If `vec(a), vec(b) , vec(c )` are three vectors such that `| vec(a) + vec(b) + vec( c )| = 1, vec( c)= lambda( vec( a) xx vec(b)), lambda` is non-zero scalar and `|vec(a)| = (1)/( sqrt( 6)), | vec(b) | = (1)/(sqrt(2)) , | vec(c )| = (1)/( sqrt(3))` , then the angle between `vec (a)` and `vec(b)` is

Options

A. `(pi )/( 4)`
B. `(pi )/( 3)`
C. `(pi)/(2)`
D. `(2pi )/( 3)`

Answer

To solve the problem step-by-step, we will analyze the given information and derive the angle between the vectors \(\vec{a}\) and \(\vec{b}\).

### Given:
1. \(|\vec{a} + \vec{b} + \vec{c}| = 1\)
2. \(\vec{c} = \lambda (\vec{a} \times \vec{b})\) where \(\lambda\) is a non-zero scalar.
3. \(|\vec{a}| = \frac{1}{\sqrt{6}}\)
4. \(|\vec{b}| = \frac{1}{\sqrt{2}}\)
5. \(|\vec{c}| = \frac{1}{\sqrt{3}}\)

### Step 1: Use the magnitude equation
From the first equation, we can express the magnitude squared:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = 1^2 = 1 \]
Expanding this using the properties of vectors:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]

### Step 2: Substitute the magnitudes
Substituting the magnitudes:
\[ |\vec{a}|^2 = \left(\frac{1}{\sqrt{6}}\right)^2 = \frac{1}{6}, \quad |\vec{b}|^2 = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}, \quad |\vec{c}|^2 = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \]
Now substituting these values into the equation:
\[ 1 = \frac{1}{6} + \frac{1}{2} + \frac{1}{3} + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]

### Step 3: Calculate the sum of squares
Calculating the sum of the squares:
\[ \frac{1}{6} + \frac{1}{2} + \frac{1}{3} = \frac{1}{6} + \frac{3}{6} + \frac{2}{6} = \frac{6}{6} = 1 \]
Thus, we have:
\[ 1 = 1 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]
This simplifies to:
\[ 0 = 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]
This implies:
\[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = 0 \]

### Step 4: Analyze \(\vec{c}\)
Since \(\vec{c} = \lambda (\vec{a} \times \vec{b})\), it is perpendicular to both \(\vec{a}\) and \(\vec{b}\). Therefore:
\[ \vec{b} \cdot \vec{c} = 0 \quad \text{and} \quad \vec{a} \cdot \vec{c} = 0 \]
Thus, we have:
\[ \vec{a} \cdot \vec{b} = 0 \]
This means that the angle \(\theta\) between \(\vec{a}\) and \(\vec{b}\) is:
\[ \cos \theta = 0 \]
This implies:
\[ \theta = \frac{\pi}{2} \quad \text{(90 degrees)} \]

### Conclusion
The angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{2}\).