If `veca, vecb, vecc` are unit vectors such that `veca. vecb=0, (veca-vecc).(vecb+vecc)=0` and `vecc=lambdaveca+muvecb+omega(veca xx vecb)`, where `lambda, mu, omega` are scalars, then

Options

A. `mu^(2)+omega^(2)=1`
B. `lambda+mu=1`
C. `(mu+1)^(2)+mu^(2)+omega^(2)=1`
D. `lambda^(2)+mu^(2)=1`

Explanation

To solve the given problem step by step, we will analyze the conditions provided and derive the necessary relationships among the vectors.

### Step 1: Understand the Given Conditions
We are given three unit vectors \( \vec{a}, \vec{b}, \vec{c} \) such that:
1. \( \vec{a} \cdot \vec{b} = 0 \) (indicating that \( \vec{a} \) and \( \vec{b} \) are perpendicular).
2. \( (\vec{a} - \vec{c}) \cdot (\vec{b} + \vec{c}) = 0 \).
3. \( \vec{c} = \lambda \vec{a} + \mu \vec{b} + \omega (\vec{a} \times \vec{b}) \), where \( \lambda, \mu, \omega \) are scalars.

### Step 2: Expand the Second Condition
We start with the second condition:
\[ (\vec{a} - \vec{c}) \cdot (\vec{b} + \vec{c}) = 0 \]
Expanding this, we have:
\[ \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} - \vec{c} \cdot \vec{b} - \vec{c} \cdot \vec{c} = 0 \]
Since \( \vec{a} \cdot \vec{b} = 0 \), this simplifies to:
\[ \vec{a} \cdot \vec{c} - \vec{c} \cdot \vec{b} - \vec{c} \cdot \vec{c} = 0 \]

### Step 3: Substitute \( \vec{c} \)
Substituting \( \vec{c} = \lambda \vec{a} + \mu \vec{b} + \omega (\vec{a} \times \vec{b}) \) into the equation:
\[ \vec{a} \cdot (\lambda \vec{a} + \mu \vec{b} + \omega (\vec{a} \times \vec{b})) - (\lambda \vec{a} + \mu \vec{b} + \omega (\vec{a} \times \vec{b})) \cdot \vec{b} - \|\vec{c}\|^2 = 0 \]

### Step 4: Calculate Each Dot Product
1. \( \vec{a} \cdot \vec{c} = \lambda \|\vec{a}\|^2 + \mu (\vec{a} \cdot \vec{b}) + \omega (\vec{a} \cdot (\vec{a} \times \vec{b})) = \lambda \) (since \( \|\vec{a}\|^2 = 1 \) and the other two terms are zero).
2. \( \vec{c} \cdot \vec{b} = \lambda (\vec{a} \cdot \vec{b}) + \mu \|\vec{b}\|^2 + \omega ((\vec{a} \times \vec{b}) \cdot \vec{b}) = \mu \) (since \( \|\vec{b}\|^2 = 1 \) and the other two terms are zero).
3. \( \|\vec{c}\|^2 = \lambda^2 + \mu^2 + \omega^2 \) (since \( \vec{c} \) is a unit vector).

### Step 5: Substitute Back into the Equation
Now substituting back, we get:
\[ \lambda - \mu - (\lambda^2 + \mu^2 + \omega^2) = 0 \]
Rearranging gives:
\[ \lambda - \mu = \lambda^2 + \mu^2 + \omega^2 \]

### Step 6: Analyzing the Magnitude
Since \( \vec{c} \) is a unit vector, we have:
\[ \lambda^2 + \mu^2 + \omega^2 = 1 \]

### Step 7: Solve for Relationships
From \( \lambda - \mu = \lambda^2 + \mu^2 + \omega^2 \) and \( \lambda^2 + \mu^2 + \omega^2 = 1 \), we can conclude:
\[ \lambda - \mu = 1 \]
This implies:
\[ \lambda = \mu + 1 \]

### Conclusion
Thus, we have established the relationships among \( \lambda, \mu, \) and \( \omega \) based on the conditions given in the problem.