Let `hata and hatb ` be mutually perpendicular unit vectors. Then for ant arbitrary `vecr`.

Options

A. `vecr= (vecr.hata)hata+(vecr.hatb)hatb+(vecr.(vecaxxhatb)) (hataxxhatb)`
B. `vecr= (vecr.hata)-(vecr.hatb)hatb-(vecr.(vecaxxhatb)) (hataxxhatb)`
C. `vecr= (vecr.hata)hata-(vecr.hatb)hatb-(vecr.(vecaxxhatb)) (hataxxhatb)`
D. none of these

Explanation

To solve the problem, we need to express an arbitrary vector \(\vec{r}\) in terms of the mutually perpendicular unit vectors \(\hat{a}\) and \(\hat{b}\). Let's go through the steps systematically.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We have two mutually perpendicular unit vectors \(\hat{a}\) and \(\hat{b}\).
- We need to express an arbitrary vector \(\vec{r}\) in terms of these unit vectors.

2. **Expressing the Vector**:
- Since \(\hat{a}\) and \(\hat{b}\) are unit vectors, we can express any vector \(\vec{r}\) in the form:
\[ \vec{r} = x_1 \hat{a} + x_2 \hat{b} + x_3 (\hat{a} \times \hat{b}) \]
- Here, \(x_1\), \(x_2\), and \(x_3\) are the scalar components of \(\vec{r}\) along the directions of \(\hat{a}\), \(\hat{b}\), and \(\hat{a} \times \hat{b}\) respectively.

3. **Dot Product with \(\hat{a}\)**:
- Taking the dot product of \(\vec{r}\) with \(\hat{a}\):
\[ \vec{r} \cdot \hat{a} = (x_1 \hat{a} + x_2 \hat{b} + x_3 (\hat{a} \times \hat{b})) \cdot \hat{a} \]
- This simplifies to:
\[ \vec{r} \cdot \hat{a} = x_1 + x_2 (\hat{b} \cdot \hat{a}) + x_3 ((\hat{a} \times \hat{b}) \cdot \hat{a}) \]
- Since \(\hat{a} \cdot \hat{b} = 0\) and \((\hat{a} \times \hat{b}) \cdot \hat{a} = 0\), we have:
\[ \vec{r} \cdot \hat{a} = x_1 \]

4. **Dot Product with \(\hat{b}\)**:
- Similarly, taking the dot product of \(\vec{r}\) with \(\hat{b}\):
\[ \vec{r} \cdot \hat{b} = (x_1 \hat{a} + x_2 \hat{b} + x_3 (\hat{a} \times \hat{b})) \cdot \hat{b} \]
- This simplifies to:
\[ \vec{r} \cdot \hat{b} = x_1 (\hat{a} \cdot \hat{b}) + x_2 + x_3 ((\hat{a} \times \hat{b}) \cdot \hat{b}) \]
- Again, since \(\hat{a} \cdot \hat{b} = 0\) and \((\hat{a} \times \hat{b}) \cdot \hat{b} = 0\), we have:
\[ \vec{r} \cdot \hat{b} = x_2 \]

5. **Cross Product with \(\hat{a} \times \hat{b}\)**:
- Now, taking the cross product of \(\vec{r}\) with \(\hat{a} \times \hat{b}\):
\[ \vec{r} \times (\hat{a} \times \hat{b}) = (x_1 \hat{a} + x_2 \hat{b} + x_3 (\hat{a} \times \hat{b})) \times (\hat{a} \times \hat{b}) \]
- Using the vector triple product identity:
\[ \vec{r} \times (\hat{a} \times \hat{b}) = (\vec{r} \cdot \hat{b}) \hat{a} - (\vec{r} \cdot \hat{a}) \hat{b} \]
- This gives us:
\[ \vec{r} \times (\hat{a} \times \hat{b}) = x_2 \hat{a} - x_1 \hat{b} \]

6. **Final Expression**:
- We can now express \(\vec{r}\) in terms of its components:
\[ \vec{r} = (\vec{r} \cdot \hat{a}) \hat{a} + (\vec{r} \cdot \hat{b}) \hat{b} + (\vec{r} \cdot (\hat{a} \times \hat{b})) (\hat{a} \times \hat{b}) \]
- Therefore, the final expression for \(\vec{r}\) is:
\[ \vec{r} = \vec{r} \cdot \hat{a} \hat{a} + \vec{r} \cdot \hat{b} \hat{b} + \vec{r} \cdot (\hat{a} \times \hat{b}) (\hat{a} \times \hat{b}) \]

### Conclusion:
The expression for any arbitrary vector \(\vec{r}\) in terms of the mutually perpendicular unit vectors \(\hat{a}\) and \(\hat{b}\) is:
\[ \vec{r} = (\vec{r} \cdot \hat{a}) \hat{a} + (\vec{r} \cdot \hat{b}) \hat{b} + (\vec{r} \cdot (\hat{a} \times \hat{b})) (\hat{a} \times \hat{b}) \]