`veca,vecb,vecc` are three pairwise non-collinear vectors. `veca+ 6vecb` is collinear with `vecc, vecb+5vecc` is collinear with `veca`, then `veca+alphavecb+betavecc=0`, then `alpha+beta=` ?

Options

A. 36
B. 30
C. 25
D. 39

Answer

To solve the problem, we need to analyze the given conditions and derive the necessary equations step by step.

### Step 1: Set up the equations from the collinearity conditions
We know that:
1. \(\vec{a} + 6\vec{b}\) is collinear with \(\vec{c}\), which means:
\[ \vec{a} + 6\vec{b} = \lambda \vec{c} \quad \text{(for some scalar } \lambda\text{)} \]
2. \(\vec{b} + 5\vec{c}\) is collinear with \(\vec{a}\), which means:
\[ \vec{b} + 5\vec{c} = \mu \vec{a} \quad \text{(for some scalar } \mu\text{)} \]

### Step 2: Rearranging the equations
From the first equation:
\[ \vec{a} + 6\vec{b} - \lambda \vec{c} = 0 \quad \text{(1)} \]
From the second equation:
\[ \vec{b} + 5\vec{c} - \mu \vec{a} = 0 \quad \text{(2)} \]

### Step 3: Multiply the second equation by 6
To eliminate \(\vec{b}\), we multiply the second equation by 6:
\[ 6\vec{b} + 30\vec{c} - 6\mu \vec{a} = 0 \quad \text{(3)} \]

### Step 4: Subtract equation (3) from equation (1)
Now, we will subtract equation (3) from equation (1):
\[ (\vec{a} + 6\vec{b} - \lambda \vec{c}) - (6\vec{b} + 30\vec{c} - 6\mu \vec{a}) = 0 \]
This simplifies to:
\[ \vec{a} - \lambda \vec{c} - 30\vec{c} + 6\mu \vec{a} = 0 \]
Rearranging gives:
\[ \vec{a} + 6\mu \vec{a} = \lambda \vec{c} + 30\vec{c} \]
Factoring out the vectors:
\[ (1 + 6\mu)\vec{a} = (30 + \lambda)\vec{c} \quad \text{(4)} \]

### Step 5: Use the non-collinearity condition
Since \(\vec{a}\) and \(\vec{c}\) are non-collinear, the coefficients must be equal to zero:
1. \(1 + 6\mu = 0\)
2. \(30 + \lambda = 0\)

### Step 6: Solve for \(\mu\) and \(\lambda\)
From \(1 + 6\mu = 0\):
\[ 6\mu = -1 \implies \mu = -\frac{1}{6} \]
From \(30 + \lambda = 0\):
\[ \lambda = -30 \]

### Step 7: Substitute \(\mu\) and \(\lambda\) back
Substituting \(\lambda\) and \(\mu\) back into the original equations:
From \(\vec{a} + 6\vec{b} = -30\vec{c}\):
\[ \vec{a} + 6\vec{b} + 30\vec{c} = 0 \]
This means:
\[ \vec{a} + 6\vec{b} + 30\vec{c} = 0 \quad \text{(5)} \]

### Step 8: Relate to the given equation
We are given that:
\[ \vec{a} + \alpha \vec{b} + \beta \vec{c} = 0 \]
From equation (5), we can identify:
\(\alpha = 6\) and \(\beta = 30\).

### Step 9: Find \(\alpha + \beta\)
Thus:
\[ \alpha + \beta = 6 + 30 = 36 \]

### Final Answer
\(\alpha + \beta = 36\)

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