Vectors ` vec aa n d vec b` are non-collinear. Find for what value of `x` vectors ` vec c=(x-2) vec a+ vec b and vec d=(2x+1) vec a- vec b` are collinear?

Options

A. `1//3`
B. `1//2`
C. 1
D. 0

Answer

To determine the value of \( x \) for which the vectors \( \vec{c} = (x-2) \vec{a} + \vec{b} \) and \( \vec{d} = (2x+1) \vec{a} - \vec{b} \) are collinear, we can follow these steps:

### Step 1: Understand the condition for collinearity
Two vectors \( \vec{c} \) and \( \vec{d} \) are collinear if one is a scalar multiple of the other. This means there exists a scalar \( \lambda \) such that:
\[ \vec{c} = \lambda \vec{d} \]

### Step 2: Set up the equation
Given:
\[ \vec{c} = (x-2) \vec{a} + \vec{b} \]
\[ \vec{d} = (2x+1) \vec{a} - \vec{b} \]
We can write the collinearity condition as:
\[ (x-2) \vec{a} + \vec{b} = \lambda \left( (2x+1) \vec{a} - \vec{b} \right) \]

### Step 3: Expand the right side
Expanding the right side gives:
\[ (x-2) \vec{a} + \vec{b} = \lambda (2x+1) \vec{a} - \lambda \vec{b} \]

### Step 4: Rearranging the equation
Rearranging the equation, we have:
\[ (x-2) \vec{a} + \vec{b} + \lambda \vec{b} = \lambda (2x+1) \vec{a} \]
This can be rewritten as:
\[ (x-2) \vec{a} + (1 + \lambda) \vec{b} = \lambda (2x+1) \vec{a} \]

### Step 5: Compare coefficients
Now, we can compare the coefficients of \( \vec{a} \) and \( \vec{b} \) on both sides:
1. Coefficient of \( \vec{a} \):
\[ x - 2 = \lambda (2x + 1) \]
2. Coefficient of \( \vec{b} \):
\[ 1 + \lambda = 0 \implies \lambda = -1 \]

### Step 6: Substitute \( \lambda \) into the first equation
Substituting \( \lambda = -1 \) into the first equation:
\[ x - 2 = -1(2x + 1) \]
This simplifies to:
\[ x - 2 = -2x - 1 \]

### Step 7: Solve for \( x \)
Now, we can solve for \( x \):
\[ x + 2x = -1 + 2 \]
\[ 3x = 1 \]
\[ x = \frac{1}{3} \]

### Conclusion
Thus, the value of \( x \) for which the vectors \( \vec{c} \) and \( \vec{d} \) are collinear is:
\[ \boxed{\frac{1}{3}} \]