দেওয়া আছে, রৈখিক বেগ $\vec{v} = \vec{\omega} \times \vec{r}$। ধরি, কৌণিক বেগের উপাংশসমূহ $\vec{\omega} = \omega_x \hat{i} + \omega_y \hat{j} + \omega_z \hat{k}$ এবং অবস্থান ভেক্টর $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$। যেহেতু $\vec{\omega}$ একটি ধ্রুব রাশি, তাই $\omega_x, \omega_y$ ও $\omega_z$ ধ্রুবক। এখন, ভেক্টর গুণন থেকে পাই, $$ \begin{aligned} \vec{v} &= \vec{\omega} \times \vec{r} \\ &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \omega_x & \omega_y & \omega_z \\ x & y & z \end{vmatrix} \\ &= \hat{i}(\omega_y z - \omega_z y) - \hat{j}(\omega_x z - \omega_z x) + \hat{k}(\omega_x y - \omega_y x) \end{aligned} $$ এখন, কার্ল (curl) নির্ণয় করি: $$ \begin{aligned} \vec{\nabla} \times \vec{v} &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \omega_y z - \omega_z y & -(\omega_x z - \omega_z x) & \omega_x y - \omega_y x \end{vmatrix} \end{aligned} $$ উপাংশ অনুযায়ী বিস্তার করলে, $$ \begin{aligned} \vec{\nabla} \times \vec{v} &= \hat{i} \left[ \frac{\partial}{\partial y}(\omega_x y - \omega_y x) - \frac{\partial}{\partial z}(-(\omega_x z - \omega_z x)) \right] \\ &- \hat{j} \left[ \frac{\partial}{\partial x}(\omega_x y - \omega_y x) - \frac{\partial}{\partial z}(\omega_y z - \omega_z y) \right] \\ &+ \hat{k} \left[ \frac{\partial}{\partial x}(-(\omega_x z - \omega_z x)) - \frac{\partial}{\partial y}(\omega_y z - \omega_z y) \right] \end{aligned} $$ আংশিক অন্তরীকরণ করে পাই, $$ \begin{aligned} \vec{\nabla} \times \vec{v} &= \hat{i} [ \omega_x - (-\omega_x) ] - \hat{j} [ -\omega_y - \omega_y ] + \hat{k} [ \omega_z - (-\omega_z) ] \\ &= \hat{i} (2\omega_x) - \hat{j} (-2\omega_y) + \hat{k} (2\omega_z) \\ &= 2\omega_x \hat{i} + 2\omega_y \hat{j} + 2\omega_z \hat{k} \\ &= 2(\omega_x \hat{i} + \omega_y \hat{j} + \omega_z \hat{k}) \\ &= 2\vec{\omega} \end{aligned} $$ সুতরাং, $$ \vec{\omega} = \frac{1}{2} (\vec{\nabla} \times \vec{v}) $$ প্রমাণিত।