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(C) The volume $V$ of a parallelepiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by the absolute value of the scalar triple product:

Vedclass · Accepted Answer

$\frac{1}{\sqrt{3}}$

Vedclass · Answer

(C) The volume $V$ of a parallelepiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by the absolute value of the scalar triple product: $V = |[\vec{a} \vec{b} \vec{c}]|$.Calculating the determinant:$V = \left| \det \begin{bmatrix} 1 & a & 1 \ 0 & 1 & a \ a & 0 & 1 \end{bmatrix} ight| = |1(1-0) - a(0-a^2) + 1(0-a)| = |1 + a^3 - a|$.To find the minimum volume,we analyze $f(a) = a^3 - a + 1$.Taking the derivative with respect to $a$: $f'(a) = 3a^2 - 1$.Setting $f'(a) = 0$,we get $3a^2 = 1$,which implies $a = \pm \frac{1}{\sqrt{3}}$.Using the second derivative test: $f''(a) = 6a$.For $a = \frac{1}{\sqrt{3}}$,$f''(\frac{1}{\sqrt{3}}) = 6(\frac{1}{\sqrt{3}}) = 2\sqrt{3} > 0$,which indicates a local minimum.Thus,the volume is minimum at $a = \frac{1}{\sqrt{3}}$.

If the vectors $\vec{a} = \hat{i} + a\hat{j} + \hat{k}$,$\vec{b} = \hat{j} + a\hat{k}$,and $\vec{c} = a\hat{i} + \hat{k}$ are given,find the value of $a$ for which the volume of the parallelepiped formed by these three vectors as coterminous edges is minimum.

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