Given two orthogonal vectors `vec(A)` and `vec(B)` each of length unity . Let `vec(P)` be the vector satisfying the equation `vec(P) xx vec(B) = vec(A) - vec(P)` . Then
`vec(P)` is equal to .

Answer

To solve the problem, we need to find the vector \(\vec{P}\) that satisfies the equation:

\[ \vec{P} \times \vec{B} = \vec{A} - \vec{P} \]

where \(\vec{A}\) and \(\vec{B}\) are orthogonal unit vectors.

### Step 1: Cross Product Both Sides with \(\vec{B}\)

We start by taking the cross product of both sides of the equation with \(\vec{B}\):

\[ \vec{P} \times \vec{B} \times \vec{B} = (\vec{A} - \vec{P}) \times \vec{B} \]

### Step 2: Simplify Left Side

Using the vector triple product identity, we can simplify the left side:

\[ \vec{P} \times \vec{B} \times \vec{B} = (\vec{P} \cdot \vec{B}) \vec{B} - (\vec{B} \cdot \vec{B}) \vec{P} \]

Since \(\vec{B} \cdot \vec{B} = 1\):

\[ \vec{P} \times \vec{B} \times \vec{B} = (\vec{P} \cdot \vec{B}) \vec{B} - \vec{P} \]

### Step 3: Simplify Right Side

Now, we simplify the right side:

\[ (\vec{A} - \vec{P}) \times \vec{B} = \vec{A} \times \vec{B} - \vec{P} \times \vec{B} \]

### Step 4: Substitute Back into the Equation

Now we substitute these results back into our equation:

\[ (\vec{P} \cdot \vec{B}) \vec{B} - \vec{P} = \vec{A} \times \vec{B} - \vec{P} \times \vec{B} \]

### Step 5: Rearranging the Equation

Rearranging gives us:

\[ \vec{P} + \vec{P} \times \vec{B} = \vec{A} \times \vec{B} + (\vec{P} \cdot \vec{B}) \vec{B} \]

### Step 6: Isolate \(\vec{P}\)

Now we can isolate \(\vec{P}\):

\[ 2\vec{P} = \vec{A} \times \vec{B} + (\vec{P} \cdot \vec{B}) \vec{B} \]

### Step 7: Divide by 2

Finally, we divide by 2 to find \(\vec{P}\):

\[ \vec{P} = \frac{1}{2} \left( \vec{A} \times \vec{B} + (\vec{P} \cdot \vec{B}) \vec{B} \right) \]

### Step 8: Dot Product with \(\vec{B}\)

Now, we take the dot product of both sides with \(\vec{B}\):

\[ \vec{P} \cdot \vec{B} = \frac{1}{2} \left( \vec{A} \times \vec{B} \cdot \vec{B} + (\vec{P} \cdot \vec{B}) (\vec{B} \cdot \vec{B}) \right) \]

### Step 9: Evaluate Dot Products

Since \(\vec{A} \times \vec{B} \cdot \vec{B} = 0\) (because they are orthogonal) and \(\vec{B} \cdot \vec{B} = 1\):

\[ \vec{P} \cdot \vec{B} = \frac{1}{2} (\vec{P} \cdot \vec{B}) \]

### Step 10: Solve for \(\vec{P} \cdot \vec{B}\)

This leads to:

\[ \frac{1}{2} (\vec{P} \cdot \vec{B}) = 0 \implies \vec{P} \cdot \vec{B} = 0 \]

### Conclusion

Thus, \(\vec{P}\) is orthogonal to \(\vec{B}\). Finally, we can express \(\vec{P}\):

\[ \vec{P} = \frac{1}{2} \left( \vec{A} - \vec{A} \times \vec{B} \right) \]

### Final Answer

\[ \vec{P} = \frac{1}{2} \left( \vec{A} - \vec{A} \times \vec{B} \right) \]