The value of `a` so that the volume of the paralelopiped formed by `hati+ahatj+hatk, hatj+ahatk `and `ahati+hatk` becomes minimum is

Options

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

Explanation

To find the value of \( a \) such that the volume of the parallelepiped formed by the vectors \( \hat{i} + a\hat{j} + \hat{k} \), \( \hat{j} + a\hat{k} \), and \( a\hat{i} + \hat{j} \) becomes minimum, we can follow these steps:

### Step 1: Define the vectors
Let:
- \( \mathbf{A} = \hat{i} + a\hat{j} + \hat{k} \)
- \( \mathbf{B} = \hat{j} + a\hat{k} \)
- \( \mathbf{C} = a\hat{i} + \hat{j} \)

### Step 2: Calculate the volume of the parallelepiped
The volume \( V \) of the parallelepiped formed by the vectors \( \mathbf{A}, \mathbf{B}, \mathbf{C} \) is given by the scalar triple product:

\[ V = |\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C})| \]

### Step 3: Compute \( \mathbf{B} \times \mathbf{C} \)
First, we need to find the cross product \( \mathbf{B} \times \mathbf{C} \).

\[ \mathbf{B} = \begin{pmatrix} 0 \\ 1 \\ a \end{pmatrix}, \quad \mathbf{C} = \begin{pmatrix} a \\ 1 \\ 0 \end{pmatrix} \]

Using the determinant method for the cross product:

\[ \mathbf{B} \times \mathbf{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & a \\ a & 1 & 0 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} 1 & a \\ 1 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & a \\ a & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & 1 \\ a & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

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

So, we have:

\[ \mathbf{B} \times \mathbf{C} = -a\hat{i} + a^2\hat{j} - a\hat{k} \]

### Step 4: Compute \( \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) \)
Now we compute the dot product:

\[ \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = \left( \hat{i} + a\hat{j} + \hat{k} \right) \cdot \left( -a\hat{i} + a^2\hat{j} - a\hat{k} \right) \]

Calculating the dot product:

\[ = (1)(-a) + (a)(a^2) + (1)(-a) = -a + a^3 - a = a^3 - 2a \]

### Step 5: Set up the volume expression
Thus, the volume \( V \) is:

\[ V = |a^3 - 2a| \]

### Step 6: Find the minimum volume
To find the minimum volume, we need to minimize the expression \( |a^3 - 2a| \). We can analyze the function \( f(a) = a^3 - 2a \).

### Step 7: Differentiate and find critical points
Differentiating \( f(a) \):

\[ f'(a) = 3a^2 - 2 \]

Setting the derivative to zero to find critical points:

\[ 3a^2 - 2 = 0 \implies a^2 = \frac{2}{3} \implies a = \pm \sqrt{\frac{2}{3}} = \pm \frac{\sqrt{6}}{3} \]

### Step 8: Evaluate the second derivative
To determine the nature of these critical points, we can use the second derivative test:

\[ f''(a) = 6a \]

Evaluating at \( a = \sqrt{\frac{2}{3}} \):

\[ f''\left(\sqrt{\frac{2}{3}}\right) = 6\sqrt{\frac{2}{3}} > 0 \quad \text{(local minimum)} \]

Evaluating at \( a = -\sqrt{\frac{2}{3}} \):

\[ f''\left(-\sqrt{\frac{2}{3}}\right) = -6\sqrt{\frac{2}{3}} < 0 \quad \text{(local maximum)} \]

### Step 9: Conclusion
The minimum volume occurs at \( a = \sqrt{\frac{2}{3}} \) and the corresponding value of \( a \) that minimizes the volume of the parallelepiped is:

\[ \boxed{\frac{1}{\sqrt{3}}} \]