Sholve the simultasneous vector equations for `vecx and vecy: vecx+veccxxvecy=veca and vecy+veccxxvecx=vecb, vecc!=0`

Options

A. `vecx=(vecbxxvecc+veca+(vecc.veca)vecc)/(1+ vecc.vecc)`
B. `vecx=(veccxxvecb+vecb+(vecc.veca)vecc)/(1+vecc.vecc)`
C. `vecy=(vecaxxvecc+vecb+(vecc.vecb)vecc)/(1+vecc.vecc)`
D. none of these

Explanation

To solve the simultaneous vector equations for \(\vec{x}\) and \(\vec{y}\):

1. **Write the Given Equations**:
\[ \vec{x} + \vec{c} \times \vec{y} = \vec{a} \tag{1} \]
\[ \vec{y} + \vec{c} \times \vec{x} = \vec{b} \tag{2} \]

2. **Cross Product with \(\vec{c}\)**:
Take the cross product of equation (1) with \(\vec{c}\):
\[ \vec{c} \times \vec{x} + \vec{c} \times (\vec{c} \times \vec{y}) = \vec{c} \times \vec{a} \]
Using the vector triple product identity \(\vec{c} \times (\vec{c} \times \vec{y}) = \vec{c}(\vec{c} \cdot \vec{y}) - \vec{c} \cdot \vec{c} \vec{y}\):
\[ \vec{c} \times \vec{x} + \vec{c}(\vec{c} \cdot \vec{y}) - (\vec{c} \cdot \vec{c}) \vec{y} = \vec{c} \times \vec{a} \tag{3} \]

3. **Substitute \(\vec{c} \times \vec{x}\)**:
From equation (2), we can express \(\vec{c} \times \vec{x}\) as:
\[ \vec{c} \times \vec{x} = \vec{c} \times (\vec{b} - \vec{c} \times \vec{y}) = \vec{c} \times \vec{b} - \vec{c} \times (\vec{c} \times \vec{y}) \]
Substitute this back into equation (3):
\[ \vec{c} \times \vec{b} - \vec{c} \times (\vec{c} \times \vec{y}) + \vec{c}(\vec{c} \cdot \vec{y}) - (\vec{c} \cdot \vec{c}) \vec{y} = \vec{c} \times \vec{a} \]

4. **Simplify the Equation**:
Rearranging gives:
\[ \vec{c} \times \vec{b} + \vec{c}(\vec{c} \cdot \vec{y}) - (\vec{c} \cdot \vec{c}) \vec{y} = \vec{c} \times \vec{a} \]
This can be simplified to find \(\vec{y}\):
\[ \vec{y}(1 + \vec{c} \cdot \vec{c}) = \vec{c} \times \vec{a} - \vec{c} \times \vec{b} \]
Thus,
\[ \vec{y} = \frac{\vec{c} \times \vec{a} - \vec{c} \times \vec{b}}{1 + \vec{c} \cdot \vec{c}} \tag{4} \]

5. **Finding \(\vec{x}\)**:
Now, we can find \(\vec{x}\) using equation (1):
\[ \vec{x} = \vec{a} - \vec{c} \times \vec{y} \]
Substitute \(\vec{y}\) from equation (4):
\[ \vec{x} = \vec{a} - \vec{c} \times \left(\frac{\vec{c} \times \vec{a} - \vec{c} \times \vec{b}}{1 + \vec{c} \cdot \vec{c}}\right) \]

6. **Final Expressions**:
After simplification, we can express \(\vec{x}\) and \(\vec{y}\) in terms of \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\).

### Final Results:
\[ \vec{y} = \frac{\vec{c} \times \vec{a} - \vec{c} \times \vec{b}}{1 + \vec{c} \cdot \vec{c}} \]
\[ \vec{x} = \vec{a} - \vec{c} \times \left(\frac{\vec{c} \times \vec{a} - \vec{c} \times \vec{b}}{1 + \vec{c} \cdot \vec{c}}\right) \]