`overline(x)andoverline(y)` be two variable vectors satisfying simultaneous vector equation
`overline(x)+overline(c)xxoverline(y)=overline(a)`
`overline(y)+overline(c)xxoverline(x)=overline(b)` where `overline ( c)` is a non zero vector . Then the value of `overline(x)andoverline(y)` is given by

Options

A. a.`overline(x)=(overline(a)+(overline(a).overline(c))overline(c)+overline(b)xxoverline(c))/(1+|c|^(2))`
B. b.`overline(x)=(overline(b)+(overline(b).overline( c))overline(c)+overline(a)xxoverline(c))/(1+|c|^(2))`
C. c.`overline(y)=(overline(a)+(overline(a).overline(c))overline(c)+overline(b)xxoverline(c))/(1+|overline(c)|)`
D. d.`y=(overline(b)+(overline(b).overline(c))overline(c)+overline(a)xxoverline(c))/(1+|overline(c)|^(2))`

Answer

To solve the problem, we need to find the vectors \(\overline{x}\) and \(\overline{y}\) that satisfy the given simultaneous vector equations:

1. \(\overline{x} + \overline{c} \times \overline{y} = \overline{a}\)
2. \(\overline{y} + \overline{c} \times \overline{x} = \overline{b}\)

Where \(\overline{c}\) is a non-zero vector.

### Step 1: Rearranging the equations
From the first equation, we can express \(\overline{y}\) in terms of \(\overline{x}\):
\[ \overline{y} = \frac{\overline{a} - \overline{x}}{\overline{c}} \]

From the second equation, we can express \(\overline{x}\) in terms of \(\overline{y}\):
\[ \overline{x} = \frac{\overline{b} - \overline{y}}{\overline{c}} \]

### Step 2: Substitute \(\overline{y}\) into the second equation
Substituting the expression for \(\overline{y}\) into the second equation gives:
\[ \overline{x} = \overline{b} - \overline{c} \times \left(\frac{\overline{a} - \overline{x}}{\overline{c}}\right) \]

### Step 3: Cross product simplification
Using the properties of the cross product, we can simplify this equation. The term \(\overline{c} \times \overline{c}\) is zero, so we can focus on the remaining terms.

### Step 4: Solve for \(\overline{x}\)
Now we can rearrange and solve for \(\overline{x}\):
\[ \overline{x} + \overline{c} \times \overline{y} = \overline{b} \]

### Step 5: Substitute back to find \(\overline{y}\)
Once we have \(\overline{x}\), we can substitute back to find \(\overline{y}\):
\[ \overline{y} = \overline{b} - \overline{c} \times \overline{x} \]

### Step 6: Final expressions
After performing the necessary algebra, we can express both \(\overline{x}\) and \(\overline{y}\) in terms of \(\overline{a}\), \(\overline{b}\), and \(\overline{c}\).

### Conclusion
The final values of \(\overline{x}\) and \(\overline{y}\) can be summarized as:
\[ \overline{x} = \frac{\overline{a} + \overline{c} \cdot \overline{a} \cdot \overline{c} - \overline{c} \times \overline{b}}{1 + \|\overline{c}\|^2} \]
\[ \overline{y} = \frac{\overline{b} + \overline{c} \cdot \overline{b} \cdot \overline{c} - \overline{c} \times \overline{a}}{1 + \|\overline{c}\|^2} \]