The value of a, such that the volume of the parallelopiped formed by the vectors `hati+hatj+hatk, hatj+ahatk and ahati+hatk` becomes minimum, is

Options

A. 3
B. `-(1)/(3)`
C. `(1)/(2)`
D. `-2`

Answer

To find the value of \( a \) such that the volume of the parallelepiped formed by the vectors \( \mathbf{v_1} = \hat{i} + \hat{j} + \hat{k} \), \( \mathbf{v_2} = \hat{j} + a\hat{k} \), and \( \mathbf{v_3} = a\hat{i} + \hat{k} \) becomes minimum, we can follow these steps:

### Step 1: Write the vectors
We have the following vectors:
- \( \mathbf{v_1} = (1, 1, 1) \)
- \( \mathbf{v_2} = (0, 1, a) \)
- \( \mathbf{v_3} = (a, 0, 1) \)

### Step 2: Set up the volume formula
The volume \( V \) of the parallelepiped formed by these vectors can be calculated using the scalar triple product, which is given by the determinant of the matrix formed by these vectors:

\[ V = \left| \begin{vmatrix} 1 & 1 & 1 \\ 0 & 1 & a \\ a & 0 & 1 \end{vmatrix} \right| \]

### Step 3: Calculate the determinant
We can calculate the determinant using cofactor expansion. Expanding along the first row gives:

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

Calculating each of these 2x2 determinants:

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

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

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

Putting it all together:

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

### Step 4: Find the minimum volume
To find the value of \( a \) that minimizes \( V \), we can take the derivative of \( V \) with respect to \( a \) and set it to zero:

\[ \frac{dV}{da} = 2a - 1 \]

Setting the derivative equal to zero:

\[ 2a - 1 = 0 \implies a = \frac{1}{2} \]

### Step 5: Verify it's a minimum
To confirm that this is a minimum, we can check the second derivative:

\[ \frac{d^2V}{da^2} = 2 \]

Since the second derivative is positive, this indicates that we have a minimum.

### Conclusion
The value of \( a \) such that the volume of the parallelepiped becomes minimum is:

\[ \boxed{\frac{1}{2}} \]