If `hati, hatj, hatk` are unit orthonormal vectors and `veca` is a vector, If `veca xx vecr = hatj`, then `veca.vecr`:

Options

A. 0
B. `in [-1,1]`
C. `in (0,1)`
D. `in R`

Answer

To solve the problem, we need to find the value of \( \vec{a} \cdot \vec{r} \) given that \( \vec{a} \times \vec{r} = \hat{j} \).

### Step-by-step Solution:

1. **Understanding the Given Information**:
- We have unit orthonormal vectors \( \hat{i}, \hat{j}, \hat{k} \).
- The cross product \( \vec{a} \times \vec{r} = \hat{j} \).

2. **Using the Properties of Cross Product**:
- The magnitude of the cross product can be expressed as:
\[ |\vec{a} \times \vec{r}| = |\vec{a}| |\vec{r}| \sin \theta \]
- Since \( \hat{j} \) is a unit vector, its magnitude is 1:
\[ |\vec{a} \times \vec{r}| = 1 \]
- Therefore, we have:
\[ |\vec{a}| |\vec{r}| \sin \theta = 1 \]

3. **Using the Dot Product**:
- The dot product is given by:
\[ \vec{a} \cdot \vec{r} = |\vec{a}| |\vec{r}| \cos \theta \]

4. **Relating the Cross Product and Dot Product**:
- We can use the identity:
\[ |\vec{a} \times \vec{r}|^2 + |\vec{a} \cdot \vec{r}|^2 = |\vec{a}|^2 |\vec{r}|^2 \]
- Substituting the known values:
\[ 1^2 + (\vec{a} \cdot \vec{r})^2 = |\vec{a}|^2 |\vec{r}|^2 \]
- This simplifies to:
\[ 1 + (\vec{a} \cdot \vec{r})^2 = |\vec{a}|^2 |\vec{r}|^2 \]

5. **Solving for \( \vec{a} \cdot \vec{r} \)**:
- Let \( |\vec{a}| = a \) and \( |\vec{r}| = r \):
\[ 1 + (\vec{a} \cdot \vec{r})^2 = a^2 r^2 \]
- Rearranging gives:
\[ (\vec{a} \cdot \vec{r})^2 = a^2 r^2 - 1 \]
- Taking the square root:
\[ \vec{a} \cdot \vec{r} = \sqrt{a^2 r^2 - 1} \]

6. **Final Expression**:
- Therefore, the final expression for \( \vec{a} \cdot \vec{r} \) is:
\[ \vec{a} \cdot \vec{r} = \sqrt{|\vec{a}|^2 |\vec{r}|^2 - 1} \]