If `a = i +j + k, b = 4i + 3j + 4k and c = i + alphaj + betak` are linearly dependent vectors and `|c| = sqrt(3)` , then

Options

A. `alpha =1, beta =-1`
B. `alpha =1, beta =pm 1`
C. `alpha = -1, beta = pm1`
D. `alpha = pm 1, beta =1`

Answer

To solve the problem step by step, we need to analyze the given vectors and their conditions for linear dependence and magnitude.

### Step 1: Understand the vectors
We are given three vectors:
- \( \mathbf{a} = \mathbf{i} + \mathbf{j} + \mathbf{k} \)
- \( \mathbf{b} = 4\mathbf{i} + 3\mathbf{j} + 4\mathbf{k} \)
- \( \mathbf{c} = \mathbf{i} + \alpha \mathbf{j} + \beta \mathbf{k} \)

### Step 2: Set up the condition for linear dependence
The vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) are linearly dependent if the determinant of the matrix formed by their components is zero:
\[ \begin{vmatrix} 1 & 1 & 1 \\ 4 & 3 & 4 \\ 1 & \alpha & \beta \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
We can calculate the determinant using cofactor expansion:
\[ \begin{vmatrix} 1 & 1 & 1 \\ 4 & 3 & 4 \\ 1 & \alpha & \beta \end{vmatrix} = 1 \cdot \begin{vmatrix} 3 & 4 \\ \alpha & \beta \end{vmatrix} - 1 \cdot \begin{vmatrix} 4 & 4 \\ 1 & \beta \end{vmatrix} + 1 \cdot \begin{vmatrix} 4 & 3 \\ 1 & \alpha \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix} 3 & 4 \\ \alpha & \beta \end{vmatrix} = 3\beta - 4\alpha \)
2. \( \begin{vmatrix} 4 & 4 \\ 1 & \beta \end{vmatrix} = 4\beta - 4 \)
3. \( \begin{vmatrix} 4 & 3 \\ 1 & \alpha \end{vmatrix} = 4\alpha - 3 \)

Putting it all together:
\[ 3\beta - 4\alpha - (4\beta - 4) + (4\alpha - 3) = 0 \]
Simplifying this gives:
\[ 3\beta - 4\alpha - 4\beta + 4 + 4\alpha - 3 = 0 \]
Combining like terms:
\[ -\beta + \alpha + 1 = 0 \]
Thus, we have:
\[ \alpha - \beta + 1 = 0 \quad \text{(1)} \]

### Step 4: Use the magnitude condition
We are also given that the magnitude of vector \( \mathbf{c} \) is \( \sqrt{3} \):
\[ |\mathbf{c}| = \sqrt{1^2 + \alpha^2 + \beta^2} = \sqrt{3} \]
Squaring both sides:
\[ 1 + \alpha^2 + \beta^2 = 3 \]
This simplifies to:
\[ \alpha^2 + \beta^2 = 2 \quad \text{(2)} \]

### Step 5: Solve the equations
From equation (1), we can express \( \alpha \) in terms of \( \beta \):
\[ \alpha = \beta - 1 \]
Substituting this into equation (2):
\[ (\beta - 1)^2 + \beta^2 = 2 \]
Expanding:
\[ \beta^2 - 2\beta + 1 + \beta^2 = 2 \]
Combining terms:
\[ 2\beta^2 - 2\beta + 1 - 2 = 0 \]
This simplifies to:
\[ 2\beta^2 - 2\beta - 1 = 0 \]
Dividing by 2:
\[ \beta^2 - \beta - \frac{1}{2} = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula:
\[ \beta = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-\frac{1}{2})}}{2 \cdot 1} \]
\[ \beta = \frac{1 \pm \sqrt{1 + 2}}{2} = \frac{1 \pm \sqrt{3}}{2} \]

### Step 7: Find \( \alpha \)
Using \( \alpha = \beta - 1 \):
1. If \( \beta = \frac{1 + \sqrt{3}}{2} \):
\[ \alpha = \frac{1 + \sqrt{3}}{2} - 1 = \frac{-1 + \sqrt{3}}{2} \]
2. If \( \beta = \frac{1 - \sqrt{3}}{2} \):
\[ \alpha = \frac{1 - \sqrt{3}}{2} - 1 = \frac{-1 - \sqrt{3}}{2} \]

### Final Results
Thus, the values of \( \alpha \) and \( \beta \) are:
- \( \beta = 1 \) and \( \alpha = 0 \)
- \( \beta = \frac{1 + \sqrt{3}}{2} \) and \( \alpha = \frac{-1 + \sqrt{3}}{2} \)
- \( \beta = \frac{1 - \sqrt{3}}{2} \) and \( \alpha = \frac{-1 - \sqrt{3}}{2} \)