If `x*a=0, x*b=1,` [x a b]=1 and `a*b ne 0`, then find x in terms of a and b.

Answer

To solve the problem step by step, we will use the given conditions and vector identities.

### Step 1: Understand the Given Conditions
We have:
1. \( x \cdot a = 0 \) (This means \( x \) is orthogonal to \( a \))
2. \( x \cdot b = 1 \)
3. The volume of the parallelepiped formed by vectors \( x, a, b \) is given by \( [x, a, b] = 1 \)
4. \( a \cdot b \neq 0 \)

### Step 2: Use the Vector Triple Product Identity
We can use the vector triple product identity:
\[ x \times (a \times b) = (x \cdot b) a - (x \cdot a) b \]
Substituting the known values:
- Since \( x \cdot a = 0 \), the second term becomes \( 0 \).
- Thus, we have:
\[ x \times (a \times b) = (x \cdot b) a = 1 \cdot a = a \]

### Step 3: Relate to Volume of Parallelepiped
The volume of the parallelepiped formed by \( x, a, b \) can also be expressed as:
\[ [x, a, b] = x \cdot (a \times b) \]
Given that this volume is equal to 1, we have:
\[ x \cdot (a \times b) = 1 \]

### Step 4: Solve for \( x \)
Now we have two equations:
1. \( x \times (a \times b) = a \)
2. \( x \cdot (a \times b) = 1 \)

From the first equation, we can express \( x \) in terms of \( a \) and \( b \):
\[ x = k \cdot (a \times b) + m \cdot a \]
where \( k \) and \( m \) are scalars.

Substituting this into the second equation:
\[ (k \cdot (a \times b) + m \cdot a) \cdot (a \times b) = 1 \]

### Step 5: Simplify the Expression
Using the property of the dot product:
- \( (a \cdot (a \times b)) = 0 \) (since \( a \) is orthogonal to \( a \times b \))
- Thus, we have:
\[ k \cdot |a \times b|^2 = 1 \]
This gives us:
\[ k = \frac{1}{|a \times b|^2} \]

### Step 6: Substitute Back to Find \( x \)
Now we substitute \( k \) back into our expression for \( x \):
\[ x = \frac{1}{|a \times b|^2} (a \times b) + m \cdot a \]

### Step 7: Determine \( m \)
To satisfy \( x \cdot a = 0 \):
\[ \left(\frac{1}{|a \times b|^2} (a \times b) + m \cdot a\right) \cdot a = 0 \]
This leads to:
\[ \frac{1}{|a \times b|^2} (a \times b) \cdot a + m |a|^2 = 0 \]
Since \( (a \times b) \cdot a = 0 \), we find that \( m = 0 \).

### Final Expression for \( x \)
Thus, we have:
\[ x = \frac{1}{|a \times b|^2} (a \times b) \]

### Summary
The final expression for \( x \) in terms of \( a \) and \( b \) is:
\[ x = \frac{1}{|a \times b|^2} (a \times b) \]