Let `vec(a), vec(b) and vec(c)` be three vectors such that ` vec(a) = vec(b) xx ( vec(b) xx vec(c))` . If magnitudes of the vectors `vec(a), vec(b) and vec(c) " are" sqrt(2) , 1 ` and 2 respectively and the angle between ` vec(b) and vec(c) " is " theta ( 0 lt theta lt (pi)/(2))` , then the value of `1 + tan theta ` is equal to :

Options

A. `sqrt(3) + 1`
B. 2
C. 1
D. `(sqrt(3)+1)/(sqrt(3))`

Answer

To solve the problem, we need to analyze the given vectors and their relationships step by step.

### Step 1: Understand the relationship between the vectors
We are given that:
\[ \vec{a} = \vec{b} \times (\vec{b} \times \vec{c}) \]
This can be simplified using the vector triple product identity:
\[ \vec{b} \times (\vec{b} \times \vec{c}) = (\vec{b} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{b}) \vec{c} \]
Thus, we can rewrite:
\[ \vec{a} = (\vec{b} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{b}) \vec{c} \]

### Step 2: Calculate the magnitudes
We know the magnitudes of the vectors:
- \(|\vec{a}| = \sqrt{2}\)
- \(|\vec{b}| = 1\)
- \(|\vec{c}| = 2\)

### Step 3: Substitute the magnitudes into the equation
Using the magnitudes, we can express the dot products:
- \(\vec{b} \cdot \vec{b} = |\vec{b}|^2 = 1^2 = 1\)
- Let \(\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \theta = 1 \cdot 2 \cos \theta = 2 \cos \theta\)

Substituting these into our expression for \(\vec{a}\):
\[ \vec{a} = (2 \cos \theta) \vec{b} - (1) \vec{c} \]

### Step 4: Find the dot product of \(\vec{a}\) with itself
Now we calculate \(|\vec{a}|^2\):
\[ |\vec{a}|^2 = \left|(2 \cos \theta) \vec{b} - \vec{c}\right|^2 \]
Using the formula for the magnitude of a vector:
\[ |\vec{a}|^2 = |(2 \cos \theta) \vec{b}|^2 + |\vec{c}|^2 - 2 \cdot (2 \cos \theta) \vec{b} \cdot \vec{c} \]
Substituting the known values:
\[ |\vec{a}|^2 = (2 \cos \theta)^2 |\vec{b}|^2 + |\vec{c}|^2 - 2 \cdot (2 \cos \theta) \cdot (2 \cos \theta) \]
This simplifies to:
\[ |\vec{a}|^2 = 4 \cos^2 \theta + 4 - 8 \cos^2 \theta = 4 - 4 \cos^2 \theta \]

### Step 5: Set the equation equal to \(|\vec{a}|^2\)
Since \(|\vec{a}|^2 = 2\):
\[ 4 - 4 \cos^2 \theta = 2 \]
Solving for \(\cos^2 \theta\):
\[ 4 \cos^2 \theta = 2 \implies \cos^2 \theta = \frac{1}{2} \implies \cos \theta = \frac{1}{\sqrt{2}} \]

### Step 6: Determine \(\tan \theta\)
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\[ \sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{1}{2} = \frac{1}{2} \implies \sin \theta = \frac{1}{\sqrt{2}} \]
Thus,
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = 1 \]

### Step 7: Calculate \(1 + \tan \theta\)
Finally, we find:
\[ 1 + \tan \theta = 1 + 1 = 2 \]

### Final Answer
The value of \(1 + \tan \theta\) is:
\[ \boxed{2} \]