If vec{OA}=6 hat{i}+3 hat{j}-4 hat{k}, vec{OB | vedclass

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(D) The volume of a parallelepiped with coterminous edges $\vec{OA}, \vec{OB}, \vec{OC}$ is given by the scalar triple product $|\vec{OA} \cdot (\vec{

Vedclass · Accepted Answer

$\frac{17}{\sqrt{6}}$

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(D) The volume of a parallelepiped with coterminous edges $\vec{OA}, \vec{OB}, \vec{OC}$ is given by the scalar triple product $|\vec{OA} \cdot (\vec{OB} 	imes \vec{OC})|$.First, we calculate the scalar triple product:$\vec{OA} \cdot (\vec{OB} 	imes \vec{OC}) = \begin{vmatrix} 6 & 3 & -4 \ 0 & 2 & 1 \ 5 & -1 & 2 \end{vmatrix} = 6(4 - (-1)) - 3(0 - 5) - 4(0 - 10) = 6(5) - 3(-5) - 4(-10) = 30 + 15 + 40 = 85$.The area of the base formed by $\vec{OB}$ and $\vec{OC}$ is $|\vec{OB} 	imes \vec{OC}|$.$\vec{OB} 	imes \vec{OC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 2 & 1 \ 5 & -1 & 2 \end{vmatrix} = \hat{i}(4 - (-1)) - \hat{j}(0 - 5) + \hat{k}(0 - 10) = 5 \hat{i} + 5 \hat{j} - 10 \hat{k}$.The magnitude of the base area is $|\vec{OB} 	imes \vec{OC}| = \sqrt{5^2 + 5^2 + (-10)^2} = \sqrt{25 + 25 + 100} = \sqrt{150} = 5 \sqrt{6}$.The height $h$ of the parallelepiped from vertex $A$ is given by $\frac{|	ext{Volume}|}{	ext{Area of base}} = \frac{85}{5 \sqrt{6}} = \frac{17}{\sqrt{6}}$.

If $\vec{OA}=6 \hat{i}+3 \hat{j}-4 \hat{k}$, $\vec{OB}=2 \hat{j}+\hat{k}$, and $\vec{OC}=5 \hat{i}-\hat{j}+2 \hat{k}$ are the coterminous edges of a parallelepiped, then the height of the parallelepiped drawn from the vertex $A$ is

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