Two vectors `vec(a) and vec(b)` are non-zero and non-collinear. What is the value of x for which the vectors `vec(p)(x-2)vec(a)+vec(b) and vec(q)=(x+1)vec(a)-vec(b)` are collinear?

Options

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

Explanation

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

### Step 1: Set up the collinearity condition
Two vectors \( \vec{p} \) and \( \vec{q} \) are collinear if there exists a scalar \( k \) such that:
\[ \vec{p} = k \vec{q} \]
This means we can express the equation as:
\[ (x-2)\vec{a} + \vec{b} = k \left( (x+1)\vec{a} - \vec{b} \right) \]

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

### Step 3: Rearranging the equation
Rearranging the equation, we have:
\[ (x-2)\vec{a} + \vec{b} = k(x+1)\vec{a} - k\vec{b} \]
This can be rewritten as:
\[ (x-2)\vec{a} + \vec{b} + k\vec{b} = k(x+1)\vec{a} \]

### Step 4: Grouping terms
Now, we can group the terms involving \( \vec{a} \) and \( \vec{b} \):
\[ (x-2)\vec{a} - k(x+1)\vec{a} = -k\vec{b} - \vec{b} \]
This simplifies to:
\[ \left( (x-2) - k(x+1) \right)\vec{a} = (-k - 1)\vec{b} \]

### Step 5: Equating coefficients
Since \( \vec{a} \) and \( \vec{b} \) are non-collinear, the coefficients of \( \vec{a} \) and \( \vec{b} \) must be equal to zero:
1. For \( \vec{a} \):
\[ (x-2) - k(x+1) = 0 \]
2. For \( \vec{b} \):
\[ -k - 1 = 0 \]

### Step 6: Solve for \( k \)
From the second equation, we can solve for \( k \):
\[ -k - 1 = 0 \implies k = -1 \]

### Step 7: Substitute \( k \) into the first equation
Now substitute \( k = -1 \) into the first equation:
\[ (x-2) - (-1)(x+1) = 0 \]
This simplifies to:
\[ (x-2) + (x+1) = 0 \]
\[ 2x - 1 = 0 \]

### Step 8: Solve for \( x \)
Now solve for \( x \):
\[ 2x = 1 \implies x = \frac{1}{2} \]

### Final Answer
The value of \( x \) for which the vectors \( \vec{p} \) and \( \vec{q} \) are collinear is:
\[ \boxed{\frac{1}{2}} \]