Let ` vecu = u_(1) hati + u_(2) hatj` be a unit vector in xy plane 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` , then

Options

A. `|u_(1)|=|u_(2)|`
B. `|u_(2)|=2|u_(2)|`
C. ` 2|u_(1)|=|u_(2)|`
D. ` |u_(1)|-3|u_(2)|`

Explanation

To solve the problem, we need to analyze the given conditions involving vectors \(\vec{u}\), \(\vec{w}\), and \(\vec{v}\). Let's break down the solution step by step.

### Step 1: Understand the given vectors
We have:
- \(\vec{u} = u_1 \hat{i} + u_2 \hat{j}\) (a unit vector in the xy-plane)
- \(\vec{w} = \frac{1}{\sqrt{6}} (\hat{i} + \hat{j} + 2\hat{k})\)

Since \(\vec{u}\) is a unit vector, we have:
\[ u_1^2 + u_2^2 = 1 \]

### Step 2: Analyze the conditions
We are given two conditions:
1. \(|\vec{u} \times \vec{v}| = 1\)
2. \(\vec{w} \cdot (\vec{u} \times \vec{v}) = 1\)

### Step 3: Use the first condition
From the first condition, we know:
\[ |\vec{u} \times \vec{v}| = |\vec{u}| |\vec{v}| \sin \theta = 1 \]
Since \(|\vec{u}| = 1\), this simplifies to:
\[ |\vec{v}| \sin \theta = 1 \]
This implies that \(|\vec{v}| \geq 1\) (since \(\sin \theta \leq 1\)).

### Step 4: Use the second condition
For the second condition, we can express it as:
\[ \vec{w} \cdot (\vec{u} \times \vec{v}) = |\vec{w}| |\vec{u} \times \vec{v}| \cos \alpha = 1 \]
where \(\alpha\) is the angle between \(\vec{w}\) and \(\vec{u} \times \vec{v}\).

### Step 5: Calculate the magnitude of \(\vec{w}\)
Calculating the magnitude of \(\vec{w}\):
\[ |\vec{w}| = \left|\frac{1}{\sqrt{6}} (1, 1, 2)\right| = \frac{1}{\sqrt{6}} \sqrt{1^2 + 1^2 + 2^2} = \frac{1}{\sqrt{6}} \sqrt{6} = 1 \]

### Step 6: Substitute back into the second condition
Substituting the magnitudes back into the equation:
\[ 1 \cdot 1 \cdot \cos \alpha = 1 \implies \cos \alpha = 1 \]
This means \(\alpha = 0\), indicating that \(\vec{w}\) is parallel to \(\vec{u} \times \vec{v}\).

### Step 7: Determine the relationship between \(\vec{u}\) and \(\vec{v}\)
Since \(\vec{w}\) is perpendicular to the plane containing \(\vec{u}\) and \(\vec{v}\), we can express \(\vec{u} \times \vec{v}\) as:
\[ \vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_1 & u_2 & 0 \\ v_1 & v_2 & v_3 \end{vmatrix} = (u_2 v_3 - 0) \hat{i} - (u_1 v_3 - 0) \hat{j} + (u_1 v_2 - u_2 v_1) \hat{k} \]

### Step 8: Set up equations from the coefficients
From the expression for \(\vec{w}\):
\[ \frac{1}{\sqrt{6}} = u_2 v_3 \]
\[ -\frac{1}{\sqrt{6}} = u_1 v_3 \]
\[ \frac{2}{\sqrt{6}} = u_1 v_2 - u_2 v_1 \]

### Step 9: Solve the equations
From the first two equations, we can equate:
\[ u_2 v_3 = \frac{1}{\sqrt{6}} \quad \text{and} \quad -u_1 v_3 = -\frac{1}{\sqrt{6}} \]
This gives us:
\[ u_2 = -u_1 \]

### Step 10: Find the relationship between \(u_1\) and \(u_2\)
Taking the magnitudes, we have:
\[ |u_1| = |u_2| \]

### Conclusion
Thus, we conclude that:
\[ u_1 = \pm u_2 \]
This means the answer is that the magnitudes are equal:
\[ |u_1| = |u_2| \]