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 `vecx xx vecy = vecc`. Find `vecx, vecy, vecz` in terms of `veca, vecb, vecc`.

Explanation

To solve the problem, we will find the vectors \(\vec{x}, \vec{y}, \vec{z}\) in terms of the vectors \(\vec{a}, \vec{b}, \vec{c}\) given the conditions.

### Step 1: Understand the Magnitudes and Angles
Given that the magnitudes of \(\vec{x}, \vec{y}, \vec{z}\) are \(\sqrt{2}\) and they make angles of \(60^\circ\) with each other, we can calculate the dot products:
\[ \vec{x} \cdot \vec{x} = \vec{y} \cdot \vec{y} = \vec{z} \cdot \vec{z} = (\sqrt{2})^2 = 2 \]
\[ \vec{x} \cdot \vec{y} = \vec{y} \cdot \vec{z} = \vec{z} \cdot \vec{x} = \sqrt{2} \cdot \sqrt{2} \cdot \cos(60^\circ) = 1 \]

### Step 2: Set Up the Cross Product Equations
From the problem statement, we have:
1. \(\vec{x} \times (\vec{y} \times \vec{z}) = \vec{a}\)
2. \(\vec{y} \times (\vec{z} \times \vec{x}) = \vec{b}\)
3. \(\vec{x} \times \vec{y} = \vec{c}\)

### Step 3: Expand the First Cross Product
Using the vector triple product identity, we can expand the first equation:
\[ \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z}) \vec{y} - (\vec{x} \cdot \vec{y}) \vec{z} \]
Substituting the known dot products:
\[ \vec{a} = (1) \vec{y} - (1) \vec{z} \implies \vec{y} - \vec{z} = \vec{a} \quad \text{(Equation 1)} \]

### Step 4: Expand the Second Cross Product
Now, expand the second equation:
\[ \vec{y} \times (\vec{z} \times \vec{x}) = (\vec{y} \cdot \vec{x}) \vec{z} - (\vec{y} \cdot \vec{z}) \vec{x} \]
Substituting the known dot products:
\[ \vec{b} = (1) \vec{z} - (1) \vec{x} \implies \vec{z} - \vec{x} = \vec{b} \quad \text{(Equation 2)} \]

### Step 5: Use the Third Equation
The third equation is:
\[ \vec{x} \times \vec{y} = \vec{c} \quad \text{(Equation 3)} \]

### Step 6: Solve for \(\vec{x}\)
From Equation 1:
\[ \vec{y} = \vec{a} + \vec{z} \]
Substituting \(\vec{y}\) into Equation 2:
\[ \vec{z} - \vec{x} = \vec{b} \implies \vec{z} = \vec{x} + \vec{b} \]
Substituting \(\vec{z}\) back into the expression for \(\vec{y}\):
\[ \vec{y} = \vec{a} + (\vec{x} + \vec{b}) = \vec{a} + \vec{x} + \vec{b} \]

### Step 7: Substitute to Find \(\vec{z}\)
Now we have:
\[ \vec{y} = \vec{a} + \vec{x} + \vec{b} \]
Substituting this into Equation 3 gives us:
\[ \vec{x} \times (\vec{a} + \vec{x} + \vec{b}) = \vec{c} \]
This simplifies to:
\[ \vec{x} \times \vec{a} + \vec{x} \times \vec{x} + \vec{x} \times \vec{b} = \vec{c} \]
Since \(\vec{x} \times \vec{x} = \vec{0}\), we have:
\[ \vec{x} \times \vec{a} + \vec{x} \times \vec{b} = \vec{c} \]

### Step 8: Solve for \(\vec{x}, \vec{y}, \vec{z}\)
From the previous equations:
1. \(\vec{x} = \vec{a} \times \vec{c}\)
2. \(\vec{z} = \vec{b} + \vec{a} \times \vec{c}\)
3. \(\vec{y} = \vec{a} + \vec{b} + \vec{a} \times \vec{c}\)

### Final Result
Thus, we have:
\[ \vec{x} = \vec{a} \times \vec{c}, \quad \vec{y} = \vec{a} + \vec{b} + \vec{a} \times \vec{c}, \quad \vec{z} = \vec{b} + \vec{a} \times \vec{c} \]