Given three vectors e`veca, vecb and vecc` two of which are non-collinear. Futrther if `(veca + vecb)` is collinear with `vecc, (vecb +vecc)` is collinear with `veca, |veca|=|vecb|=|vecc|=sqrt2` find the value of `veca. Vecb + vecb.vecc+vecc.veca`

Options

A. 3
B. `-3`
C. 0
D. cannot of these

Explanation

To solve the problem, we need to find the value of \( \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \) given the conditions about the vectors. Let's go through the solution step by step.

### Step 1: Understand the collinearity conditions
We know that:
1. \( \vec{a} + \vec{b} \) is collinear with \( \vec{c} \).
2. \( \vec{b} + \vec{c} \) is collinear with \( \vec{a} \).

This means we can express these relationships as:
- \( \vec{a} + \vec{b} = \lambda \vec{c} \) for some scalar \( \lambda \).
- \( \vec{b} + \vec{c} = \mu \vec{a} \) for some scalar \( \mu \).

### Step 2: Rearranging the equations
From the first equation, we can express \( \vec{b} \):
\[ \vec{b} = \lambda \vec{c} - \vec{a} \]

Substituting this into the second equation gives:
\[ \lambda \vec{c} - \vec{a} + \vec{c} = \mu \vec{a} \]
Rearranging this, we have:
\[ (\lambda + 1) \vec{c} = (\mu + 1) \vec{a} \]

### Step 3: Analyze the implications
Since \( \vec{a} \) and \( \vec{c} \) are non-collinear, the coefficients must be equal:
\[ \lambda + 1 = k(\mu + 1) \]
for some scalar \( k \). However, we can also explore the case when \( \lambda = \mu = -1 \) as a specific solution.

### Step 4: Substitute back into the equations
If \( \lambda = -1 \) and \( \mu = -1 \), we have:
\[ \vec{a} + \vec{b} + \vec{c} = 0 \]
This implies:
\[ \vec{c} = -(\vec{a} + \vec{b}) \]

### Step 5: Use the magnitudes
Given that \( |\vec{a}| = |\vec{b}| = |\vec{c}| = \sqrt{2} \), we can calculate:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = 0 \]
Expanding this, we get:
\[ |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \]

### Step 6: Substitute the magnitudes
Since \( |\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = 2 \):
\[ 2 + 2 + 2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \]
This simplifies to:
\[ 6 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \]

### Step 7: Solve for the dot products
Rearranging gives:
\[ 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -6 \]
Thus,
\[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -3 \]

### Final Answer
The value of \( \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \) is \( -3 \).

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