If the volume of the parallelepiped formed by the vectors $\vec{a} = \hat{i} + \lambda \hat{j} + \hat{k}$,…
If the volume of the parallelepiped formed by the vectors $\vec{a} = \hat{i} + \lambda \hat{j} + \hat{k}$, $\vec{b} = \hat{j} + \lambda \hat{k}$, and $\vec{c} = \lambda \hat{i} + \hat{k}$ is minimum, then $\lambda$ is equal to:
Solution
The volume of the parallelepiped can be expressed as $\text{Volume} = [\vec{A}, \vec{B}, \vec{C}]$, where $\vec{A}$, $\vec{B}$, and $\vec{C}$ are the given vectors.
Let $f(\lambda) = \begin{vmatrix} 1 & \lambda & 1 \ 0 & 1 & \lambda \ \lambda & 0 & 1 \end{vmatrix} = \lambda^3 - \lambda + 1$.
The derivative of $f(\lambda)$ is $f'(\lambda) = 3\lambda^2 - 1$.
The roots of $f'(\lambda)$ determine the critical points. Solving $f'(\lambda) = 0$, we find $\lambda = \pm \frac{1}{\sqrt{3}}$.
At $\lambda = \frac{1}{\sqrt{3}}$, $f'(\lambda)$ changes sign from negative to positive, indicating a point of local minima.
Thus, at $\lambda = \frac{1}{\sqrt{3}}$, the volume of the parallelepiped is minimized.