1} = \vec{OD} = \hat{i} + 2\hat{j} + 6\hat{k}$ and $\vec{d_2} = \vec{CB} = \vec{OB} - \vec{OC} = -3

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$9$

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(C) In parallelogram $OBDC$, diagonals are $\vec{d_1} = \vec{OD} = \hat{i} + 2\hat{j} + 6\hat{k}$ and $\vec{d_2} = \vec{CB} = \vec{OB} - \vec{OC} = -3\hat{i} - 2\hat{k}$.We know $\vec{OD} = \vec{OB} + \vec{OC}$.Adding the two equations: $2\vec{OB} = (\hat{i} + 2\hat{j} + 6\hat{k}) + (-3\hat{i} - 2\hat{k}) = -2\hat{i} + 2\hat{j} + 4\hat{k} \implies \vec{OB} = -\hat{i} + \hat{j} + 2\hat{k}$.Subtracting the two equations: $2\vec{OC} = (\hat{i} + 2\hat{j} + 6\hat{k}) - (-3\hat{i} - 2\hat{k}) = 4\hat{i} + 2\hat{j} + 8\hat{k} \implies \vec{OC} = 2\hat{i} + \hat{j} + 4\hat{k}$.The volume of the parallelepiped is given by the scalar triple product $|\vec{OA} \cdot (\vec{OB} 	imes \vec{OC})|$.$\vec{OA} \cdot (\vec{OB} 	imes \vec{OC}) = \begin{vmatrix} 1 & 2 & 3 \ -1 & 1 & 2 \ 2 & 1 & 4 \end{vmatrix}$.$= 1(4 - 2) - 2(-4 - 4) + 3(-1 - 2) = 1(2) - 2(-8) + 3(-3) = 2 + 16 - 9 = 9$.Thus, the volume is $9$ cubic units.

Let $\vec{OD} = \hat{i} + 2\hat{j} + 6\hat{k}$ and $\vec{CB} = -3\hat{i} - 2\hat{k}$ be the diagonals of the parallelogram $OBDC$. If $\vec{OA} = \hat{i} + 2\hat{j} + 3\hat{k}$, then the volume of the parallelepiped determined by vectors $\vec{OA}, \vec{OB}$, and $\vec{OC}$ (in cubic units) is:

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