Let ` vecu = u_(1)hati + u_(2)hatj +u_(3)hatk` be a unit vector in ` R^(3) and vecw = 1/sqrt6 ( hati + hatj + 2hatk)` , Given that there exists a vector `vecv " in " R^(3)` such that ` | vecu xx vecv| =1 and vecw . ( vecu xx vecv) =1` which of the following statements is correct ? (a)there is exactly one choice for such `vecv` (b)there are infinitely many choices for such `vecv` (c)if `hatu` lies in the xy - plane then `|u_(1)|=|u_(2)|` (d)if `hatu` lies in the xz-plane then `2|u_(1)|=|u_(3)|`

Options

A. there is exactly one choice for such `vecv`
B. there are infinitely many choices for such `vecv`
C. if `hatu` lies in the xy - plane then `|u_(1)|=|u_(2)|`
D. if `hatu` lies in the xz-plane then `2|u_(1)|=|u_(3)|`

Explanation

To solve the problem, we start with the given information about the vectors \( \vec{u} \) and \( \vec{w} \).

1. **Understanding the Vectors**:
- The unit vector \( \vec{u} = u_1 \hat{i} + u_2 \hat{j} + u_3 \hat{k} \) has a magnitude of 1, which implies:
\[ u_1^2 + u_2^2 + u_3^2 = 1 \]
- The vector \( \vec{w} = \frac{1}{\sqrt{6}} (\hat{i} + \hat{j} + 2\hat{k}) \) is also a unit vector since:
\[ |\vec{w}| = \frac{1}{\sqrt{6}} \sqrt{1^2 + 1^2 + 2^2} = \frac{1}{\sqrt{6}} \sqrt{6} = 1 \]

2. **Condition on Cross Product**:
- We are given that there exists a vector \( \vec{v} \) such that:
\[ |\vec{u} \times \vec{v}| = 1 \]
- This means that \( \vec{u} \) and \( \vec{v} \) are not parallel, and the magnitude of the cross product is equal to the area of the parallelogram formed by \( \vec{u} \) and \( \vec{v} \).

3. **Dot Product Condition**:
- We also have the condition:
\[ \vec{w} \cdot (\vec{u} \times \vec{v}) = 1 \]
- Taking the magnitude on both sides, we can express this as:
\[ |\vec{w}| \cdot |\vec{u} \times \vec{v}| \cdot \cos \alpha = 1 \]
- Since \( |\vec{w}| = 1 \) and \( |\vec{u} \times \vec{v}| = 1 \), we have:
\[ 1 \cdot 1 \cdot \cos \alpha = 1 \implies \cos \alpha = 1 \implies \alpha = 0^\circ \]
- This indicates that \( \vec{w} \) is parallel to \( \vec{u} \times \vec{v} \).

4. **Perpendicularity**:
- Since \( \vec{w} \) is parallel to \( \vec{u} \times \vec{v} \), it follows that \( \vec{w} \) is perpendicular to both \( \vec{u} \) and \( \vec{v} \). Therefore:
\[ \vec{u} \cdot \vec{w} = 0 \]
- This leads to:
\[ u_1 + u_2 + 2u_3 = 0 \]

5. **Analyzing the Statements**:
- **(a)** There is exactly one choice for such \( \vec{v} \): **False**. There are infinitely many choices for \( \vec{v} \) since \( \vec{u} \) can vary while satisfying the dot product condition.
- **(b)** There are infinitely many choices for such \( \vec{v} \): **True**. As shown, for different values of \( u_1, u_2, u_3 \), there are infinite vectors \( \vec{v} \).
- **(c)** If \( \vec{u} \) lies in the xy-plane, then \( |u_1| = |u_2| \): **True**. If \( u_3 = 0 \), then \( u_1 + u_2 = 0 \) implies \( u_1 = -u_2 \).
- **(d)** If \( \vec{u} \) lies in the xz-plane, then \( 2|u_1| = |u_3| \): **True**. If \( u_2 = 0 \), then from \( u_1 + 2u_3 = 0 \), we get \( |u_3| = 2|u_1| \).

Thus, the correct statements are (b), (c), and (d).