If `barb` and `barc` are two non-collinear vectors, then number of solutions (x,y) in `x,yin[0,10]` satisfying the equation `baracdot[barb+barc]=5` and `baraxx(barbxxbarc)=(x^(2)-2x+7)barb+(siny)barc` is

A. one
B. Two
C. Zero
D> Infinite

Options

A. one
B. two
C. zero
D. infinite

Explanation

To solve the problem, we need to analyze the given equations and find the number of solutions (x, y) in the range [0, 10].

### Step 1: Understand the Given Equations
We have two equations:
1. \( \mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = 5 \)
2. \( \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (x^2 - 2x + 7) \mathbf{b} + \sin(y) \mathbf{c} \)

### Step 2: Use the Triple Product Identity
Using the identity for the triple product:
\[ \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c} \]
We can equate this to the right-hand side of the second equation:
\[ (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c} = (x^2 - 2x + 7) \mathbf{b} + \sin(y) \mathbf{c} \]

### Step 3: Compare Coefficients
From the above equation, we can compare coefficients of \(\mathbf{b}\) and \(\mathbf{c}\):
1. \( \mathbf{a} \cdot \mathbf{c} = x^2 - 2x + 7 \)
2. \( -(\mathbf{a} \cdot \mathbf{b}) = \sin(y) \)

### Step 4: Substitute into the First Equation
Substituting the first equation into the second:
\[ \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} = 5 \]
Substituting \(\mathbf{a} \cdot \mathbf{c}\):
\[ \mathbf{a} \cdot \mathbf{b} + (x^2 - 2x + 7) = 5 \]
This simplifies to:
\[ \mathbf{a} \cdot \mathbf{b} = 5 - (x^2 - 2x + 7) = -x^2 + 2x - 2 \]

### Step 5: Substitute into the Second Equation
Now substituting \(\mathbf{a} \cdot \mathbf{b}\) into the second equation:
\[ -(-x^2 + 2x - 2) = \sin(y) \]
This leads to:
\[ \sin(y) = x^2 - 2x + 2 \]

### Step 6: Analyze the Range of \(\sin(y)\)
Since \(\sin(y)\) must be in the range \([-1, 1]\), we set up the inequality:
\[ -1 \leq x^2 - 2x + 2 \leq 1 \]

### Step 7: Solve the Inequalities
1. **Lower Bound**:
\[ x^2 - 2x + 2 \geq -1 \implies x^2 - 2x + 3 \geq 0 \]
The discriminant is negative, so this is always true.

2. **Upper Bound**:
\[ x^2 - 2x + 2 \leq 1 \implies x^2 - 2x + 1 \leq 0 \implies (x - 1)^2 \leq 0 \]
This implies \(x = 1\).

### Step 8: Find Corresponding \(y\) Values
Substituting \(x = 1\):
\[ \sin(y) = 1^2 - 2(1) + 2 = 1 \]
This gives:
\[ y = \frac{\pi}{2} + 2k\pi \quad (k \in \mathbb{Z}) \]
The valid values of \(y\) in the range \([0, 10]\) are:
- \(y = \frac{\pi}{2}\)
- \(y = \frac{5\pi}{2}\)

### Step 9: Count the Solutions
Thus, we have two valid pairs \((1, \frac{\pi}{2})\) and \((1, \frac{5\pi}{2})\).

### Conclusion
The number of solutions \((x, y)\) in the range \([0, 10]\) is **two**.

### Final Answer
**B. Two**

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