Find the value of `lambda` for which the volume of parallelepiped formed by the vector `hati+lambda hat j+hat k. hatj+lambda hatk` and `lambda hti + hatk` is minimum.

Options

A. `-1/sqrt 3`
B. `-sqrt3`
C. `sqrt 3`
D. `1/sqrt 3`

Answer

To find the value of \( \lambda \) for which the volume of the parallelepiped formed by the vectors \( \mathbf{a} = \hat{i} + \lambda \hat{j} + \hat{k} \), \( \mathbf{b} = \hat{j} + \lambda \hat{k} \), and \( \mathbf{c} = \lambda \hat{i} + \hat{k} \) is minimized, we will follow these steps:

### Step 1: Write the vectors
We have the following vectors:
- \( \mathbf{a} = \hat{i} + \lambda \hat{j} + \hat{k} \)
- \( \mathbf{b} = \hat{j} + \lambda \hat{k} \)
- \( \mathbf{c} = \lambda \hat{i} + \hat{k} \)

### Step 2: Calculate the volume of the parallelepiped
The volume \( V \) of the parallelepiped formed by the vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) can be calculated using the scalar triple product, which is given by the determinant of the matrix formed by these vectors:

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

This can be computed as the determinant of the matrix:

\[ \begin{vmatrix} 1 & \lambda & 1 \\ 0 & 1 & \lambda \\ \lambda & 0 & 1 \end{vmatrix} \]

### Step 3: Compute the determinant
Calculating the determinant, we have:

\[ V = \begin{vmatrix} 1 & \lambda & 1 \\ 0 & 1 & \lambda \\ \lambda & 0 & 1 \end{vmatrix} = 1 \cdot \begin{vmatrix} 1 & \lambda \\ 0 & 1 \end{vmatrix} - \lambda \cdot \begin{vmatrix} 0 & \lambda \\ \lambda & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 0 & 1 \\ \lambda & 0 \end{vmatrix} \]

Calculating the minors:

1. \( \begin{vmatrix} 1 & \lambda \\ 0 & 1 \end{vmatrix} = 1 \)

2. \( \begin{vmatrix} 0 & \lambda \\ \lambda & 1 \end{vmatrix} = 0 \cdot 1 - \lambda \cdot \lambda = -\lambda^2 \)

3. \( \begin{vmatrix} 0 & 1 \\ \lambda & 0 \end{vmatrix} = 0 \cdot 0 - 1 \cdot \lambda = -\lambda \)

Substituting these back into the determinant:

\[ V = 1 \cdot 1 - \lambda \cdot (-\lambda^2) + 1 \cdot (-\lambda) = 1 + \lambda^3 - \lambda \]

Thus, the volume is:

\[ V = \lambda^3 - \lambda + 1 \]

### Step 4: Differentiate the volume with respect to \( \lambda \)
To find the minimum volume, we differentiate \( V \):

\[ \frac{dV}{d\lambda} = 3\lambda^2 - 1 \]

### Step 5: Set the derivative to zero
Setting the derivative equal to zero to find critical points:

\[ 3\lambda^2 - 1 = 0 \]

Solving for \( \lambda \):

\[ 3\lambda^2 = 1 \implies \lambda^2 = \frac{1}{3} \implies \lambda = \pm \frac{1}{\sqrt{3}} \]

### Step 6: Determine the minimum value
To find which of these values gives a minimum volume, we can check the second derivative:

\[ \frac{d^2V}{d\lambda^2} = 6\lambda \]

At \( \lambda = \frac{1}{\sqrt{3}} \), \( \frac{d^2V}{d\lambda^2} = 6 \cdot \frac{1}{\sqrt{3}} > 0 \) (indicating a local minimum).

At \( \lambda = -\frac{1}{\sqrt{3}} \), \( \frac{d^2V}{d\lambda^2} = 6 \cdot -\frac{1}{\sqrt{3}} < 0 \) (indicating a local maximum).

Thus, the value of \( \lambda \) for which the volume is minimized is:

\[ \lambda = \frac{1}{\sqrt{3}} \]