If `veca and vecb` are othogonal unit vectors, then for a vector `vecr` non - coplanar with `veca and vecb` vector `vecr xx veca` is equal to

Options

A. `[vecr vecavecb]vecb - (vecr.vecb)(vecbxxveca)`
B. `[vecrvecavecb] (veca+vecb)`
C. `[vecr vecavecb]veca + (vecr.veca) vecaxxvecb`
D. none of these

Explanation

To solve the problem, we need to find the expression for the vector \( \vec{r} \times \vec{a} \) given that \( \vec{a} \) and \( \vec{b} \) are orthogonal unit vectors, and \( \vec{r} \) is a vector that is non-coplanar with \( \vec{a} \) and \( \vec{b} \).

### Step-by-Step Solution:

1. **Understand the Setup**:
- We know that \( \vec{a} \) and \( \vec{b} \) are orthogonal unit vectors. This means:
\[ \vec{a} \cdot \vec{b} = 0 \]
- The magnitude of both vectors is 1:
\[ |\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 1 \]

2. **Express \( \vec{r} \)**:
- Since \( \vec{r} \) is non-coplanar with \( \vec{a} \) and \( \vec{b} \), we can express \( \vec{r} \) in terms of \( \vec{a} \), \( \vec{b} \), and their cross product:
\[ \vec{r} = \lambda \vec{a} + \mu \vec{b} + \gamma (\vec{a} \times \vec{b}) \]
- Here, \( \lambda \), \( \mu \), and \( \gamma \) are scalars.

3. **Calculate \( \vec{r} \times \vec{a} \)**:
- We can compute the cross product:
\[ \vec{r} \times \vec{a} = (\lambda \vec{a} + \mu \vec{b} + \gamma (\vec{a} \times \vec{b})) \times \vec{a} \]
- Using the distributive property of the cross product:
\[ \vec{r} \times \vec{a} = \lambda (\vec{a} \times \vec{a}) + \mu (\vec{b} \times \vec{a}) + \gamma ((\vec{a} \times \vec{b}) \times \vec{a}) \]
- Since \( \vec{a} \times \vec{a} = \vec{0} \):
\[ \vec{r} \times \vec{a} = \mu (\vec{b} \times \vec{a}) + \gamma ((\vec{a} \times \vec{b}) \times \vec{a}) \]

4. **Simplify \( (\vec{a} \times \vec{b}) \times \vec{a} \)**:
- Using the vector triple product identity:
\[ \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z}) \vec{y} - (\vec{x} \cdot \vec{y}) \vec{z} \]
- Applying it here:
\[ (\vec{a} \times \vec{b}) \times \vec{a} = (\vec{a} \cdot \vec{a}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{a} \]
- Since \( \vec{a} \cdot \vec{b} = 0 \) and \( \vec{a} \cdot \vec{a} = 1 \):
\[ (\vec{a} \times \vec{b}) \times \vec{a} = \vec{b} \]

5. **Final Expression**:
- Now substituting back:
\[ \vec{r} \times \vec{a} = \mu (\vec{b} \times \vec{a}) + \gamma \vec{b} \]
- Since \( \vec{b} \times \vec{a} = -(\vec{a} \times \vec{b}) \):
\[ \vec{r} \times \vec{a} = -\mu (\vec{a} \times \vec{b}) + \gamma \vec{b} \]

### Conclusion:
Thus, the expression for \( \vec{r} \times \vec{a} \) is:
\[ \vec{r} \times \vec{a} = -\mu (\vec{a} \times \vec{b}) + \gamma \vec{b} \]