Question

The value of 'a' for which the scaler triple product formed by the vectors \(\vec\alpha=\hat{i}+a\hat{j}+\hat{k}\),  \(\vec{\beta}=\hat{j}+a\hat{k}\)  and  \(\vec{\gamma}=a\hat{i}+\hat{k}\)  is maximum, is

• A. 3
• B. -3
• C. \(\frac{1}{\sqrt3}\)
• D. \(-\frac{1}{\sqrt3}\)

Answer 1

The Correct Option is D

We are given a determinant: \[ \begin{vmatrix} 1 & a & 1 \\ 0 & 1 & a \\ a & 0 & 1 \end{vmatrix} \]

 Step 1: Expand the determinant
Using the first row: \[ = 1 \cdot (1 \cdot 1 - a \cdot 0) - a \cdot (0 \cdot 1 - a \cdot a) + 1 \cdot (0 \cdot 0 - 1 \cdot a) \] \[ = 1(1) - a(-a^2) + 1(-a) = 1 + a^3 - a \] 

Step 2: Differentiate with respect to \(a\)
\[ \frac{d}{da}(1 + a^3 - a) = 3a^2 - 1 \] 

Step 3: Set derivative to 0 for critical point
\[ 3a^2 - 1 = 0 \Rightarrow a^2 = \frac{1}{3} \Rightarrow a = \boxed{\pm\frac{1}{\sqrt{3}}} \] 

Correct Option: (D) \(\boxed{\frac{-1}{\sqrt{3}}}\)

Answer 2

Given:
\[ \alpha = \hat{i} + a\hat{j} + \hat{k},\quad \beta = \hat{j} + a\hat{k},\quad \gamma = a\hat{i} + \hat{k} \] 

Scalar triple product:
\[ [\alpha, \beta, \gamma] = \begin{vmatrix} 1 & a & 1 \\ 0 & 1 & a \\ a & 0 & 1 \end{vmatrix} \] 

Step 1: Evaluate the determinant
\[ = 1(1 \cdot 1 - a \cdot 0) - a(0 \cdot 1 - a \cdot a) + 1(0 \cdot 0 - 1 \cdot a) \] \[ = 1 + a^3 - a = f(a) \] 

Step 2: Find critical points
\[ f'(a) = 3a^2 - 1 = 0 \Rightarrow a^2 = \frac{1}{3} \Rightarrow a = \pm\frac{1}{\sqrt{3}} \] 

Step 3: Second derivative test
\[ f''(a) = 6a \Rightarrow f''\left(\frac{1}{\sqrt{3}}\right) > 0 \text{ (min)},\quad f''\left(-\frac{1}{\sqrt{3}}\right) < 0 \text{ (max)} \] Therefore, maximum at: \(\boxed{a = -\frac{1}{\sqrt{3}}}\)
Correct Option: (D) \(\boxed{\frac{-1}{\sqrt{3}}}\)