If `veca=hati+6hatj+3hatk, vecb=3hati+2hatj+hatk and vec c=(alpha+1)hati+(beta-1)hatj+hatk` are linearly dependent vectors and `|vec c|=sqrt(6)`, then the possible value(s) of `(alpha+beta)` can be : (a) 1 (b) 2 (c) 3 (d) 4

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the possible values of \( \alpha + \beta \) given that the vectors \( \vec{a}, \vec{b}, \) and \( \vec{c} \) are linearly dependent and the magnitude of \( \vec{c} \) is \( \sqrt{6} \).

### Step 1: Write down the vectors
Given:
\[ \vec{a} = \hat{i} + 6\hat{j} + 3\hat{k} \]
\[ \vec{b} = 3\hat{i} + 2\hat{j} + \hat{k} \]
\[ \vec{c} = (\alpha + 1)\hat{i} + (\beta - 1)\hat{j} + \hat{k} \]

### Step 2: Set up the linear dependence condition
Since the vectors are linearly dependent, we can express \( \vec{c} \) as a linear combination of \( \vec{a} \) and \( \vec{b} \):
\[ \vec{c} = l \vec{a} + m \vec{b} \]
Expanding this, we have:
\[ (\alpha + 1)\hat{i} + (\beta - 1)\hat{j} + \hat{k} = l(\hat{i} + 6\hat{j} + 3\hat{k}) + m(3\hat{i} + 2\hat{j} + \hat{k}) \]

### Step 3: Equate coefficients
From the equation above, we can equate the coefficients of \( \hat{i}, \hat{j}, \) and \( \hat{k} \):
1. Coefficient of \( \hat{i} \):
\[ \alpha + 1 = l + 3m \tag{1} \]
2. Coefficient of \( \hat{j} \):
\[ \beta - 1 = 6l + 2m \tag{2} \]
3. Coefficient of \( \hat{k} \):
\[ 1 = 3l + m \tag{3} \]

### Step 4: Solve the equations
From equation (3), we can express \( m \) in terms of \( l \):
\[ m = 1 - 3l \tag{4} \]

Substituting equation (4) into equation (1):
\[ \alpha + 1 = l + 3(1 - 3l) \]
\[ \alpha + 1 = l + 3 - 9l \]
\[ \alpha + 1 = 3 - 8l \]
\[ \alpha = 2 - 8l \tag{5} \]

Now substituting equation (4) into equation (2):
\[ \beta - 1 = 6l + 2(1 - 3l) \]
\[ \beta - 1 = 6l + 2 - 6l \]
\[ \beta - 1 = 2 \]
\[ \beta = 3 \tag{6} \]

### Step 5: Find \( \alpha \)
Now substituting equation (6) into equation (5):
\[ \alpha = 2 - 8l \]
We need to find \( l \) now.

### Step 6: Use the magnitude condition
We know that the magnitude of \( \vec{c} \) is \( \sqrt{6} \):
\[ |\vec{c}| = \sqrt{(\alpha + 1)^2 + (\beta - 1)^2 + 1^2} = \sqrt{6} \]
Squaring both sides:
\[ (\alpha + 1)^2 + (\beta - 1)^2 + 1 = 6 \]
Substituting \( \beta = 3 \):
\[ (\alpha + 1)^2 + (3 - 1)^2 + 1 = 6 \]
\[ (\alpha + 1)^2 + 4 + 1 = 6 \]
\[ (\alpha + 1)^2 + 5 = 6 \]
\[ (\alpha + 1)^2 = 1 \]
Taking square roots:
\[ \alpha + 1 = 1 \quad \text{or} \quad \alpha + 1 = -1 \]
This gives:
1. \( \alpha = 0 \)
2. \( \alpha = -2 \)

### Step 7: Calculate \( \alpha + \beta \)
Now we can find \( \alpha + \beta \):
1. If \( \alpha = 0 \):
\[ \alpha + \beta = 0 + 3 = 3 \]
2. If \( \alpha = -2 \):
\[ \alpha + \beta = -2 + 3 = 1 \]

### Conclusion
The possible values of \( \alpha + \beta \) are \( 3 \) and \( 1 \).