Let $A_1, A_2,..., A_n (n>2)$ be the vertices of a regular polygon of n sides with its centre at the origin. Let $\vec{a_k}$ be the position vector of the point $A_k, k = 1, 2,..., n$. If $\left|\sum\limits_{k=1}^{n-1}(\vec a_k×\vec a_{k+1})\right|=\left|\sum\limits_{k=1}^{n-1}(\vec a_k.\vec a_{k+1})\right|$, then the minimum value of n is

8

Let $OA_1 =OA_2 =... =O A_n=r$.

Clearly, $∠A_{k-1} OA_K=\frac{2π}{n}$ for $k = 2, 3,..., n$.

Now,

$\left|\sum\limits_{k=1}^{n-1}(\vec a_k×\vec a_{k+1})\right|=\left|\sum\limits_{k=1}^{n-1}(\vec a_k.\vec a_{k+1})\right|$

$⇒\sum\limits_{k=1}^{n-1}\left|\vec a_k\right|\left|\vec a_{k+1}\right|\sin\frac{2π}{n}=\sum\limits_{k=1}^{n-1}\left|\vec a_k\right|\left|\vec a_{k+1}\right|\cos\frac{2π}{n}$

$⇒\sum\limits_{k=1}^{n-1}r^2\sin\frac{2π}{n}=\sum\limits_{k=1}^{n-1}r^2\cos\frac{2π}{n}$

$⇒r^2(n-1)\sin\frac{2π}{n}=r^2(n-1)\cos\frac{2π}{n}$

$⇒\tan\frac{2π}{n}=1⇒\frac{2π}{n}=\frac{π}{4}⇒n=8$