If the volume of parallelepiped formed by the vectors $$\hat{i} + \lambda\hat{j} + \hat{k}$$, $$\hat{j} + \lambda\hat{k}$$ and $$\lambda\hat{i} + \hat{k}$$ is minimum, then $$\lambda$$ is

Let the three vectors forming the parallelepiped be:

$$\vec{a} = \hat{i} + \lambda\hat{j} + \hat{k}$$ $$\vec{b} = 0\hat{i} + \hat{j} + \lambda\hat{k}$$ $$\vec{c} = \lambda\hat{i} + 0\hat{j} + \hat{k}$$

The volume $$V$$ of a parallelepiped formed by three vectors $$\vec{a}, \vec{b}, \vec{c}$$ is given by the magnitude of their scalar triple product:

$$V = | [\vec{a} \cdot (\vec{b} \times \vec{c})] |$$

This can be calculated using the determinant of the matrix formed by their components:

$$V = \begin{vmatrix} 1 & \lambda & 1 \\ 0 & 1 & \lambda \\ \lambda & 0 & 1 \end{vmatrix}$$

Expanding the determinant along the first row:

$$V = 1(1 \cdot 1 - 0 \cdot \lambda) - \lambda(0 \cdot 1 - \lambda \cdot \lambda) + 1(0 \cdot 0 - \lambda \cdot 1)$$ $$V = 1(1) - \lambda(-\lambda^2) + 1(-\lambda)$$ $$V = 1 + \lambda^3 - \lambda$$

Thus, the volume function is $$f(\lambda) = \lambda^3 - \lambda + 1$$.

To find the minimum value, we differentiate $f(\lambda)$ with respect to $$\lambda$$:

$$f'(\lambda) = \frac{d}{d\lambda}(\lambda^3 - \lambda + 1) = 3\lambda^2 - 1$$

Setting the first derivative to zero for critical points:

$$3\lambda^2 - 1 = 0 \implies \lambda^2 = \frac{1}{3} \implies \lambda = \pm \frac{1}{\sqrt{3}}$$

To determine whether these points are a local minimum or maximum, we find the second derivative:

$$f''(\lambda) = 6\lambda$$

• For $$\lambda = \frac{1}{\sqrt{3}}$$:$$f''\left(\frac{1}{\sqrt{3}}\right) = \frac{6}{\sqrt{3}} > 0$$

  Since the second derivative is positive, $$\lambda = \frac{1}{\sqrt{3}}$$ is a point of local minimum.

• For $$\lambda = -\frac{1}{\sqrt{3}}$$:$$f''\left(-\frac{1}{\sqrt{3}}\right) = -\frac{6}{\sqrt{3}} < 0$$

  Since the second derivative is negative, $$\lambda = -\frac{1}{\sqrt{3}}$$ is a point of local maximum.

The volume is minimum when:

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