We will numerically verify the vector identity: \[ \mathbf{\omega} \times (\mathbf{\omega} \times \mathbf{r}) = -\omega^2 \mathbf{r} \] for a particle undergoing circular motion. ### Step 1: Define Given Quantities - Let the angular velocity vector be \( \mathbf{\omega} = (0, 0, \omega) \), assuming rotation about the \( z \)-axis. - Let the position vector of the particle be \( \mathbf{r} = (r_x, r_y, 0) \), since it moves in the \( xy \)-plane. ### Step 2: Compute \( \mathbf{\omega} \times \mathbf{r} \) The cross product formula states: \[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix} \] Substituting \( \mathbf{\omega} = (0, 0, \omega) \) and \( \mathbf{r} = (r_x, r_y, 0) \): \[ \mathbf{\omega}