3D Vector Rotation of Point Mass

Triangle defined by points OA, OB and OC : (-i,3 j,-4 k), (i,2 j,2 k) and (3 i,7 j,- k) where i, j, k are unit vectors along x,y,z axis. Point mass is placed at OA. Triangle rotates with angular velocity ω about BC axis. Calculate linear velocity of mass as it passes OA.

My Work: $$v=\omega \times r$$ Does r have to be the vector perpendicular to BC to the point OA or can it be any vector from the line BC to OA??? If so can you explain? I know the answer as my tutor told me I got it right but I didn't need to find the vector perpendicular to BC and I forgot to ask why...

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It could be any vector from BC to OA.

Let's assume that $r$ is the vector perpendicular to BC to the point OA. Any vector from BC to OA (whether or not it is perpendicular to BC) has the form $r + s$, where $s$ is some vector parallel to BC. Of course, you've told us that $\omega$ is also parallel to BC, so we could also write that as $r+\alpha\, \omega$, for some number $\alpha$. So let's take the cross product: $$\omega \times (r+\alpha\, \omega) = \omega \times r + \omega \times (\alpha\, \omega) = \omega \times r + \alpha\, \omega \times \omega = \omega \times r~,$$ since $\omega \times \omega = 0$. So $\omega$ cross any vector from BC to OA will equal $\omega$ cross the perpendicular vector. So you don't need to specifically find the perpendicular.

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$$\vec{v}_A = \vec{v}_B + \vec{\omega} \times (\vec{r}_A -\vec{r}_B )$$

with $\vec{v}_B =0$.

Any components of $(\vec{r}_A -\vec{r}_B )$ along $\vec{\omega}$ cancel out, so you don't have to worry about finding the perpendicular.

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