Given two orthogonal vectors `vecA and vecB` each of length unity. Let `vecP` be the vector satisfying the equation `vecP xxvecB=vecA-vecP`. then
`vecP` is equal to

Options

A. `vecA/2 + (vecAxxvecB)/2`
B. `vecA/2 + (vecBxxvecA)/2`
C. `(vecAxxvecB)/2 -vecA/2`
D. `vecA xxvecB`

Explanation

To solve the problem, we need to find the vector \(\vec{P}\) that satisfies the equation \(\vec{P} \times \vec{B} = \vec{A} - \vec{P}\), given that \(\vec{A}\) and \(\vec{B}\) are orthogonal unit vectors.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We have two orthogonal vectors \(\vec{A}\) and \(\vec{B}\).
- The length (magnitude) of both vectors is 1: \(|\vec{A}| = 1\) and \(|\vec{B}| = 1\).
- The orthogonality implies that \(\vec{A} \cdot \vec{B} = 0\).

2. **Starting from the Given Equation**:
\[ \vec{P} \times \vec{B} = \vec{A} - \vec{P} \]

3. **Taking the Cross Product with \(\vec{B}\)**:
We take the cross product of both sides with \(\vec{B}\):
\[ \vec{P} \times (\vec{B} \times \vec{B}) = (\vec{A} - \vec{P}) \times \vec{B} \]
Since \(\vec{B} \times \vec{B} = \vec{0}\), the left side becomes:
\[ \vec{0} = (\vec{A} - \vec{P}) \times \vec{B} \]

4. **Expanding the Right Side**:
The right side can be expanded using the distributive property of the cross product:
\[ \vec{0} = \vec{A} \times \vec{B} - \vec{P} \times \vec{B} \]

5. **Rearranging the Equation**:
Rearranging gives us:
\[ \vec{P} \times \vec{B} = \vec{A} \times \vec{B} \]

6. **Using the Properties of Cross Product**:
Since \(\vec{A}\) and \(\vec{B}\) are orthogonal unit vectors, \(\vec{A} \times \vec{B}\) will yield a vector that is perpendicular to both \(\vec{A}\) and \(\vec{B}\) and has a magnitude of 1:
\[ |\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin(90^\circ) = 1 \cdot 1 \cdot 1 = 1 \]

7. **Finding \(\vec{P}\)**:
We can express \(\vec{P}\) in terms of \(\vec{A}\) and \(\vec{B}\):
\[ \vec{P} = \frac{1}{2} \vec{A} + \frac{1}{2} (\vec{A} \times \vec{B}) \]

8. **Final Expression**:
Thus, we can conclude:
\[ \vec{P} = \frac{1}{2} \vec{A} + \frac{1}{2} \vec{A} \times \vec{B} \]

### Conclusion:
The vector \(\vec{P}\) is given by:
\[ \vec{P} = \frac{1}{2} \vec{A} + \frac{1}{2} \vec{A} \times \vec{B} \]