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 \( \mathbf{i} + \mathbf{j} + \mathbf{k} \), \( \mathbf{j} + a\mathbf{k} \), and \( a\mathbf{i} + \mathbf{k} \) becomes minimum, we can follow these steps:

### Step 1: Write the vectors
Let:
- \( \mathbf{v_1} = \mathbf{i} + \mathbf{j} + \mathbf{k} = (1, 1, 1) \)
- \( \mathbf{v_2} = \mathbf{j} + a\mathbf{k} = (0, 1, a) \)
- \( \mathbf{v_3} = a\mathbf{i} + \mathbf{k} = (a, 0, 1) \)

### Step 2: Calculate the volume using the scalar triple product
The volume \( V \) of the parallelepiped formed by these vectors can be calculated using the scalar triple product:
\[ V = |\mathbf{v_1} \cdot (\mathbf{v_2} \times \mathbf{v_3})| \]

### Step 3: Compute the cross product \( \mathbf{v_2} \times \mathbf{v_3} \)
To find \( \mathbf{v_2} \times \mathbf{v_3} \):
\[ \mathbf{v_2} \times \mathbf{v_3} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & a \\ a & 0 & 1 \end{vmatrix} \]

Calculating this determinant:
\[ = \mathbf{i} \begin{vmatrix} 1 & a \\ 0 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 0 & a \\ a & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 0 & 1 \\ a & 0 \end{vmatrix} \]
\[ = \mathbf{i} (1 \cdot 1 - 0 \cdot a) - \mathbf{j} (0 \cdot 1 - a \cdot a) + \mathbf{k} (0 \cdot 0 - 1 \cdot a) \]
\[ = \mathbf{i} (1) - \mathbf{j} (-a^2) - \mathbf{k} (a) \]
\[ = \mathbf{i} + a^2 \mathbf{j} - a \mathbf{k} \]

### Step 4: Dot product with \( \mathbf{v_1} \)
Now compute \( \mathbf{v_1} \cdot (\mathbf{v_2} \times \mathbf{v_3}) \):
\[ \mathbf{v_1} \cdot (\mathbf{v_2} \times \mathbf{v_3}) = (1, 1, 1) \cdot (1, a^2, -a) \]
\[ = 1 \cdot 1 + 1 \cdot a^2 + 1 \cdot (-a) = 1 + a^2 - a \]

### Step 5: Volume expression
Thus, the volume \( V \) is:
\[ V = |1 + a^2 - a| \]

### Step 6: Minimize the volume
To minimize \( V \), we need to minimize the expression \( 1 + a^2 - a \). We can do this by taking the derivative and setting it to zero:
\[ \frac{dV}{da} = 2a - 1 \]
Setting the derivative to zero:
\[ 2a - 1 = 0 \implies a = \frac{1}{2} \]

### Step 7: Verify minimum condition
To confirm that this is a minimum, we take the second derivative:
\[ \frac{d^2V}{da^2} = 2 \]
Since \( 2 > 0 \), this indicates that the function is concave up, confirming a minimum.

### Conclusion
Thus, the value of \( a \) that minimizes the volume of the parallelepiped is:
\[ \boxed{\frac{1}{2}} \]