Mass Moment of Inertia as a Cross Product

In Eq.(B.14) above, the mass moment of inertia was expressed in terms of orthogonal projection as $I = mR^2 = m\cdot \vert\vert\,\underline{x}-{\cal P}_{\underline{\omega}}(\underline{x})\,\vert\vert ^2 = m\cdot \vert\vert\,\underline{x}-(\underline{\tilde{\omega}}^T\underline{x})\underline{\tilde{\omega}}\,\vert\vert ^2$ , where $\underline{\tilde{\omega}}\isdeftext \underline{\omega}/ \vert\vert\,\underline{\omega}\,\vert\vert$ . In terms of the vector cross product, we can now express it as

$$I \eqsp m\cdot(\underline{\tilde{\omega}}\times \underline{x})^2 \eqsp m\cdot\left[\left\Vert\,\underline{\tilde{\omega}}\,\right\Vert\cdot\left\Vert\,\underline{x}\,\right\Vert\cdot\sin(\theta_{\underline{\tilde{\omega}}\underline{x}})\right]^2 \eqsp mR^2$$

where $R= \vert\vert\,\underline{x}\,\vert\vert \sin(\theta_{\underline{\tilde{\omega}}\underline{x}})$ is the distance from the rotation axis out to the point $\underline {x}$ (which equals the length of the vector $\underline{x}-{\cal P}_{\underline{\omega}}(\underline{x})$ ).