Let `vec(V_(1))=hati+ahatj+hatk, vec(V_(2))=hatj+ahatk` and `vec(V_(3))=ahati+hatk, AA a gt 0.` If `[(vec(V_(1)),vec(V_(2)),vec(V_(3)))]` is minimum, then the value of a is

Options

A. `sqrt3`
B. `3`
C. `(1)/(3)`
D. `(1)/(sqrt3)`

Answer

To solve the problem, we need to find the value of \( a \) such that the determinant formed by the vectors \( \vec{V_1} \), \( \vec{V_2} \), and \( \vec{V_3} \) is minimized. Let's go through the steps systematically.

### Step 1: Define the vectors
We have:
- \( \vec{V_1} = \hat{i} + a\hat{j} + \hat{k} \)
- \( \vec{V_2} = 0\hat{i} + \hat{j} + a\hat{k} \)
- \( \vec{V_3} = a\hat{i} + 0\hat{j} + \hat{k} \)

### Step 2: Set up the determinant
The determinant of the matrix formed by these vectors can be expressed as:
\[ D = \begin{vmatrix} 1 & a & 1 \\ 0 & 1 & a \\ a & 0 & 1 \end{vmatrix} \]

### Step 3: Calculate the determinant
To compute the determinant, we can use the rule of Sarrus or cofactor expansion. Let's expand along the first row:
\[ D = 1 \cdot \begin{vmatrix} 1 & a \\ 0 & 1 \end{vmatrix} - a \cdot \begin{vmatrix} 0 & a \\ a & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 0 & 1 \\ a & 0 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix} 1 & a \\ 0 & 1 \end{vmatrix} = 1 \)
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 these back into the determinant:
\[ D = 1 \cdot 1 - a \cdot (-a^2) + 1 \cdot (-a) = 1 + a^3 - a \]
Thus, we have:
\[ D = a^3 - a + 1 \]

### Step 4: Find the critical points
To find the minimum value of \( D \), we take the derivative with respect to \( a \):
\[ D' = 3a^2 - 1 \]
Setting the derivative equal to zero to find critical points:
\[ 3a^2 - 1 = 0 \implies 3a^2 = 1 \implies a^2 = \frac{1}{3} \implies a = \frac{1}{\sqrt{3}} \text{ (since } a > 0\text{)} \]

### Step 5: Verify if it's a minimum
To confirm that this critical point is a minimum, we can find the second derivative:
\[ D'' = 6a \]
Substituting \( a = \frac{1}{\sqrt{3}} \):
\[ D'' = 6 \cdot \frac{1}{\sqrt{3}} > 0 \]
Since \( D'' > 0 \), this indicates that \( D \) has a local minimum at \( a = \frac{1}{\sqrt{3}} \).

### Conclusion
Thus, the value of \( a \) that minimizes the determinant is:
\[ \boxed{\frac{1}{\sqrt{3}}} \]