Let `vecb` and `vecc` be non-collinear vectors. If `veca` is a vector such that `veca.(vecb + vecc)=4` and `veca xx (vecb xx vecc)=(x^(2) - 2x + 6)vecb + sin y.vecc`, then (x,y) lies on the line:

Options

A. x+y=0
B. x-y=0
C. x=1
D. `y=pi/2`

Answer

To solve the given problem step by step, we will use the properties of vectors and the vector triple product.

### Step 1: Understand the given equations
We have two equations:
1. \(\vec{a} \cdot (\vec{b} + \vec{c}) = 4\)
2. \(\vec{a} \times (\vec{b} \times \vec{c}) = (x^2 - 2x + 6)\vec{b} + \sin y \vec{c}\)

### Step 2: Apply the vector triple product identity
Using the vector triple product identity, we have:
\[ \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \]
Thus, we can rewrite the second equation as:
\[ (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = (x^2 - 2x + 6)\vec{b} + \sin y \vec{c} \]

### Step 3: Compare coefficients
From the above equation, we can compare the coefficients of \(\vec{b}\) and \(\vec{c}\):
1. Coefficient of \(\vec{b}\): \(\vec{a} \cdot \vec{c} = x^2 - 2x + 6\)
2. Coefficient of \(\vec{c}\): \(-\vec{a} \cdot \vec{b} = \sin y\)

### Step 4: Use the first equation
From the first equation, we can expand it:
\[ \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 4 \]
Substituting the values from the coefficients we found:
\[ -\sin y + (x^2 - 2x + 6) = 4 \]
Rearranging gives:
\[ x^2 - 2x + 6 - \sin y = 4 \]
This simplifies to:
\[ x^2 - 2x + 2 = \sin y \]

### Step 5: Analyze the equation
The left-hand side \(x^2 - 2x + 2\) is a quadratic function. To find its minimum value, we can complete the square:
\[ x^2 - 2x + 2 = (x - 1)^2 + 1 \]
The minimum value occurs when \((x - 1)^2 = 0\), which gives \(x = 1\). Thus, the minimum value of \(x^2 - 2x + 2\) is \(1\).

### Step 6: Relate to \(\sin y\)
Since \(\sin y\) has a maximum value of \(1\), we have:
\[ 1 \geq \sin y \geq 0 \]
This implies:
\[ 1 = \sin y \]
The only angle \(y\) for which \(\sin y = 1\) is:
\[ y = \frac{\pi}{2} \]

### Step 7: Conclusion
Thus, we find that:
\[ (x, y) = (1, \frac{\pi}{2}) \]
This means that the pair \((x, y)\) lies on the line represented by the point \((1, \frac{\pi}{2})\).