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(B) The volume $V$ of a parallelepiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by the scalar triple product $|\vec{a} \cdot (\vec{

Vedclass · Accepted Answer

$\frac{2}{3 \sqrt{3}}+1, \frac{-2}{3 \sqrt{3}}+1$

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(B) The volume $V$ of a parallelepiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by the scalar triple product $|\vec{a} \cdot (\vec{b} 	imes \vec{c})|$.Given $\vec{a} = \hat{i}+x \hat{j}+\hat{k}$,$\vec{b} = \hat{j}+x \hat{k}$,and $\vec{c} = x \hat{i}+\hat{k}$.The scalar triple product is the determinant:$V = |\det \begin{bmatrix} 1 & x & 1 \ 0 & 1 & x \ x & 0 & 1 \end{bmatrix}|$Expanding along the first row:$V = |1(1-0) - x(0-x^2) + 1(0-x)| = |1 + x^3 - x|$.Let $f(x) = x^3 - x + 1$. To find the extrema,we set $f'(x) = 3x^2 - 1 = 0$,which gives $x^2 = \frac{1}{3}$,so $x = \pm \frac{1}{\sqrt{3}}$.At $x = \frac{1}{\sqrt{3}}$,$f(\frac{1}{\sqrt{3}}) = (\frac{1}{\sqrt{3}})^3 - \frac{1}{\sqrt{3}} + 1 = \frac{1}{3\sqrt{3}} - \frac{3}{3\sqrt{3}} + 1 = 1 - \frac{2}{3\sqrt{3}}$.At $x = -\frac{1}{\sqrt{3}}$,$f(-\frac{1}{\sqrt{3}}) = -\frac{1}{3\sqrt{3}} + \frac{1}{\sqrt{3}} + 1 = 1 + \frac{2}{3\sqrt{3}}$.Since volume must be non-negative,the maximum value is $1 + \frac{2}{3\sqrt{3}}$ and the minimum value is $1 - \frac{2}{3\sqrt{3}}$.

The maximum value and minimum value of the volume of the parallelepiped having coterminous edges $\hat{i}+x \hat{j}+\hat{k}$,$\hat{j}+x \hat{k}$,and $x \hat{i}+\hat{k}$ are respectively:

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