The altitude of the parallelepiped, whose cot | vedclass

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(A) The volume of a parallelepiped is given by the scalar triple product $|\vec{a} \cdot (\vec{b} 	imes \vec{c})|$.First, calculate the cross product

Vedclass · Accepted Answer

$2\sqrt{2}/\sqrt{19} 	ext{ units}$

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(A) The volume of a parallelepiped is given by the scalar triple product $|\vec{a} \cdot (\vec{b} 	imes \vec{c})|$.First, calculate the cross product $\vec{b} 	imes \vec{c}$:$\vec{b} 	imes \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 4 & -1 \ 1 & 1 & 3 \end{vmatrix} = \hat{i}(12 - (-1)) - \hat{j}(6 - (-1)) + \hat{k}(2 - 4) = 13\hat{i} - 7\hat{j} - 2\hat{k}$.The volume $V = |\vec{a} \cdot (\vec{b} 	imes \vec{c})| = |(1, 1, 1) \cdot (13, -7, -2)| = |13 - 7 - 2| = 4 	ext{ cubic units}$.The base area $A$ is the magnitude of the cross product of the base vectors $\vec{a}$ and $\vec{b}$:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 2 & 4 & -1 \end{vmatrix} = \hat{i}(-1 - 4) - \hat{j}(-1 - 2) + \hat{k}(4 - 2) = -5\hat{i} + 3\hat{j} + 2\hat{k}$.$A = |\vec{a} 	imes \vec{b}| = \sqrt{(-5)^2 + 3^2 + 2^2} = \sqrt{25 + 9 + 4} = \sqrt{38}$.The altitude $h = V / A = 4 / \sqrt{38} = 4 / (\sqrt{2} \cdot \sqrt{19}) = 2\sqrt{2} / \sqrt{19} 	ext{ units}$.

The altitude of the parallelepiped, whose coterminous edges are the vectors $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} + 4\hat{j} - \hat{k}$, and $\vec{c} = \hat{i} + \hat{j} + 3\hat{k}$, where $\vec{a}$ and $\vec{b}$ are the sides of the base of the parallelepiped, is

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