Given two orthogonal vectors `vecA and VecB` each of length unity. Let `vecP` be the vector satisfying the equation `vecP xxvecB=vecA-vecP`. then
which of the following statements is false ?

Options

A. vectors `vecP, vecA and vecP xx vecB` ar linearly dependent.
B. vectors `vecP, vecB and vecP xx vecB` ar linearly independent
C. `vecP` is orthogonal to `vecB` and has length `1/sqrt2`.
D. none of these

Explanation

To solve the problem step by step, we will analyze the given equation and the properties of the vectors involved.

### Step 1: Understand the Given Information
We have two orthogonal vectors, \(\vec{A}\) and \(\vec{B}\), both of unit length:
- \(|\vec{A}| = 1\)
- \(|\vec{B}| = 1\)
- \(\vec{A} \cdot \vec{B} = 0\) (since they are orthogonal)

We also have the equation:
\[ \vec{P} \times \vec{B} = \vec{A} - \vec{P} \]

### Step 2: Rearranging the Equation
Rearranging the equation gives us:
\[ \vec{P} \times \vec{B} + \vec{P} = \vec{A} \]

### Step 3: Taking the Cross Product with \(\vec{B}\)
Now, we take the cross product of both sides with \(\vec{B}\):
\[ \vec{P} \times (\vec{B} \times \vec{B}) + \vec{P} \times \vec{B} = \vec{A} \times \vec{B} \]
Since \(\vec{B} \times \vec{B} = \vec{0}\), the equation simplifies to:
\[ \vec{0} + \vec{P} \times \vec{B} = \vec{A} \times \vec{B} \]
Thus, we have:
\[ \vec{P} \times \vec{B} = \vec{A} \times \vec{B} \]

### Step 4: Analyzing the Result
From the above, we can conclude that:
\[ \vec{P} \times \vec{B} = \vec{A} \times \vec{B} \]
This indicates that \(\vec{P}\) and \(\vec{A}\) have a similar cross product with \(\vec{B}\).

### Step 5: Finding the Magnitude of \(\vec{P}\)
To find the magnitude of \(\vec{P}\), we can use the properties of cross products:
\[ |\vec{P}| = |\vec{A} - \vec{P}| \]
Squaring both sides:
\[ |\vec{P}|^2 = |\vec{A} - \vec{P}|^2 \]
Expanding the right-hand side:
\[ |\vec{P}|^2 = |\vec{A}|^2 + |\vec{P}|^2 - 2 \vec{A} \cdot \vec{P} \]
Since \(|\vec{A}| = 1\), we have:
\[ |\vec{P}|^2 = 1 + |\vec{P}|^2 - 2 \vec{A} \cdot \vec{P} \]
This simplifies to:
\[ 0 = 1 - 2 \vec{A} \cdot \vec{P} \]
Thus:
\[ \vec{A} \cdot \vec{P} = \frac{1}{2} \]

### Step 6: Evaluating the Statements
Now we analyze which of the statements is false:
1. Vectors are linearly dependent.
2. Vectors are linearly independent.
3. \(\vec{P}\) is orthogonal to \(\vec{B}\).
4. \(|\vec{P}| = \frac{1}{\sqrt{2}}\).
5. None of the above.

Given that \(\vec{A}\) and \(\vec{B}\) are orthogonal and unit vectors, \(\vec{P}\) cannot be orthogonal to both \(\vec{A}\) and \(\vec{B}\) simultaneously while still maintaining the conditions provided. Therefore, the statement that \(\vec{P}\) is orthogonal to \(\vec{B}\) is false.

### Conclusion
The false statement is:
- \(\vec{P}\) is orthogonal to \(\vec{B}\).