Let ` A_(1) , A_(2) ,…..,A_(n) ( n lt 2)` be the vertices of regular polygon of n sides with its centre at he origin. Let `veca_(k)` be the position vector of the point ` A_(k) ,k = 1,2,….,n`
if `|sum_(k=1)^(n-1) (veca_(k) xx veca_(k) +1)|=|sum_(k=1)^(n-1) (vecak.vecak+1)|` then the minimum value of n is

Options

A. 1
B. 2
C. 8
D. 9

Explanation

To solve the problem, we need to analyze the given condition involving the position vectors of the vertices of a regular polygon and the relationships between their cross and dot products. Let's break down the solution step by step.

### Step 1: Understanding the Position Vectors
The vertices of a regular polygon with \( n \) sides centered at the origin can be represented by position vectors:
\[ \vec{A_k} = r (\cos \theta_k, \sin \theta_k) \]
where \( \theta_k = \frac{2\pi k}{n} \) for \( k = 1, 2, \ldots, n \).

### Step 2: Setting Up the Sums
We need to evaluate the sums:
\[ \sum_{k=1}^{n-1} (\vec{A_k} \times \vec{A_{k+1}}) \quad \text{and} \quad \sum_{k=1}^{n-1} (\vec{A_k} \cdot \vec{A_{k+1}}) \]

### Step 3: Cross Product Calculation
The cross product \( \vec{A_k} \times \vec{A_{k+1}} \) can be expressed in terms of the sine of the angle between them:
\[ \vec{A_k} \times \vec{A_{k+1}} = |\vec{A_k}| |\vec{A_{k+1}}| \sin(\theta_{k+1} - \theta_k) \hat{n} \]
Since the angle between consecutive vectors is \( \frac{2\pi}{n} \), we have:
\[ \sin(\theta_{k+1} - \theta_k) = \sin\left(\frac{2\pi}{n}\right) \]
Thus, the sum becomes:
\[ \sum_{k=1}^{n-1} (\vec{A_k} \times \vec{A_{k+1}}) = r^2 (n-1) \sin\left(\frac{2\pi}{n}\right) \hat{n} \]

### Step 4: Dot Product Calculation
The dot product \( \vec{A_k} \cdot \vec{A_{k+1}} \) can be expressed in terms of the cosine of the angle:
\[ \vec{A_k} \cdot \vec{A_{k+1}} = |\vec{A_k}| |\vec{A_{k+1}}| \cos(\theta_{k+1} - \theta_k) = r^2 \cos\left(\frac{2\pi}{n}\right) \]
Thus, the sum becomes:
\[ \sum_{k=1}^{n-1} (\vec{A_k} \cdot \vec{A_{k+1}}) = (n-1) r^2 \cos\left(\frac{2\pi}{n}\right) \]

### Step 5: Setting Up the Equation
From the problem statement, we have:
\[ \left| \sum_{k=1}^{n-1} (\vec{A_k} \times \vec{A_{k+1}}) \right| = \left| \sum_{k=1}^{n-1} (\vec{A_k} \cdot \vec{A_{k+1}}) \right| \]
Substituting the results from Steps 3 and 4:
\[ r^2 (n-1) \sin\left(\frac{2\pi}{n}\right) = (n-1) r^2 \cos\left(\frac{2\pi}{n}\right) \]

### Step 6: Simplifying the Equation
Assuming \( r^2 \neq 0 \) and \( n-1 \neq 0 \):
\[ \sin\left(\frac{2\pi}{n}\right) = \cos\left(\frac{2\pi}{n}\right) \]
This implies:
\[ \tan\left(\frac{2\pi}{n}\right) = 1 \]
Thus:
\[ \frac{2\pi}{n} = \frac{\pi}{4} \]
Solving for \( n \):
\[ n = 8 \]

### Conclusion
The minimum value of \( n \) such that the condition holds is:
\[ \boxed{8} \]