If the volume of parallelopiped formed by the vectors `hati+lamdahatj+hatk,hatj+lamdahatk` and `lamdahati+hatk` is minimum then `lamda` is equal to

Options

A. `-(1)/(sqrt(3))`
B. `(1)/(sqrt(3))`
C. `sqrt(3)`
D. `-sqrt(3)`

Answer

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

### Step 1: Write the vectors
The vectors are:
1. \(\mathbf{A} = \hat{i} + \lambda \hat{j} + \hat{k}\)
2. \(\mathbf{B} = \hat{j} + \lambda \hat{k}\)
3. \(\mathbf{C} = \lambda \hat{i} + \hat{k}\)

### Step 2: Form the matrix
The volume \(V\) of the parallelepiped formed by these vectors can be calculated using the determinant of the matrix formed by these vectors:
\[ V = \begin{vmatrix} 1 & \lambda & 1 \\ 0 & 1 & \lambda \\ \lambda & 0 & 1 \end{vmatrix} \]

### Step 3: Calculate the determinant
To find the determinant, we can expand it:
\[ V = 1 \cdot \begin{vmatrix} 1 & \lambda \\ 0 & 1 \end{vmatrix} - \lambda \cdot \begin{vmatrix} 0 & \lambda \\ \lambda & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 0 & 1 \\ \lambda & 0 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
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. \(\begin{vmatrix} 0 & 1 \\ \lambda & 0 \end{vmatrix} = 0 \cdot 0 - 1 \cdot \lambda = -\lambda\)

Putting it all together:
\[ V = 1 - \lambda(-\lambda^2) - \lambda = 1 + \lambda^3 - \lambda \]
Thus, we have:
\[ V = \lambda^3 - \lambda + 1 \]

### Step 4: Differentiate the volume with respect to \(\lambda\)
To find the minimum volume, we differentiate \(V\) with respect to \(\lambda\):
\[ \frac{dV}{d\lambda} = 3\lambda^2 - 1 \]

### Step 5: Set the derivative to zero
Setting the derivative equal to zero for 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 nature of critical points
To determine whether these points are minima or maxima, we find the second derivative:
\[ \frac{d^2V}{d\lambda^2} = 6\lambda \]
Evaluating at \(\lambda = \frac{1}{\sqrt{3}}\):
\[ \frac{d^2V}{d\lambda^2} = 6 \cdot \frac{1}{\sqrt{3}} > 0 \quad (\text{indicating a minimum}) \]
Evaluating at \(\lambda = -\frac{1}{\sqrt{3}}\):
\[ \frac{d^2V}{d\lambda^2} = 6 \cdot -\frac{1}{\sqrt{3}} < 0 \quad (\text{indicating a maximum}) \]

### Conclusion
Thus, the value of \(\lambda\) for which the volume of the parallelepiped is minimized is:
\[ \lambda = \frac{1}{\sqrt{3}} \]