Solve: `vecrxxvecb=veca, where veca and vecb` are given vectors such that `veca.vecb=0`.

Answer

To solve the equation \( \vec{R} \times \vec{B} = \vec{A} \) where \( \vec{A} \) and \( \vec{B} \) are given vectors such that \( \vec{A} \cdot \vec{B} = 0 \), we can follow these steps:

### Step 1: Understand the Given Information
We know that \( \vec{A} \cdot \vec{B} = 0 \), which implies that the vectors \( \vec{A} \) and \( \vec{B} \) are perpendicular to each other.

### Step 2: Start with the Given Equation
The equation we need to solve is:
\[ \vec{R} \times \vec{B} = \vec{A} \]

### Step 3: Cross Multiply with \( \vec{A} \)
To manipulate the equation, we can cross multiply both sides with \( \vec{A} \):
\[ \vec{R} \times (\vec{B} \times \vec{A}) = \vec{A} \times \vec{A} \]
Since \( \vec{A} \times \vec{A} = \vec{0} \) (the cross product of any vector with itself is zero), we have:
\[ \vec{R} \times (\vec{B} \times \vec{A}) = \vec{0} \]

### Step 4: Use the Scalar Triple Product Identity
Using the scalar triple product identity, we can expand \( \vec{R} \times (\vec{B} \times \vec{A}) \):
\[ \vec{R} \cdot \vec{A} \vec{B} - \vec{B} \cdot \vec{A} \vec{R} = \vec{0} \]
Since \( \vec{A} \cdot \vec{B} = 0 \), the second term vanishes:
\[ \vec{R} \cdot \vec{A} \vec{B} = \vec{0} \]

### Step 5: Analyze the Result
From \( \vec{R} \cdot \vec{A} \vec{B} = \vec{0} \), we can conclude:
1. Since \( \vec{B} \neq \vec{0} \), it follows that \( \vec{R} \cdot \vec{A} = 0 \). This means \( \vec{R} \) is perpendicular to \( \vec{A} \).

### Step 6: Cross Multiply with \( \vec{B} \)
Now, we will cross multiply the original equation \( \vec{R} \times \vec{B} = \vec{A} \) with \( \vec{B} \):
\[ \vec{R} \times (\vec{B} \times \vec{B}) = \vec{A} \times \vec{B} \]
The left side becomes zero since \( \vec{B} \times \vec{B} = \vec{0} \):
\[ \vec{0} = \vec{A} \times \vec{B} \]

### Step 7: Expand Using the Scalar Triple Product
Using the identity again, we have:
\[ \vec{R} \cdot \vec{B} \vec{B} - |\vec{B}|^2 \vec{R} = \vec{A} \times \vec{B} \]
Since \( \vec{R} \cdot \vec{B} = 0 \) from earlier, we simplify to:
\[ -|\vec{B}|^2 \vec{R} = \vec{A} \times \vec{B} \]

### Step 8: Solve for \( \vec{R} \)
Rearranging gives:
\[ \vec{R} = -\frac{\vec{A} \times \vec{B}}{|\vec{B}|^2} \]
This can also be written as:
\[ \vec{R} = \frac{\vec{B} \times \vec{A}}{|\vec{B}|^2} \]

### Final Solution
Thus, the solution for \( \vec{R} \) is:
\[ \vec{R} = \frac{\vec{B} \times \vec{A}}{|\vec{B}|^2} \]