`vecb and vecc` are non- collinear if `veca xx (vecb xx vecc) + (veca .vecb) vecb = ( 4-2x- sin y) vecb + ( x^(2) -1) vecc andd (vec. vecc) veca =veca ` then

Options

A. x =1
B. x = -1
C. `y = (4 n+1) pi/2, n in I `
D. `y ( 2n + 1) pi/2, n in I`

Explanation

To solve the problem step by step, we will analyze the given vector equation and derive the necessary conditions for the vectors to be non-collinear.

### Given:
1. \( \vec{b} \) and \( \vec{c} \) are non-collinear vectors.
2. The equation is:
\[ \vec{a} \times (\vec{b} \times \vec{c}) + (\vec{a} \cdot \vec{b}) \vec{b} = (4 - 2x - \sin y) \vec{b} + (x^2 - 1) \vec{c} \]
3. Another condition is:
\[ (\vec{c} \cdot \vec{c}) \vec{a} = \vec{c} \]

### Step 1: Rewrite the Cross Product
Using the vector triple product identity, we can rewrite \( \vec{a} \times (\vec{b} \times \vec{c}) \):
\[ \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \]
Substituting this into the equation gives:
\[ (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} + (\vec{a} \cdot \vec{b}) \vec{b} = (4 - 2x - \sin y) \vec{b} + (x^2 - 1) \vec{c} \]

### Step 2: Combine Like Terms
Now, combine the terms:
\[ (\vec{a} \cdot \vec{c}) \vec{b} + (\vec{a} \cdot \vec{b}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = (4 - 2x - \sin y) \vec{b} + (x^2 - 1) \vec{c} \]
This simplifies to:
\[ \left((\vec{a} \cdot \vec{c}) + (\vec{a} \cdot \vec{b})\right) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = (4 - 2x - \sin y) \vec{b} + (x^2 - 1) \vec{c} \]

### Step 3: Compare Coefficients
Now, we can compare coefficients of \( \vec{b} \) and \( \vec{c} \):
1. Coefficient of \( \vec{b} \):
\[ (\vec{a} \cdot \vec{c}) + (\vec{a} \cdot \vec{b}) = 4 - 2x - \sin y \]
2. Coefficient of \( \vec{c} \):
\[ -(\vec{a} \cdot \vec{b}) = x^2 - 1 \]

### Step 4: Solve for \( \vec{a} \cdot \vec{b} \)
From the coefficient of \( \vec{c} \):
\[ \vec{a} \cdot \vec{b} = 1 - x^2 \]
Substituting this into the coefficient of \( \vec{b} \):
\[ (\vec{a} \cdot \vec{c}) + (1 - x^2) = 4 - 2x - \sin y \]
Thus,
\[ \vec{a} \cdot \vec{c} = 4 - 2x - \sin y - (1 - x^2) \]
This simplifies to:
\[ \vec{a} \cdot \vec{c} = 3 - 2x + x^2 - \sin y \]

### Step 5: Use the Second Condition
Using the second condition \( (\vec{c} \cdot \vec{c}) \vec{a} = \vec{c} \), we can multiply both sides by \( \vec{c} \):
\[ (\vec{c} \cdot \vec{c})(\vec{a} \cdot \vec{c}) = \vec{c} \cdot \vec{c} \]
Assuming \( \vec{c} \cdot \vec{c} \neq 0 \), we can divide both sides by \( \vec{c} \cdot \vec{c} \):
\[ \vec{a} \cdot \vec{c} = 1 \]

### Step 6: Set Up the Equation
Now we have:
\[ 3 - 2x + x^2 - \sin y = 1 \]
This simplifies to:
\[ x^2 - 2x + 2 - \sin y = 0 \]
Rearranging gives:
\[ \sin y = x^2 - 2x + 2 \]

### Step 7: Analyze the Expression
The expression \( x^2 - 2x + 2 \) is always greater than or equal to 1 (as it is a quadratic that opens upwards). Thus, for \( \sin y \) to be valid, we need:
\[ x^2 - 2x + 2 \leq 1 \]
This leads to:
\[ x^2 - 2x + 1 \leq 0 \implies (x - 1)^2 \leq 0 \]
This implies \( x = 1 \).

### Step 8: Find \( y \)
Substituting \( x = 1 \) into the equation for \( \sin y \):
\[ \sin y = 1^2 - 2(1) + 2 = 1 \]
Thus, \( y = \frac{\pi}{2} + 2n\pi \) for \( n \in \mathbb{Z} \).

### Final Answer
The values of \( x \) and \( y \) that satisfy the conditions are:
- \( x = 1 \)
- \( y = \frac{\pi}{2} + 2n\pi \) for \( n \in \mathbb{Z} \)