Question

If \[ \vec a=2\hat i+3\mu\hat j-\hat k, \qquad \vec b=\mu\hat i-2\hat j+3\hat k, \qquad \vec c=\hat i+3\hat j-2\mu\hat k \] are three vectors such that \[ \alpha\vec a+\beta\vec b+\gamma\vec c=\vec 0 \] only when \[ \alpha=\beta=\gamma=0, \] then the set of all real values of \(\mu\) is

• A. \(\mathbb{R}-\left\{9,1,-\frac76\right\}\)
• B. \(\mathbb{R}-\{1\}\)
• C. \(\mathbb{R}-\left\{1,-\frac53\right\}\)
• D. \(\mathbb{R}-\{0\}\)

Hint:

Three vectors in \(\mathbb R^3\) are linearly independent if and only if the determinant formed by their components is non-zero.

Answer 1

The Correct Option is B

Concept:
The condition \[ \alpha \vec a + \beta \vec b + \gamma \vec c = \vec 0 \] having only the trivial solution \[ \alpha = \beta = \gamma = 0 \] means that the vectors \( \vec a, \vec b, \vec c \) are linearly independent. For three vectors in \( \mathbb{R}^3 \), linear independence is equivalent to the determinant of the matrix formed by their components being non-zero. Thus we form the matrix \[ A= \begin{bmatrix} 2 & \mu & 1 \\ 3\mu & -2 & 3 \\ -1 & 3 & -2\mu \end{bmatrix} \] and require \[ |A| \neq 0. \] 

Step 1: Form the determinant.
\[ \Delta= \begin{vmatrix} 2 & \mu & 1 \\ 3\mu & -2 & 3 \\ -1 & 3 & -2\mu \end{vmatrix}. \] Expanding along the first row, \[ \Delta = 2 \begin{vmatrix} -2 & 3 \\ 3 & -2\mu \end{vmatrix} -\mu \begin{vmatrix} 3\mu & 3 \\ -1 & -2\mu \end{vmatrix} + \begin{vmatrix} 3\mu & -2 \\ -1 & 3 \end{vmatrix}. \] 

Step 2: Evaluate each minor carefully.
First minor: \[ (-2)(-2\mu) - 3(3) = 4\mu - 9 \] Contribution: \[ 2(4\mu - 9) = 8\mu - 18 \] Second minor: \[ (3\mu)(-2\mu) - 3(-1) = -6\mu^2 + 3 \] Contribution: \[ -\mu(-6\mu^2 + 3) = 6\mu^3 - 3\mu \] Third minor: \[ (3\mu)(3) - (-2)(-1) = 9\mu - 2 \] Contribution: \[ 9\mu - 2 \] 

Step 3: Combine all terms.
\[ \Delta = (8\mu - 18) + (6\mu^3 - 3\mu) + (9\mu - 2) \] \[ = 6\mu^3 + 14\mu - 20 \] Factorizing, \[ \Delta = 2(3\mu^3 + 7\mu - 10) \] Now check \( \mu = 1 \): \[ 3(1)^3 + 7(1) - 10 = 0 \] Hence \( (\mu - 1) \) is a factor. Dividing, \[ 3\mu^3 + 7\mu - 10 = (\mu - 1)(3\mu^2 + 3\mu + 10) \] Therefore, \[ \Delta = 2(\mu - 1)(3\mu^2 + 3\mu + 10) \] 

Step 4: Analyze the quadratic factor.
\[ 3\mu^2 + 3\mu + 10 \] Its discriminant is \[ D = 3^2 - 4(3)(10) = 9 - 120 = -111 < 0 \] Hence this quadratic has no real roots. Therefore, the determinant becomes zero only when \[ \mu = 1 \] 

Step 5: Apply the linear independence condition.
For linear independence, \[ \Delta \neq 0 \] Thus, \[ \mu \neq 1 \] Hence all real values except \(1\) are allowed. 

Conclusion: \[ \boxed{\mu \in \mathbb{R} - \{1\}} \]