If `vecx * xvecy=veca, vecy xx vecz=vecb, vecx.vecb=gamma, vecx.vecy=1 and vecy.vecz=1 ` then find x,y,z in terms of `veca,vecb and gamma.

Options

A. `gamma/(|veca xxvecb|^(2))[veca+vecbxx(vecaxxvecb)]`
B. `gamma/(|veca xxvecb|^(2))[veca+vecb-vecaxx(vecaxxvecb)]`
C. `gamma/(|veca xxvecb|^(2))[veca+vecb+vecaxx(vecaxxvecb)]`
D. none of these

Explanation

To solve the problem, we need to express the vectors \( \vec{x}, \vec{y}, \vec{z} \) in terms of \( \vec{a}, \vec{b}, \) and \( \gamma \) based on the given equations.

### Step 1: Start with the given equations
We have the following equations:
1. \( \vec{x} \times \vec{y} = \vec{a} \) (Equation 1)
2. \( \vec{y} \times \vec{z} = \vec{b} \) (Equation 2)
3. \( \vec{x} \cdot \vec{b} = \gamma \) (Equation 3)
4. \( \vec{x} \cdot \vec{y} = 1 \) (Equation 4)
5. \( \vec{y} \cdot \vec{z} = 1 \) (Equation 5)

### Step 2: Find \( \vec{y} \)
From Equation 1, we can express \( \vec{y} \) in terms of \( \vec{x} \) and \( \vec{a} \):
\[ \vec{y} = \frac{\vec{a} \times \vec{b}}{\gamma} \]
This is derived from rearranging the cross product and using the properties of dot products.

### Step 3: Substitute \( \vec{y} \) into Equation 4
Using Equation 4, we can substitute \( \vec{y} \):
\[ \vec{x} \cdot \left(\frac{\vec{a} \times \vec{b}}{\gamma}\right) = 1 \]
This implies:
\[ \vec{x} \cdot (\vec{a} \times \vec{b}) = \gamma \]

### Step 4: Find \( \vec{x} \)
Now, we can express \( \vec{x} \) in terms of \( \vec{a} \) and \( \vec{b} \):
\[ \vec{x} = \frac{\vec{a} \times \vec{b}}{\gamma} \]
This gives us the first vector \( \vec{x} \).

### Step 5: Find \( \vec{z} \)
Now, we can use Equation 2 to find \( \vec{z} \):
\[ \vec{y} \times \vec{z} = \vec{b} \]
Substituting \( \vec{y} \):
\[ \left(\frac{\vec{a} \times \vec{b}}{\gamma}\right) \times \vec{z} = \vec{b} \]
Rearranging gives us:
\[ \vec{z} = \frac{\vec{b} \times \gamma}{\vec{a} \times \vec{b}} \]

### Final Expressions
Thus, we have:
\[ \vec{x} = \frac{\vec{a} \times \vec{b}}{\gamma}, \quad \vec{y} = \frac{\vec{a} \times \vec{b}}{\gamma}, \quad \vec{z} = \frac{\vec{b} \times \gamma}{\vec{a} \times \vec{b}} \]

### Summary of Results
- \( \vec{x} = \frac{\vec{a} \times \vec{b}}{\gamma} \)
- \( \vec{y} = \frac{\vec{a} \times \vec{b}}{\gamma} \)
- \( \vec{z} = \frac{\vec{b} \times \gamma}{\vec{a} \times \vec{b}} \)