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. `u^(2) + w^(2)=1`
B. `2u^(2) + w^(2)=1`
C. `2u(u+1)+w^(2)=0`
D. None of these

Answer

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

### Step 1: Understand the given conditions
We have three unit vectors: \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). The following conditions are given:
1. \(\vec{a} \cdot \vec{b} = 0\) (which means \(\vec{a}\) 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 \(\lambda\), \(\mu\), and \(\omega\) are scalars.

### Step 2: Analyze the first condition
Since \(\vec{a}\) and \(\vec{b}\) are unit vectors and perpendicular, we have:
\[ |\vec{a}| = 1, \quad |\vec{b}| = 1, \quad \vec{a} \cdot \vec{b} = 0. \]

### Step 3: Analyze the second condition
Expanding the second condition:
\[ (\vec{a} - \vec{c}) \cdot (\vec{b} + \vec{c}) = 0. \]
This expands to:
\[ \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} - \vec{c} \cdot \vec{b} - \vec{c} \cdot \vec{c} = 0. \]
Since \(\vec{a} \cdot \vec{b} = 0\) and \(|\vec{c}| = 1\) (as \(\vec{c}\) is a unit vector), we can substitute:
\[ 0 + \vec{a} \cdot \vec{c} - \vec{c} \cdot \vec{b} - 1 = 0. \]
Thus, we have:
\[ \vec{a} \cdot \vec{c} - \vec{c} \cdot \vec{b} - 1 = 0. \]
Rearranging gives:
\[ \vec{a} \cdot \vec{c} - \vec{c} \cdot \vec{b} = 1. \]

### Step 4: Substitute \(\vec{c}\)
Substituting \(\vec{c} = \lambda \vec{a} + \mu \vec{b} + \omega (\vec{a} \times \vec{b})\) into the equation:
\[ \vec{a} \cdot (\lambda \vec{a} + \mu \vec{b} + \omega (\vec{a} \times \vec{b})) - (\lambda \vec{a} + \mu \vec{b} + \omega (\vec{a} \times \vec{b})) \cdot \vec{b} = 1. \]
Calculating the dot products:
1. \(\vec{a} \cdot \vec{c} = \lambda (\vec{a} \cdot \vec{a}) + \mu (\vec{a} \cdot \vec{b}) + \omega (\vec{a} \cdot (\vec{a} \times \vec{b})) = \lambda\) (since \(\vec{a} \cdot \vec{b} = 0\) and \(\vec{a} \cdot (\vec{a} \times \vec{b}) = 0\)).
2. \(\vec{c} \cdot \vec{b} = \lambda (\vec{b} \cdot \vec{a}) + \mu (\vec{b} \cdot \vec{b}) + \omega (\vec{b} \cdot (\vec{a} \times \vec{b})) = \mu\) (since \(\vec{b} \cdot \vec{a} = 0\) and \(\vec{b} \cdot (\vec{a} \times \vec{b}) = 0\)).

Substituting these results back gives:
\[ \lambda - \mu = 1. \]

### Step 5: Magnitude of \(\vec{c}\)
Since \(\vec{c}\) is a unit vector, we have:
\[ |\vec{c}|^2 = \lambda^2 + \mu^2 + \omega^2 = 1. \]

### Step 6: Solve the equations
From \(\lambda - \mu = 1\), we can express \(\lambda\) as:
\[ \lambda = \mu + 1. \]
Substituting this into the magnitude equation:
\[ (\mu + 1)^2 + \mu^2 + \omega^2 = 1. \]
Expanding gives:
\[ \mu^2 + 2\mu + 1 + \mu^2 + \omega^2 = 1. \]
This simplifies to:
\[ 2\mu^2 + 2\mu + \omega^2 + 1 = 1. \]
Thus:
\[ 2\mu^2 + 2\mu + \omega^2 = 0. \]

### Step 7: Analyze the quadratic equation
This equation can be factored or analyzed for roots. Since all terms are squared, we can conclude:
\[ \mu = 0 \quad \text{and} \quad \omega = 0. \]
Substituting \(\mu = 0\) back gives \(\lambda = 1\).

### Conclusion
Thus, we have:
\[ \lambda = 1, \quad \mu = 0, \quad \omega = 0. \]