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