A particle moves on a given straight line with a constant $v$. At a certain time it is at a point P on its straight line path. $\mathrm{O}$ is a fixed point. Show that $\overrightarrow{O P} \times \vec{v}$ is independent of the position $\mathrm{P}$.
The particle moves from PP' in a straight line with a constant speed $v$.
From the figure, we see that $O P \times v=(O P) v \sin \theta$ û, where û is a unit vector perpendicular to $v$ and OP, Now,
We know, OQ= OP $\sin \theta=$ OP' $\sin \theta$ '
So, the position of the particle may vary, but the magnitude and direction of OP $\times \mathrm{V}$ will be constant. $O P \times v$ is independent of $P$.
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