Vectors `vecx,vecy,vecz` each of magnitude `sqrt(2)` make angles of `60^0` with each other. If `vecx`` xx(vecyxxvecz)=veca` ,`vecy`` xx(veczxxvecx)=vecb` and `vecxxxvecy=vecc`, find vecx, vecy, vecz` in terms of `veca,vecb and vecc`.

Options

A. `1/2[(veca-vecc) xx vecc-vecb+veca]`
B. `1/2[(veca-vecb) xx vecc+vecb-veca]`
C. `1/2[veccxx(veca-vecb) + vecb +veca]`
D. none of these

Explanation

To solve the problem, we need to express the vectors \(\vec{x}\), \(\vec{y}\), and \(\vec{z}\) in terms of the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). We start by using the given conditions and the properties of vector operations.

### Step-by-step Solution:

1. **Understanding the Magnitudes and Angles**:
Each vector \(\vec{x}\), \(\vec{y}\), and \(\vec{z}\) has a magnitude of \(\sqrt{2}\) and makes angles of \(60^\circ\) with each other. Thus, we can calculate the dot products:
\[ \vec{x} \cdot \vec{y} = |\vec{x}| |\vec{y}| \cos(60^\circ) = \sqrt{2} \cdot \sqrt{2} \cdot \frac{1}{2} = 1. \]
Similarly, \(\vec{y} \cdot \vec{z} = 1\) and \(\vec{z} \cdot \vec{x} = 1\).

2. **Setting Up the Cross Product Equations**:
We have:
\[ \vec{x} \times (\vec{y} \times \vec{z}) = \vec{a}, \]
\[ \vec{y} \times (\vec{z} \times \vec{x}) = \vec{b}, \]
\[ \vec{x} \times \vec{y} = \vec{c}. \]

3. **Using the Vector Triple Product Identity**:
The vector triple product identity states that:
\[ \vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w}) \vec{v} - (\vec{u} \cdot \vec{v}) \vec{w}. \]
Applying this to the first equation:
\[ \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z}) \vec{y} - (\vec{x} \cdot \vec{y}) \vec{z} = 1 \cdot \vec{y} - 1 \cdot \vec{z} = \vec{y} - \vec{z}. \]
Therefore, we have:
\[ \vec{y} - \vec{z} = \vec{a} \quad \text{(Equation 1)}. \]

4. **Applying the Second Equation**:
For the second equation:
\[ \vec{y} \times (\vec{z} \times \vec{x}) = (\vec{y} \cdot \vec{x}) \vec{z} - (\vec{y} \cdot \vec{z}) \vec{x} = 1 \cdot \vec{z} - 1 \cdot \vec{x} = \vec{z} - \vec{x}. \]
Thus, we have:
\[ \vec{z} - \vec{x} = \vec{b} \quad \text{(Equation 2)}. \]

5. **Using the Third Equation**:
From the third equation:
\[ \vec{x} \times \vec{y} = \vec{c}. \]

6. **Expressing \(\vec{x}\), \(\vec{y}\), and \(\vec{z}\)**:
From Equation 1, we can express \(\vec{z}\) in terms of \(\vec{y}\):
\[ \vec{z} = \vec{y} - \vec{a}. \]
Substituting this into Equation 2 gives:
\[ \vec{y} - (\vec{y} - \vec{a}) = \vec{b} \implies \vec{a} = \vec{b}. \]
Thus, we can express \(\vec{y}\) in terms of \(\vec{b}\) and \(\vec{a}\):
\[ \vec{y} = \vec{b} + \vec{a}. \]
Now substituting back to find \(\vec{x}\):
\[ \vec{z} = \vec{b} + \vec{a} - \vec{a} = \vec{b}. \]
Finally, substituting \(\vec{z}\) into the expression for \(\vec{y}\):
\[ \vec{y} = \vec{b} + \vec{a} \quad \text{and} \quad \vec{x} = \vec{a} + \vec{c}. \]

7. **Final Expressions**:
Therefore, we have:
\[ \vec{x} = \vec{a} \times \vec{c}, \]
\[ \vec{y} = \vec{a} + \vec{b} + \vec{a} \times \vec{c}, \]
\[ \vec{z} = \vec{b} + \vec{a} \times \vec{c}. \]

### Summary of Results:
- \(\vec{x} = \vec{a} \times \vec{c}\)
- \(\vec{y} = \vec{a} + \vec{b} + \vec{a} \times \vec{c}\)
- \(\vec{z} = \vec{b} + \vec{a} \times \vec{c}\)