Let `Sigma_(i=1)^(n) vec(a_(i))=0` , where `|vec(a_(i))|=1AAi`, then `Sigma_(1leiltjlen)Sigmavec(a_(i))vec(a_(j))=`

Options

A. n
B. `-n`
C. `(n)/(2)`
D. `(-n)/(2)`

Explanation

To solve the problem, we start with the given condition:

\[ \sum_{i=1}^{n} \vec{a_i} = 0 \]

where \(|\vec{a_i}| = 1\) for all \(i\). We need to find:

\[ \sum_{1 \leq i < j \leq n} \vec{a_i} \cdot \vec{a_j} \]

### Step 1: Understanding the Dot Product
The dot product \(\vec{a_i} \cdot \vec{a_j}\) can be expressed in terms of the angle \(\theta\) between the two vectors:

\[ \vec{a_i} \cdot \vec{a_j} = |\vec{a_i}| |\vec{a_j}| \cos(\theta) = 1 \cdot 1 \cdot \cos(\theta) = \cos(\theta) \]

### Step 2: Squaring the Sum of Vectors
Since \(\sum_{i=1}^{n} \vec{a_i} = 0\), we can square this equation:

\[ \left( \sum_{i=1}^{n} \vec{a_i} \right) \cdot \left( \sum_{j=1}^{n} \vec{a_j} \right) = 0 \]

Expanding this gives:

\[ \sum_{i=1}^{n} \vec{a_i} \cdot \vec{a_i} + \sum_{i \neq j} \vec{a_i} \cdot \vec{a_j} = 0 \]

### Step 3: Evaluating the Terms
1. The term \(\sum_{i=1}^{n} \vec{a_i} \cdot \vec{a_i}\) is simply the sum of the squares of the magnitudes of the vectors:

\[ \sum_{i=1}^{n} \vec{a_i} \cdot \vec{a_i} = n \]

since each \(|\vec{a_i}| = 1\).

2. The term \(\sum_{i \neq j} \vec{a_i} \cdot \vec{a_j}\) can be rewritten as:

\[ \sum_{i=1}^{n} \sum_{j=1}^{n} \vec{a_i} \cdot \vec{a_j} - \sum_{i=1}^{n} \vec{a_i} \cdot \vec{a_i} = \sum_{i=1}^{n} \sum_{j=1}^{n} \vec{a_i} \cdot \vec{a_j} - n \]

### Step 4: Setting Up the Equation
Putting it all together, we have:

\[ n + \sum_{i \neq j} \vec{a_i} \cdot \vec{a_j} = 0 \]

This leads to:

\[ \sum_{i \neq j} \vec{a_i} \cdot \vec{a_j} = -n \]

### Step 5: Relating to the Required Sum
The sum \(\sum_{1 \leq i < j \leq n} \vec{a_i} \cdot \vec{a_j}\) can be related to \(\sum_{i \neq j} \vec{a_i} \cdot \vec{a_j}\):

\[ \sum_{i \neq j} \vec{a_i} \cdot \vec{a_j} = 2 \sum_{1 \leq i < j \leq n} \vec{a_i} \cdot \vec{a_j} \]

Thus, we can write:

\[ 2 \sum_{1 \leq i < j \leq n} \vec{a_i} \cdot \vec{a_j} = -n \]

### Step 6: Solving for the Required Sum
Dividing both sides by 2 gives:

\[ \sum_{1 \leq i < j \leq n} \vec{a_i} \cdot \vec{a_j} = -\frac{n}{2} \]

### Final Result
Thus, the final answer is:

\[ \sum_{1 \leq i < j \leq n} \vec{a_i} \cdot \vec{a_j} = -\frac{n}{2} \]

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