The altitude of the parallelepiped,whose cote | vedclass

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(NONE) The volume of a parallelepiped is given by the scalar triple product $|[\bar{a} \bar{b} \bar{c}]|$.
First,calculate the scalar triple product:

Vedclass · Accepted Answer

$2\sqrt{38}/19$

Vedclass · Answer

(NONE) The volume of a parallelepiped is given by the scalar triple product $|[\bar{a} \bar{b} \bar{c}]|$.
First,calculate the scalar triple product:
$[\bar{a} \bar{b} \bar{c}] = \begin{vmatrix} 1 & 1 & 1 \ 2 & 4 & -1 \ 1 & 1 & 3 \end{vmatrix} = 1(12 - (-1)) - 1(6 - (-1)) + 1(2 - 4) = 1(13) - 1(7) + 1(-2) = 13 - 7 - 2 = 4$.
So,the volume $V = 4$.
The area of the base formed by $\bar{a}$ and $\bar{b}$ is $|\bar{a} 	imes \bar{b}|$.
$\bar{a} 	imes \bar{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 2 & 4 & -1 \end{vmatrix} = \hat{i}(-1 - 4) - \hat{j}(-1 - 2) + \hat{k}(4 - 2) = -5\hat{i} + 3\hat{j} + 2\hat{k}$.
Area of base = $|-5\hat{i} + 3\hat{j} + 2\hat{k}| = \sqrt{(-5)^2 + 3^2 + 2^2} = \sqrt{25 + 9 + 4} = \sqrt{38}$.
The altitude $h$ is given by $V / 	ext{Area of base} = 4 / \sqrt{38} = 4\sqrt{38}/38 = 2\sqrt{38}/19$.

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

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